EDIT: Since most people seem to be raging about the wording of the problem, which is perfectly well-defined, note that there are others who can understand it properly:
It is correct that I am asking a conditional probability question: Given that at least one of my games was Zerg, what is the probability that both of my games are zerg.
It is most definitely not 1/3.
The problem can be understood as the following: Let's say IMNestea (always Z) is playing against, say, TLO, and TLO is playing random. They're going to play exactly two games (for simplicity). Your friend spoils the fun by saying "OMG Nestea's ZvZ is absolutely sick!" (implying he saw a ZvZ game).
What is the probability that you have to watch 2 mirror match games?
Those who say that is is definitely 1/3: It's true that if I specify the first game was Zerg, what is the probability that the second game was also Zerg, it is 1/3. In this case these are independent events.
In the question's case, we are not asking the above. We are asking that given that at least 1 game was Zerg, what is the probability that both games are Zerg.
If you still don't agree, I can't help you. Try wikipedia.
On June 10 2011 11:45 Whitewing wrote: Not really a brainteaser, but it looks like most people here have no idea what conditional probability is >_<.
The question isn't "What are the odds of randomly getting zerg in a match."
The question is "What are the odds of getting zerg twice in two matches, given that I got zerg at least once."
Yep. It seems a lot of people are missing the whole conditional part of the problem statement. I didn't find any ambiguity in it at all (unless it has been edited, which I don't think it was). And as someone else said before, the poll was only the last part of the full question, which may have been the cause for some confusion.
On June 10 2011 12:11 Cryllic wrote: It's like saying I flipped a coin 15 times and it was heads 15 times whats the chance of getting heads on the 16th try and the probability is 1/2. You can get zerg 30 times in a row the next game the probability of zerg will still be 1/3
That is unrelated to the question that the OP was asking. He is not saying ``I played my first game today as random and got zerg. What is the probability that I was zerg in the second game, too?''. He is saying ``I played two games today, and I will tell you that I was zerg at least once. Knowing that, what is the probability that I was zerg twice?''.
Many of the other posters in the past page or two have had the same misunderstanding.
For those of you who will say that I originally said that the problem was ambiguous, I will say that if someone gave me this problem I would first say 1/5, then say, ``but you should be careful about how you word the problem in the future, because...''. The ambiguity in this problem is very close to being nonexistent, and most of the people in this thread missed the actual ambiguity and are actually completely misunderstanding the question.
I've played two games. Then the possible combinations are: ZZ, ZP, ZT, PZ, PP, PT, TZ, TP, TT.
However, I've said I played Zerg. Then that eliminates PP, PT, TP, TT.
Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices, and that is the only which corresponds to "The other game is Zerg."
Then the correct answer is 1/5.
I posted this as a simple application of probability theory to real life.
It's counter-intuitive to a lot people, which makes for a bunch of "I don't see what's special about this problem" responses. Not everyone is familiar with this paradox, but it's fun =)
Just because you were zerg in 1 game, your odds don't change being zerg the next game. Edit: If you said you were going to play 2 games and wanted to know the odds of zerg twice in a row, the odds are 1/9, but since we know what you were in game 1, the odds of being a certain race in game 2 is 1/3
Edit for stupidity: Since it wasn't necessarily game 1 that you were zerg...it's 1/5
"Playing AGAINST" means that there is a 1/3 possibility(1/4 + 1/4*1/3 = 4/12 = 1/3, because there is the possibility to have a "Random" oponent, and that one can be Zerg)
<edit> I just read the edit.
It's still 1/3 regardless, since there are three races and they shouldn't be inclined to either side.
it should be 1/3 since being zerg the first games shouldnt change what race you will randomly get the next game. but blizzards system might be different.
so are we not supposed 2 say that the odds of getting zerg in 1 game in a row is 1/3 therefore the odds of getting ne race 2 games in a row twice must be (1/3)x(1/3)=1/9
On June 10 2011 10:15 DOMINOSC wrote: it should be 1/3 since being zerg the first games shouldnt change what race you will randomly get the next game. but blizzards system might be different.
Please don't misinterpret the question. It's simply saying:
I only played those 2 games today. I played at least one game as Zerg today.
On June 10 2011 10:14 Trezeguet wrote: Just because you were zerg in 1 game, your odds don't change being zerg the next game.
Are you sure? I'm merely stating that I played Zerg today in at least one of those games.
You actually said you were zerg in one game and asked what the odds were that you were zerg in your OTHER game. it doesn't matter which game you were zerg in really.
On June 10 2011 10:16 nealdt wrote: The correct answer is not listed. I assumed you're just regurgitating the standard Boy/Girl Paradox (http://en.wikipedia.org/wiki/Boy_or_Girl_paradox) but it's not as simple here.
There are 9 total possibilities for your two games: PP PT PZ TP TT TZ ZP ZT ZZ. Since at least one was a zerg, we cross out PP and TT. That leaves 7 options, only one of which was also Zerg, so the answer is actually 1/7.
The possibility for your race, regardless of the amount of games and sequence, is 1/3, assuming you played random both games, and one of the games has a fixed result as stated.
On June 10 2011 10:15 DOMINOSC wrote: it should be 1/3 since being zerg the first games shouldnt change what race you will randomly get the next game. but blizzards system might be different.
Please don't misinterpret the question. It's simply saying:
I only played those 2 games today. I played at least one game as Zerg today.
I'm sorry but that's not how you worded the "brain teaser" you're asking for the probability of one game only since it's the only game we don't know the exact result for since you state it';s not the one that you've already confirmed being zerg.
Oxidised would be right if the question was worded correctly.
On June 10 2011 10:16 nealdt wrote: The correct answer is not listed. I assumed you're just regurgitating the standard Boy/Girl Paradox (http://en.wikipedia.org/wiki/Boy_or_Girl_paradox) but it's not as simple here.
There are 9 total possibilities for your two games: PP PT PZ TP TT TZ ZP ZT ZZ. Since at least one was a zerg, we cross out PP and TT. That leaves 7 options, only one of which was also Zerg, so the answer is actually 1/7.
Are you trolling, or did you really not understand the question at all?
It's a famous question in probability, and your interpretation will vary strongly according to the exact wording of the problem, which is intentionally ambiguous.
On June 10 2011 10:15 DOMINOSC wrote: it should be 1/3 since being zerg the first games shouldnt change what race you will randomly get the next game. but blizzards system might be different.
Please don't misinterpret the question. It's simply saying:
I only played those 2 games today. I played at least one game as Zerg today.
I'm sorry but that's not how you worded the "brain teaser" you're asking for the probability of one game only since it's the only game we don't know the exact result for since you state it';s not the one that you've already confirmed being zerg.
Oxidised would be right if the question was worded correctly.
No, I'm only asking for the probability that that one game is played as Zerg. There's a well-defined answer to that.
On June 10 2011 10:16 nealdt wrote: The correct answer is not listed. I assumed you're just regurgitating the standard Boy/Girl Paradox (http://en.wikipedia.org/wiki/Boy_or_Girl_paradox) but it's not as simple here.
There are 9 total possibilities for your two games: PP PT PZ TP TT TZ ZP ZT ZZ. Since at least one was a zerg, we cross out PP and TT. That leaves 7 options, only one of which was also Zerg, so the answer is actually 1/7.
This except PT and TP don't have a zerg either, making it 1/5.
Hmm, i think it is 1/5 if you look at all the options, you have ZZ,ZT,ZP,PZ,PP,PT,TT,TP,TZ, so if u at least played Z once then the only options available are ZZ,ZT,ZP,PZ,TZ. So the chance of getting ZZ is 1/5?
On June 10 2011 10:22 teamsolid wrote: Probability of ZZ = 1/9 Probability of at least one Z game = 5/9
Probability of ZZ given at least one Z game = (1/9) / (5/9) = 1/5
THIS. People who answered 1/3 and said it was easy have no clue.
The thing is if he mentioned that he got FIRST game as a zerg then the answer is gonna be 1/3, but we don't know if he got a zerg in first or second game.
Well, many people don't understand how the question is formed, or much of else it seems.
The question is not asking what he played against. It's asking what the odds that HE played zerg again. I'm just... I feel like I'm being trolled by many of the responses.
Anyhow, the question is pretty poorly formed. The odds of one random event, in a series of random events, is not modified by past events. Getting zerg as random player 10 times in a row, previously, then starting up your 11th game does not change the odds of getting zerg, it's still 1/3. People often, both consciously and subconsciously do this, they feel that "my number needs to come up, it's been so long" but a die or coin doesn't remember what it rolled or flipped before.
Now, if you're looking at future events, statistically you can determine the odds of a 1/3rd rate occurrence to happen 11 times in a row, but that doesn't matter after you've gotten zerg 10 times. Statistics are future looking chances, really. As soon as the numbers start coming up, the stats change to only look at the future events, and not the past ones.
Heres the answer 1/5 Reasoning starting with a similar example but only 2 options. I have 2 children and ONE of them is a boy, what are the odds the other is a boy
I could have: 2 boys, 1 girl and 1 boy, 1 boy and 1 girl, or 2 girls. in a first and second child breakdown. HOWEVER, i have told you that one of them is a boy, thus 2 girls is impossible.
So now your options are 2 boys, 1 boy 1 girl, 1 girl 1 boy. the chances of the 2nd one being a boy is 1 out of 3.
You get confused because me telling you ONE of them is a boy vs telling you my first child is a boy. The boy could be item 1 or 2 or both, thus you have to account for the 1 boy 1 girl and 1 girl 1 boy choice.
Now if you reform this example but with 3 choices over 2 oppurtunites, there are 9 total pairings. Of these, 5 of them contain at least 1 zerg, and 1 of them is a zerg zerg combination. Thus, 1/5 chance the OTHER (note, not the second) game is a zerg
The most legitimate reason for choosing 1/3 over 1/5 would be because you asked for the probability of getting zerg twice by saying ``the other game''. This has implications about how the experiment was conducted. Essentially, the issue is that in the case that you did get zerg twice, there are two choices for ``the other game'', which could (depending on how the setup is interpreted) double the count for that case, raising the probability from 1/5 to 1/3.
I believe what you intended to ask was: We perform an experiment, where we play two games as random and throw out any sets of two games where we did not get zerg at least once. What is the probability, given that we don't throw the set out, that we got zerg twice? In this case, the answer is certainly 1/5 (although I am sure that there are people on this forum who will not believe it).
However, because of the wording of your original post, one might interpret it as the following. We perform an experiment, where we play two games as random, then choose one of those two games and throw out any sets of games where we didn't get zerg in that randomly chosen game. What is the probability that we got zerg in the game that wasn't chosen? In this case, the probability is 1/3.
The difference is that in the second setup, we'll throw out half of the sets where we got TZ, PZ, ZT, or PZ, while we won't throw out any ZZ sets.
On June 10 2011 10:25 Lochat wrote: Well, many people don't understand how the question is formed, or much of else it seems.
The question is not asking what he played against. It's asking what the odds that HE played zerg again. I'm just... I feel like I'm being trolled by many of the responses.
Anyhow, the question is pretty poorly formed. The odds of one random event, in a series of random events, is not modified by past events. Getting zerg as random player 10 times in a row, previously, then starting up your 11th game does not change the odds of getting zerg, it's still 1/3. People often, both consciously and subconsciously do this, they feel that "my number needs to come up, it's been so long" but a die or coin doesn't remember what it rolled or flipped before.
Now, if you're looking at future events, statistically you can determine the odds of a 1/3rd rate occurrence to happen 11 times in a row, but that doesn't matter after you've gotten zerg 10 times. Statistics are future looking chances, really. As soon as the numbers start coming up, the stats change to only look at the future events, and not the past ones.
You're absolutely correct here. It's a gambler's fallacy when they see 10 games as Zerg and think there's no way for another game as Zerg to happen again. However, this question is different. It is unknown to you which game was played (for sure) as Zerg.
I must have misread the question as well.. You just said, that you already played one game of random, in which you spawned as Zerg. You then queue up for another game, and ask for the probability of you spawning as Zerg in the second game, which would be 1/3
That was fun, I had to read it a few times to get it. Another fun brainteaser for those so inclined, list 50 consecutive numbers all of which aren't prime
On June 10 2011 10:29 Hamster1800 wrote: The most legitimate reason for choosing 1/3 over 1/5 would be because you asked for the probability of getting zerg twice by saying ``the other game''. This has implications about how the experiment was conducted. Essentially, the issue is that in the case that you did get zerg twice, there are two choices for ``the other game'', which could (depending on how the setup is interpreted) double the count for that case, raising the probability from 1/5 to 1/3.
I believe what you intended to ask was: We perform an experiment, where we play two games as random and throw out any sets of two games where we did not get zerg at least once. What is the probability, given that we don't throw the set out, that we got zerg twice? In this case, the answer is certainly 1/5 (although I am sure that there are people on this forum who will not believe it).
However, because of the wording of your original post, one might interpret it as the following. We perform an experiment, where we play two games as random, then choose one of those two games and throw out any sets of games where we didn't get zerg in that randomly chosen game. What is the probability that we got zerg in the game that wasn't chosen? In this case, the probability is 1/3.
The difference is that in the second setup, we'll throw out half of the sets where we got TZ, PZ, ZT, or PZ, while we won't throw out any ZZ sets.
Actually, the problem is well-defined. Read the edit at the bottom of the OP to see another scenario which makes more intuitive sense.
Throwing out games gives away the answer. By saying that I played Zerg at least once, that's equivalent to throwing out all sets that don't contain Z.
On June 10 2011 10:26 neo_sporin wrote: Heres the answer 1/5
Now if you reform this example but with 3 choices over 2 oppurtunites, there are 9 total pairings. Of these, 5 of them contain at least 1 zerg, and 1 of them is a zerg zerg combination. Thus, 1/5 chance the OTHER (note, not the second) game is a zerg
I don't mean to come out rude, but the boy/girl example just complicates manners. I think what you said at the end was the money explanation.
I myself almost voted 1/3 until I re-read the question. Re-reading = good
On June 10 2011 10:26 neo_sporin wrote: Heres the answer 1/5 Reasoning starting with a similar example but only 2 options. I have 2 children and ONE of them is a boy, what are the odds the other is a boy
I could have: 2 boys, 1 girl and 1 boy, 1 boy and 1 girl, or 2 girls. in a first and second child breakdown. HOWEVER, i have told you that one of them is a boy, thus 2 girls is impossible.
So now your options are 2 boys, 1 boy 1 girl, 1 girl 1 boy. the chances of the 2nd one being a boy is 1 out of 3.
You get confused because me telling you ONE of them is a boy vs telling you my first child is a boy. The boy could be item 1 or 2 or both, thus you have to account for the 1 boy 1 girl and 1 girl 1 boy choice.
Now if you reform this example but with 3 choices over 2 oppurtunites, there are 9 total pairings. Of these, 5 of them contain at least 1 zerg, and 1 of them is a zerg zerg combination. Thus, 1/5 chance the OTHER (note, not the second) game is a zerg
Yeah I misread, and assumed that he was asking for the probability of the second game.. Then I was reading answers and starting to wonder, wtf are all these people about? 1/5? fuck no, P(A | B) is the same as P(A) since they are independent! Then this lovely fellow previous page made a post making my realise my mistake in making an ass out of me (or however that one goes) tion.
On June 10 2011 10:29 Hamster1800 wrote: The most legitimate reason for choosing 1/3 over 1/5 would be because you asked for the probability of getting zerg twice by saying ``the other game''. This has implications about how the experiment was conducted. Essentially, the issue is that in the case that you did get zerg twice, there are two choices for ``the other game'', which could (depending on how the setup is interpreted) double the count for that case, raising the probability from 1/5 to 1/3.
I believe what you intended to ask was: We perform an experiment, where we play two games as random and throw out any sets of two games where we did not get zerg at least once. What is the probability, given that we don't throw the set out, that we got zerg twice? In this case, the answer is certainly 1/5 (although I am sure that there are people on this forum who will not believe it).
However, because of the wording of your original post, one might interpret it as the following. We perform an experiment, where we play two games as random, then choose one of those two games and throw out any sets of games where we didn't get zerg in that randomly chosen game. What is the probability that we got zerg in the game that wasn't chosen? In this case, the probability is 1/3.
The difference is that in the second setup, we'll throw out half of the sets where we got TZ, PZ, ZT, or PZ, while we won't throw out any ZZ sets.
Actually, the problem is well-defined. Read the edit at the bottom of the OP to see another scenario which makes more intuitive sense.
Throwing out games gives away the answer. By saying that I played Zerg at least once, that's equivalent to throwing out all sets that don't contain Z.
You missed my point. In the case where you get ZZ, what is ``the other game''?
On June 10 2011 10:22 teamsolid wrote: Probability of ZZ = 1/9 Probability of at least one Z game = 5/9
Probability of ZZ given at least one Z game = (1/9) / (5/9) = 1/5
Haha... I got to 5/9 and was searching for the answer. And searching. And wondering - wtf did I do wrong? Then I realised the question wasnt "what's the probability of at least one Z game"
Btw - probability of at least 1 Zerg game is basically: 1 - (probability of 0 Zerg games) which is way easier to calculate on the fly.
On June 10 2011 10:30 n.DieJokes wrote: That was fun, I had to read it a few times to get it. Another fun brainteaser for those so inclined, list 50 consecutive numbers all of which aren't prime
Surely you don't mean consecutive as in:
Let a,b be consecutive numbers, then b = a+1. Multiples of 2 will bite you in the ass =)
I believe you mean list the first 50 prime numbers?
On June 10 2011 10:29 Hamster1800 wrote: The most legitimate reason for choosing 1/3 over 1/5 would be because you asked for the probability of getting zerg twice by saying ``the other game''. This has implications about how the experiment was conducted. Essentially, the issue is that in the case that you did get zerg twice, there are two choices for ``the other game'', which could (depending on how the setup is interpreted) double the count for that case, raising the probability from 1/5 to 1/3.
I believe what you intended to ask was: We perform an experiment, where we play two games as random and throw out any sets of two games where we did not get zerg at least once. What is the probability, given that we don't throw the set out, that we got zerg twice? In this case, the answer is certainly 1/5 (although I am sure that there are people on this forum who will not believe it).
However, because of the wording of your original post, one might interpret it as the following. We perform an experiment, where we play two games as random, then choose one of those two games and throw out any sets of games where we didn't get zerg in that randomly chosen game. What is the probability that we got zerg in the game that wasn't chosen? In this case, the probability is 1/3.
The difference is that in the second setup, we'll throw out half of the sets where we got TZ, PZ, ZT, or PZ, while we won't throw out any ZZ sets.
Actually, the problem is well-defined. Read the edit at the bottom of the OP to see another scenario which makes more intuitive sense.
Throwing out games gives away the answer. By saying that I played Zerg at least once, that's equivalent to throwing out all sets that don't contain Z.
You missed my point. In the case where you get ZZ, what is ``the other game''?
The other game is Z. The question uses a confirmed Z game, and an unknown game. The order of the two is also unknown. This satisfies the constraints of the paradox.
This makes no sense. In the "riddle" you state that you played as Zerg at least once and then ask what is the chance that you play as Zerg again. In your answer you list all the possible matchups and say that you played a Zerg...
On June 10 2011 10:34 Keitzer wrote: One of my games was Zerg. I may or may not have been Zerg in the other game. What are my chances of getting Zerg twice in a row?
"I played 2 games. Under the assumption that one of my game is Zerg - what is the probability that both of my games are zerg."
unfortunately you worded the question incorrectly and the answer you provide is incorrect as a result of the way you asked the question.
The chances the "other game was Zerg as well" is 1/3
You used the phrase "as well" in your mind as meaning "in addition to" whereas colloquially "as well" means "similarly to".
Therefore the question can be justifiably read as saying that You played Zerg in one game and what are the chances that you similarly played Zerg in the second game.
Randomly selected you could play Zerg as one of 3 races a second time in a row since each event (playing a game as Random) always has the same probability (1/3 for each race). Therefore the chances you play Zerg the second game is 1/3 since each game is its own roll of the dice if you will. This is assuming that you meant "as well" as meaning "similar to" referencing the probability to roll Zerg just like you rolled Zerg in the first game.
Had you used a different phrase then the 1/5 would be correct. Unfortunately, due to English ambiguity and the colloquial meaning of the phrase "as well" your current brain teaser is flawed.
On June 10 2011 10:34 Keitzer wrote: One of my games was Zerg. I may or may not have been Zerg in the other game. What are my chances of getting Zerg twice in a row?
"I played 2 games. Under the assumption that one of my game is Zerg - what is the probability that both of my games are zerg."
On June 10 2011 10:29 Hamster1800 wrote: The most legitimate reason for choosing 1/3 over 1/5 would be because you asked for the probability of getting zerg twice by saying ``the other game''. This has implications about how the experiment was conducted. Essentially, the issue is that in the case that you did get zerg twice, there are two choices for ``the other game'', which could (depending on how the setup is interpreted) double the count for that case, raising the probability from 1/5 to 1/3.
I believe what you intended to ask was: We perform an experiment, where we play two games as random and throw out any sets of two games where we did not get zerg at least once. What is the probability, given that we don't throw the set out, that we got zerg twice? In this case, the answer is certainly 1/5 (although I am sure that there are people on this forum who will not believe it).
However, because of the wording of your original post, one might interpret it as the following. We perform an experiment, where we play two games as random, then choose one of those two games and throw out any sets of games where we didn't get zerg in that randomly chosen game. What is the probability that we got zerg in the game that wasn't chosen? In this case, the probability is 1/3.
The difference is that in the second setup, we'll throw out half of the sets where we got TZ, PZ, ZT, or PZ, while we won't throw out any ZZ sets.
Actually, the problem is well-defined. Read the edit at the bottom of the OP to see another scenario which makes more intuitive sense.
Throwing out games gives away the answer. By saying that I played Zerg at least once, that's equivalent to throwing out all sets that don't contain Z.
You missed my point. In the case where you get ZZ, what is ``the other game''?
The other game is Z. The question uses a confirmed Z game, and an unknown game. The order of the two is also unknown. This satisfies the constraints of the paradox.
I meant is ``the other game'' the first or the second game. I believe what you actually meant to ask then is this:
I play two games. If I never got zerg then I throw it out and try again. Otherwise, I choose a random game of the two in which I played as zerg. What is the chance that I was zerg in the game that I didn't choose?
In this case again the answer is 1/5. However, my point is that you asked the brainteaser poorly because in saying ``the other game'', you imply that you chose a game at one point, and you have hidden the constraints on that choice and in what manner you made that choice, which makes my second experiment a possible interpretation.
On June 10 2011 10:30 n.DieJokes wrote: That was fun, I had to read it a few times to get it. Another fun brainteaser for those so inclined, list 50 consecutive numbers all of which aren't prime
Surely you don't mean consecutive as in:
Let a,b be consecutive numbers, then b = a+1. Multiples of 2 will bite you in the ass =)
I believe you mean list the first 50 prime numbers?
Yep, like I said its a brainteaser (actually its an anecdote I read in a introductory number theory text but its more fun phrased this way ) so a1+1=a2, a2+1=a3 and so on and so forth
On June 10 2011 10:30 n.DieJokes wrote: That was fun, I had to read it a few times to get it. Another fun brainteaser for those so inclined, list 50 consecutive numbers all of which aren't prime
Surely you don't mean consecutive as in:
Let a,b be consecutive numbers, then b = a+1. Multiples of 2 will bite you in the ass =)
I believe you mean list the first 50 prime numbers?
Note that he said aren't prime.
Also, to further illustrate the issues that you can run into when wording the scenario, I'll take your OP:
I posted this as a simple application of probability theory to real life. Let's say IMNestea (always Z) is playing against, say, TLO, and TLO is playing random. They're going to play exactly two games (for simplicity). Your friend spoils the fun by saying "OMG Nestea's ZvZ is absolutely sick!" (implying he saw a ZvZ game).
Then this is the same problem as saying what is the probability that you have to watch two mirror matches!
If instead, your friend had said "So, in one of the games, it was ZvZ, and -" at which point you cut him off because you don't want spoilers.
Now your friend has chosen a game first (presumably the more exciting one), and then noted that it was ZvZ, rather than specifically choosing a game to comment on. Then in this case the probability that both are ZvZ is actually 1/3.
On June 10 2011 10:26 neo_sporin wrote: Heres the answer 1/5 Reasoning starting with a similar example but only 2 options. I have 2 children and ONE of them is a boy, what are the odds the other is a boy
I could have: 2 boys, 1 girl and 1 boy, 1 boy and 1 girl, or 2 girls. in a first and second child breakdown. HOWEVER, i have told you that one of them is a boy, thus 2 girls is impossible.
So now your options are 2 boys, 1 boy 1 girl, 1 girl 1 boy. the chances of the 2nd one being a boy is 1 out of 3.
You get confused because me telling you ONE of them is a boy vs telling you my first child is a boy. The boy could be item 1 or 2 or both, thus you have to account for the 1 boy 1 girl and 1 girl 1 boy choice.
Now if you reform this example but with 3 choices over 2 oppurtunites, there are 9 total pairings. Of these, 5 of them contain at least 1 zerg, and 1 of them is a zerg zerg combination. Thus, 1/5 chance the OTHER (note, not the second) game is a zerg
Best explanation in the thread.
Thanks, I was given the boy/girl question in my Algebra 2 class back in 2004 and got it right. I posed it to a few friends later and one of my schools BIO teachers started to rage at the wrongness of me and my algrebra teacher as he insisted it was 1/2. I ended up getting extra credit in my math class for proving the Bio teacher wrong (he then raged about he semantics of the question regarding first/second child vs one/other child)
On June 10 2011 10:30 n.DieJokes wrote: That was fun, I had to read it a few times to get it. Another fun brainteaser for those so inclined, list 50 consecutive numbers all of which aren't prime
Surely you don't mean consecutive as in:
Let a,b be consecutive numbers, then b = a+1. Multiples of 2 will bite you in the ass =)
I believe you mean list the first 50 prime numbers?
Yep, like I said its a brainteaser (actually its an anecdote I read in a introductory number theory text but its more fun phrased this way ) so a1+1=a2, a2+1=a3 and so on and so forth
Hehe, the trick here is in using factorials. At first, I was trying to do 50!, but then 50!+1 is not guaranteed to be a non-prime number.
So... you need to start on 50!+2, because that is definitely divisible by 2. But since we start on 2, we need to bump up our initial factorial by 1.
So the range will be [51!+2 , 51!+51].
Because 51! is divisible by each number between 2 and 51. so that guarantees that 51!+x is non-prime as long as x is between 2 and 51.
On June 10 2011 10:36 ZeromuS wrote: unfortunately you worded the question incorrectly and the answer you provide is incorrect as a result of the way you asked the question.
The chances the "other game was Zerg as well" is 1/3
You used the phrase "as well" in your mind as meaning "in addition to" whereas colloquially "as well" means "similarly to".
Therefore the question can be justifiably read as saying that You played Zerg in one game and what are the chances that you similarly played Zerg in the second game.
Randomly selected you could play Zerg as one of 3 races a second time in a row since each event (playing a game as Random) always has the same probability (1/3 for each race). Therefore the chances you play Zerg the second game is 1/3 since each game is its own roll of the dice if you will. This is assuming that you meant "as well" as meaning "similar to" referencing the probability to roll Zerg just like you rolled Zerg in the first game.
Had you used a different phrase then the 1/5 would be correct. Unfortunately, due to English ambiguity and the colloquial meaning of the phrase "as well" your current brain teaser is flawed.
Thank you, you explained that really well. I thought "as well" meant "similarly to".
I no longer consider everyone else morons for saying 1/5 :D
On June 10 2011 10:30 n.DieJokes wrote: That was fun, I had to read it a few times to get it. Another fun brainteaser for those so inclined, list 50 consecutive numbers all of which aren't prime
Surely you don't mean consecutive as in:
Let a,b be consecutive numbers, then b = a+1. Multiples of 2 will bite you in the ass =)
I believe you mean list the first 50 prime numbers?
Yep, like I said its a brainteaser (actually its an anecdote I read in a introductory number theory text but its more fun phrased this way ) so a1+1=a2, a2+1=a3 and so on and so forth
P(A|B) = P(B|A)*P(A) / P(B) . P(B|A) = 1 P(A) = 1/3 P(B)=1 . P(A|B) = 0.33 = 1/3 This is because you already stated that you were Zerg in one game. The probability that you were Zerg in one game is 100% because you already knew that. Because you already knew you were Zerg one game the only three possible options are: ZZ ZP or ZT. You asked what the other game was, so the order is irrelevant.
Its not a paradox, its all to do with how you frame the question. You are asking for the joint probability, which is 1/5. The people who are answering 1/3 are simply answering the wrong question correctly, which is considering the second game interdependently as a separate event.. Because you worded it in a rather ambiguous way its not surprising so many people answered 1/3.
The probability of your OTHER game being Zerg is mutually exclusive to the race of your original game. Hence the option of what race you play in your other game are: Z, P, T.
Z is 1 of 3 possible options, hence the probability is 1/3
You worded your question incorrectly.
The question you provided an answer for is: "What is the probability that both my games were played as zerg given at least one game was played as zerg" which is indeed 1/5.
The probability of your OTHER game being Zerg is mutually exclusive to the race of your original game. Hence the option of what race you play in your other game are: Z, P, T.
Z is 1 of 3 possible options, hence the probability is 1/3
You worded your question incorrectly.
The question you provided an answer for is: "What is the probability that both my games were played as zerg given at least one game was played as zerg" which is indeed 1/5.
This, this, this. I was like how the hell are people confused about this? lol
It's one thing to deliver a confusing mathematical situation, (ie. the boy/girl approach to this 'brainteaser') but it's an entirely different thing to just word it poorly and have people argue over semantics rather than the actual math behind it. The answer is 1/3, as explained by Tektos. You fooled a lot fewer people than you think you did. (Maybe it was possible that people actually saw the question as you intended, and still failed?)
Moral: Don't try to adapt 'brainteasers' to practical examples when you suck at language.
The real brainteaser is trying to figure out what you are actually asking.
Another solution to the 50 consecutive number problem is to use negative numbers since they weren't restricted in the phrasing of the question. That was my initial answer. But that comes from knowing another mathematical brain teaser.
Find 3 numbers in an arithmetic sequence that when multiplied together yield a prime number.
On June 10 2011 10:26 neo_sporin wrote: Heres the answer 1/5 Reasoning starting with a similar example but only 2 options. I have 2 children and ONE of them is a boy, what are the odds the other is a boy
I could have: 2 boys, 1 girl and 1 boy, 1 boy and 1 girl, or 2 girls. in a first and second child breakdown. HOWEVER, i have told you that one of them is a boy, thus 2 girls is impossible.
So now your options are 2 boys, 1 boy 1 girl, 1 girl 1 boy. the chances of the 2nd one being a boy is 1 out of 3.
You get confused because me telling you ONE of them is a boy vs telling you my first child is a boy. The boy could be item 1 or 2 or both, thus you have to account for the 1 boy 1 girl and 1 girl 1 boy choice.
Now if you reform this example but with 3 choices over 2 oppurtunites, there are 9 total pairings. Of these, 5 of them contain at least 1 zerg, and 1 of them is a zerg zerg combination. Thus, 1/5 chance the OTHER (note, not the second) game is a zerg
Best explanation in the thread.
I have almost no understanding of probability, so help me out with something. If I flip a coin 100 times and it lands heads every time, wouldn't each flip after that still have a 50/50 chance of coming up heads?
I'm not understanding how the fact that you got zerg 1 game has any affect on the chances of you getting zerg in the next game. I would be inclined to answer 1/3, being that in any truly random event with 3 possible outcomes the likely hood of each of those outcomes would be 1/3.
The post didn't ask "what are the chances of getting zerg in both games if you got zerg at least once." Even if it did ask that, by definition how can 1 random event change the odds of another random event? The question only asks about 1 game, so how is the first game relevant to the odds of the second game?
As for the boy girl part above I am not following. You are saying that girl girl is an impossibility of course, but (boy + girl) and (girl + boy) why would you count this possibility twice? It is making both children variables, when only 1 is actually a variable.You just don't know which one is the variable. If you count (boy + girl) twice then wouldn't you have to count (Boy + Boy) twice?
On June 10 2011 11:11 Tektos wrote: [...] The probability of your OTHER game being Zerg is mutually exclusive to the race of your original game. Hence the option of what race you play in your other game are: Z, P, T.
Z is 1 of 3 possible options, hence the probability is 1/3
You worded your question incorrectly.
The question you provided an answer for is: "What is the probability that both my games were played as zerg given at least one game was played as zerg" which is indeed 1/5.
On June 10 2011 11:22 nitdkim wrote: I have a coin. I flipped it twice. It was heads at least once. What is the probability that it was heads on the both times?
(the real teaser :p)
Fixed... but nice try. =/
"2nd flip" means you're asking for 1 flip's probability.
The answer is 1 in 3. You didn't ask what the odds were that both games were Z if at least one was. you asked what is the chance that 1 game had selected 1 option of 3. if I ate a red cupcake and another cupcake (even chances of red, green, blue) The chance that I had 2 red cupcakes is 1 in 3.
On June 10 2011 11:22 nitdkim wrote: I have a coin. I flipped it twice. It was heads at least once. What is the probability that it was heads on the both times?
(the real teaser :p)
Fixed... but nice try. =/
"2nd flip" means you're asking for 1 flip's probability.
That's indeed what he was doing, and the answer is not 1/2.
On June 10 2011 11:22 nitdkim wrote: I have a coin. I flipped it twice. It was heads at least once. What is the probability that it was heads on the both times?
(the real teaser :p)
Fixed... but nice try. =/
"2nd flip" means you're asking for 1 flip's probability.
That's indeed what he was doing, and the answer is not 1/2.
I disagree, he said 'on the 2nd flip' which excludes all other information, because he's asking for the probability of a single, specific, 50/50 flip. Had he said 'both times', or something similar, it would have been fine.
On June 10 2011 11:22 nitdkim wrote: I have a coin. I flipped it twice. It was heads at least once. What is the probability that it was heads on the both times?
(the real teaser :p)
Fixed... but nice try. =/
"2nd flip" means you're asking for 1 flip's probability.
That's indeed what he was doing, and the answer is not 1/2.
I disagree, he said 'on the 2nd flip' which excludes all other information, because he's asking for the probability of a single, specific, 50/50 flip. Had he said 'both times', or something similar, it would have been fine.
Then do it yourself. Get a coin, flip it in pairs 100 times (so 200 flips). Then cross out all pairs where neither of the ones were heads and count up how many are left and how many had the second one as heads.
On June 10 2011 11:29 Kyuukyuu wrote: Since "I played Z once" is a given it has no effect on the probability of the other one; I'm with the "you worded it wrongly" camp
The question is actually worded fine. It just sounds ambiguous, because the question is in the poll, while the given part is written earlier. If it was all put in one sentence, there'd be 0 ambiguity though.
clever brainteaser...i thought it was 1/9 at first, but with probability you have to be really careful to read the question as it was stated. 1/5 makes sense - you limit the possibilities first (must have been z for either 1st or second game), then you determine the probability of x event (being z both games). Of all math branches, I think probability is the toughest because you have to be the most precise/logical in your thinking.
On June 10 2011 11:22 nitdkim wrote: I have a coin. I flipped it twice. It was heads at least once. What is the probability that it was heads on the both times?
(the real teaser :p)
Fixed... but nice try. =/
"2nd flip" means you're asking for 1 flip's probability.
That's indeed what he was doing, and the answer is not 1/2.
I disagree, he said 'on the 2nd flip' which excludes all other information, because he's asking for the probability of a single, specific, 50/50 flip. Had he said 'both times', or something similar, it would have been fine.
I meant to say what I've said. If you think about it, you will be confused for a little while.
On June 10 2011 11:22 nitdkim wrote: I have a coin. I flipped it twice. It was heads at least once. What is the probability that it was heads on the 2nd flip?
(the real teaser :p)
Good teaser. Now, ignoring all those comments about you miswriting the question, here is what we have.
There are initially 4 scenarios possible:
HT TH HH TT
But you gave us the information that heads came up at least once. That rules out TT, leaving HT, TH, and HH.
That gives you a 2/3 probability that it was heads on the second flip.
On June 10 2011 11:45 Whitewing wrote: Not really a brainteaser, but it looks like most people here have no idea what conditional probability is >_<.
The question isn't "What are the odds of randomly getting zerg in a match."
The question is "What are the odds of getting zerg twice in two matches, given that I got zerg at least once."
Yep. It seems a lot of people are missing the whole conditional part of the problem statement. I didn't find any ambiguity in it at all (unless it has been edited, which I don't think it was). And as someone else said before, the poll was only the last part of the full question, which may have been the cause for some confusion.
The odds that you play zerg will always be 1/3. No matter if you play as zerg 6 times before, random doesn't remember. You still have 1/3 chance of getting zerg.
The probability of getting things, 'in a row' id different.
Anyways, I'm not a math major (but I will be in a few year =D ), but this thread has been an overall bad experience.
On June 10 2011 11:51 Warpath wrote: Guys, the answer is 1/2 It's either he gets Zerg, or he doesn't.
On June 10 2011 10:26 neo_sporin wrote: Heres the answer 1/5 Reasoning starting with a similar example but only 2 options. I have 2 children and ONE of them is a boy, what are the odds the other is a boy
I could have: 2 boys, 1 girl and 1 boy, 1 boy and 1 girl, or 2 girls. in a first and second child breakdown. HOWEVER, i have told you that one of them is a boy, thus 2 girls is impossible.
So now your options are 2 boys, 1 boy 1 girl, 1 girl 1 boy. the chances of the 2nd one being a boy is 1 out of 3.
You get confused because me telling you ONE of them is a boy vs telling you my first child is a boy. The boy could be item 1 or 2 or both, thus you have to account for the 1 boy 1 girl and 1 girl 1 boy choice.
Now if you reform this example but with 3 choices over 2 oppurtunites, there are 9 total pairings. Of these, 5 of them contain at least 1 zerg, and 1 of them is a zerg zerg combination. Thus, 1/5 chance the OTHER (note, not the second) game is a zerg
Best explanation in the thread.
I have almost no understanding of probability, so help me out with something. If I flip a coin 100 times and it lands heads every time, wouldn't each flip after that still have a 50/50 chance of coming up heads?
I'm not understanding how the fact that you got zerg 1 game has any affect on the chances of you getting zerg in the next game. I would be inclined to answer 1/3, being that in any truly random event with 3 possible outcomes the likely hood of each of those outcomes would be 1/3.
The post didn't ask "what are the chances of getting zerg in both games if you got zerg at least once." Even if it did ask that, by definition how can 1 random event change the odds of another random event? The question only asks about 1 game, so how is the first game relevant to the odds of the second game?
As for the boy girl part above I am not following. You are saying that girl girl is an impossibility of course, but (boy + girl) and (girl + boy) why would you count this possibility twice? It is making both children variables, when only 1 is actually a variable.You just don't know which one is the variable. If you count (boy + girl) twice then wouldn't you have to count (Boy + Boy) twice?
I will respond using the boy/girl as it only has 2 variable. The boy/girl and girl/boy are split up for clarity purposes, but really have no effect on the outcome of boy/boy being 1/3 So by using this word choice you have to account for the two very different situations of the boy coming first and the boy coming second. They do not have to statistically count for 2 different values though.
IF i told you the FIRST was a boy, then yes there is a 50% as they are exclusive events and do not depend on eachother. But since the question is being asked in such a way that the two events do depend on eachother for a combined probability, you have to look at the entire problem concurrently, not independently.
So the choices become 1st child 2nd child boy girl boy girl If you create all possible events in this scenario you could have since having a boy and girl each have 50% chance of occuring. 1st boy 2nd boy 1st boy 2nd girl 1st girl 2nd boy 1st girl 2nd boy So now you see there is a 25% boy boy, 50% boy/girl combination in some fashion. 25% girl girl.
However the question as presented has told you that there is at least one boy (but possibly 2) so you take out the girl girl option. You are now left with 25% boy/boy or 50% boy/girl combination. Since it will never be girl/girl there is now a total of 75 times out of 100 that these solutions will come up. 25 out of the 75 are solutions involving boy boy, the other 50 are boy/girl or girl/boy. 25 out of 75 is 1/3.
On June 10 2011 11:52 Steel wrote: It really depends how you word it.
Nitpick: It doesn't matter how you word it. It matters what the situation is. As it turns out there are lots of wordings that correspond to more than one situation, and this is arguably one of them (although it's better than a lot of things that I've seen).
On June 10 2011 11:52 Steel wrote: but this thread has been an overall bad experience.
I know, right?!?!?!? What the hell is it with people getting so worked up about math. Remember the whole 0.999...=1 proofs? God, I don't get why people get so worked up over this stuff. X_X
On June 10 2011 11:51 Warpath wrote: Guys, the answer is 1/2 It's either he gets Zerg, or he doesn't.
On June 10 2011 11:52 Steel wrote: but this thread has been an overall bad experience.
I know, right?!?!?!? What the hell is it with people getting so worked up about math. Remember the whole 0.999...=1 proofs? God, I don't get why people get so worked up over this stuff. X_X
On June 10 2011 11:52 Steel wrote: but this thread has been an overall bad experience.
I know, right?!?!?!? What the hell is it with people getting so worked up about math. Remember the whole 0.999...=1 proofs? God, I don't get why people get so worked up over this stuff. X_X
On June 10 2011 11:51 Warpath wrote: Guys, the answer is 1/2 It's either he gets Zerg, or he doesn't.
HAHAHA, nice one!
Actually I think mathematics are amazing...when they are used properly.
These threads ALWAYS turns into a wording argument. It's frustrating. Like that 1/2*5=? bullshit. There's no right or wrong answer because the wording the problem itself is too ambiguous so people just argue for goddamn ever!! ARHGHH!!!
Srs bzns note: I thought it was 1/3 as well rather than 1/5. It seems to be very ambiguously worded -- then again, I always sucked at the verbal part of standardized tests.
It's like saying I flipped a coin 15 times and it was heads 15 times whats the chance of getting heads on the 16th try and the probability is 1/2. You can get zerg 30 times in a row the next game the probability of zerg will still be 1/3
I think the question is all messed up man. You said you played as zerg AT LEAST once, so that's saying you played as zerg once, which is the minimum.
so at MOST the chance is 1/3 becuz you asked if you played AS zerg and the choices are Terran, Protoss, and Zerg. Correct me if I'm wrong and explain it better, lol. I want to know
You almost ask two different questions here, the phrasing is terrible. It should either be "I played as Zerg at least once. What is the probability that in both games I was Zerg?" or "I played as Zerg in one game. What is the probability that my other game was as Zerg as well?"
"My other game" implies that you are only asking about the odds of that one game *given* that you were Zerg in the first game.
On June 10 2011 12:11 Cryllic wrote: It's like saying I flipped a coin 15 times and it was heads 15 times whats the chance of getting heads on the 16th try and the probability is 1/2. You can get zerg 30 times in a row the next game the probability of zerg will still be 1/3
That is unrelated to the question that the OP was asking. He is not saying ``I played my first game today as random and got zerg. What is the probability that I was zerg in the second game, too?''. He is saying ``I played two games today, and I will tell you that I was zerg at least once. Knowing that, what is the probability that I was zerg twice?''.
Many of the other posters in the past page or two have had the same misunderstanding.
For those of you who will say that I originally said that the problem was ambiguous, I will say that if someone gave me this problem I would first say 1/5, then say, ``but you should be careful about how you word the problem in the future, because...''. The ambiguity in this problem is very close to being nonexistent, and most of the people in this thread missed the actual ambiguity and are actually completely misunderstanding the question.
On June 10 2011 12:15 jesse20ghet wrote: I think the question is all messed up man. You said you played as zerg AT LEAST once, so that's saying you played as zerg once, which is the minimum.
so at MOST the chance is 1/3 becuz you asked if you played AS zerg and the choices are Terran, Protoss, and Zerg. Correct me if I'm wrong and explain it better, lol. I want to know
He's asking what the odds are that he got Zerg BOTH games. The way the question is worded it looks like he's asking what the odds are that he'd get Zerg in a single game.
On June 10 2011 12:15 jesse20ghet wrote: I think the question is all messed up man. You said you played as zerg AT LEAST once, so that's saying you played as zerg once, which is the minimum.
so at MOST the chance is 1/3 becuz you asked if you played AS zerg and the choices are Terran, Protoss, and Zerg. Correct me if I'm wrong and explain it better, lol. I want to know
He's asking what the odds are that he got Zerg BOTH games. The way the question is worded it looks like he's asking what the odds are that he'd get Zerg in a single game.
Exactly. The way it's worded you can just totally ignore ONE of 2 games, and give the probability of the other gane, which is 1/3.
I don't understand how people can know the paradox, see that it's worded in the exact same way (i.e. purposely ambiguous) and still argue that it's for sure one or the other.
You are treating PZ and ZP separate. That's not correct. If one of your game was played as zerg, then the probability of getting any one of the races on the other is 1/3.
1/5 would be the answer if the Bnet server tells you that you are about to play two games, and it will make it so that you play zerg at least once. In that case, PZ and ZP would be different, but in this case, they are the same.
I answered 1/3. The reason why I answered 1/3 is because the wording made it seem that you played Z first and the # that you were looking for was the probability that you would get Z again. If I hadn't read your op incorrectly (which is prolly my fault) I would have answered 1/5 (only 1 ZZ option out of all Zx pairings)
the wording of the question is fine, and it is clearly 1/5. focusing too much on the order of reading question leads to problems in understanding....basic probablity = HS education
pleaseee restore the original wording of the Q and put ur rebuttal in spoilers or at the bottom of the post...I enjoyed it as a Q
I spent about 20-30 minutes mulling it over in my mind, thinking, it has to be 1/3. But then why would he post it? I couldn't get behind the idea of it being anything else though, so I decided to brute force it in a slightly different way than OP.
OP got Zerg at least once, so possible permutations are: ZP,ZT,ZZ; PZ,TZ,ZZ. The duplicate "ZZ" is crossed out. Now we have one "ZZ" left and the others are ZP,ZT,PZ,TZ -- meaning the chance of getting Zerg in the other game as well once already getting Zerg once has to be 1/5.
Probability will stick you in the egg boiler and make you delicious.
On June 10 2011 12:51 Beez wrote: the answer doesnt rely on knowing the race of your opponent so you can be T, P, or Z. theres no reason to even think about matchups.
indeed you are correct, but everybody knows this and nobody is arguing about matchups it is simply "i play two games as random, i spawn as zerg atleast once, what's the probability of me spawning as zerg in both games"
You are taking an incredibly simple problem, obscuring it with imprecise language and presenting it to a casual populace, then defending yourself with "but three people understood it".
That is not how you should perform a serious survey.
On June 10 2011 12:39 DragonDefonce wrote: You are treating PZ and ZP separate. That's not correct. If one of your game was played as zerg, then the probability of getting any one of the races on the other is 1/3.
1/5 would be the answer if the Bnet server tells you that you are about to play two games, and it will make it so that you play zerg at least once. In that case, PZ and ZP would be different, but in this case, they are the same.
Pretty sure this is correct.
The order of the games has absolutely no effect. At least one game is the same as there is a game, is it not?
On June 10 2011 13:01 Jinsho wrote: You are taking an incredibly simple problem, obscuring it with imprecise language and presenting it to a casual populace, then defending yourself with "but three people understood it".
That is not how you should perform a serious survey.
Yea, if you are going to post a brainteaser, I'd rather they be fun riddles where you have to think outside the box (like those prisoner/island/village population ones). Re-visiting math isn't fun for me. Anyone got some good riddles to share?
This isn't even a brain teaser it is a simple probability question with added convoluted and ambiguous use of the English language to throw people off.
Example of an actual brain teaser:
During a visit to a mental asylum, a visitor asked the Director what the criteria is that defines if a patient should be institutionalized.
"Well," said the Director, "we fill up a bathtub. Then we offer a teaspoon, a teacup, and a bucket to the patient and ask the patient to empty the bathtub."
Okay, here's your test: 1. Would you use the spoon? 2. Would you use the teacup? 3. Would you use the bucket?
"Oh, I understand," said the visitor. "A normal person would choose the bucket, as it is larger than the spoon." What was the director's response?
I love how this is at least the third time this problem has come up on a Teamliquid thread. I think a very small minority of people got this question wrong because they misunderstood probability theory. This isn't a brainteaser. You've just managed to convince yourself that it is. This is a very straightforward question pushed into muddled oblivion with semantics. It's very simple. If at least one of the games is zerg then it is 1/5. If you're asking the probability the second game is zerg it's 1/3. EDIT: In b4 "it's already posted" several people got here first.
On June 10 2011 13:55 d.o.c wrote: If you're asking the probability the second game is zerg it's 1/3.
That is not true. If I said ``I played two games as random today. In at least one of them, I was zerg. What is the probability that I was zerg in the second game?'' then the answer is 3/5, not 1/3. You changed the setup of the problem (to ``I was zerg in the first game'' instead of ``I was zerg at least once''), not the question that it is asking.
"In the question's case, we are not asking the above. We are asking that given that at least 1 game was Zerg, what is the probability that both games are Zerg."
"What is the probability that my other game was as Zerg as well?"
These are two different things, that's not how you ask this question properly in English.
Reading the original post, it seems perfectly clear. It says he played as Zerg at least once. What is the probability that the other game he played is as Zerg as well, it has no mention of what is first or what is second, so the answer is 1/5.
On June 10 2011 14:08 terr13 wrote: Reading the original post, it seems perfectly clear. It says he played as Zerg at least once. What is the probability that the other game he played is as Zerg as well, it has no mention of what is first or what is second, so the answer is 1/5.
He re-worded the question...
The initial wording implied that one game was locked as zerg, then what is the probability that the other was zerg. Giving only three posibilities zerg/terran or protoss - therefore 1/3.
The OP's question is just a variation of the "brainteaser" presented above (3 choices instead of 2). The wiki article helps explain why some people might view the question as ambiguous and presents explanations for both sides.
What is the probability that you have to watch 2 mirror match games?
This question is still ambiguous. Did your friend, who commented on the ZvZ, specifically set out searching for a ZvZ game? Did he simply pick a two-game series at random which may or may not have had a zerg in it? If its the former then yes, there are 5 possible permutations and the answer is 1/5. If its the latter, then there is no longer any relation between the first and second game and the answer is 1/3.
From your wording, it seems as if your friend simply wanted to see Nestea vs TLO, and so this series very well may not have had TLO as zerg at all. Therefore the answer is 1/3.
The wikipedia page for the boy/girl paradox words this reasoning rather nicely:
From all families with two children, at least one of whom is a boy, a family is chosen at random. This would yield the answer of 1/3. From all families with two children, one child is selected at random, and the sex of that child is specified. This would yield an answer of 1/2.
There is some further explaining of the problem on the wikipedia page, I suggest you read it.
If you still don't agree, I can't help you. Try wikipedia.
In the wikipedia entry on the boy and girl problem the question is worded "what is the probability that BOTH children are the same gender". You say "the other game" which is a single game (as evidenced by the fact that the noun is singular). If you're talking about a single game, the race you draw is not impacted by what you drew in another game. Had you said "the other game as well", THEN it would have been clear you were asking the chances that both were zerg. The only people who are saying it's fine are the ones who were familiar with the problem prior to looking at this post and so immediately knew what you were going for regardless of the wording.
Just say "both games" instead of implying that we're all stupid for reading what you wrote instead of psychically knowing what you meant without prior knowledge of the problem.
EDIT: Sorry didn't catch where you already changed the wording. The poll still says "the other game" so you're still going to get people saying 1/3.
The two games are independant events, and their order doesn't matter. Saying that it's twice as likely you'll get P because ZP and PZ are two different possibilities should intuitively make no sense to anyone. PZ and ZP are the same thing because the order of the games played has no effect on the solution. The odds of getting zerg in "the other game," REGARDLESS of what happens in "a game" are 1/3.
On June 10 2011 10:11 theDreamStick wrote: It is correct that I am asking a conditional probability question: Given that at least one of my games was Zerg, what is the probability that both of my games are zerg.
It's either 100%, or 0%, depending on whether or not both games were zerg.
I'll give you a puzzle: I either ate a sandwich today, or I didn't.
I've never learned about conditional probability before, so I ran a few calculations, and interesting to find out, 1/5 was indeed the correct answer.
For the couple posts above me that still don't understand, 1/3 would be the answer if the two games are individual events, but they are not. They are linked with the condition that if one game is P or T, then the other cannot be. If you factory in this condition into your calculations, you end up with 1/5.
On June 10 2011 10:11 theDreamStick wrote: It is correct that I am asking a conditional probability question: Given that at least one of my games was Zerg, what is the probability that both of my games are zerg.
It's either 100%, or 0%, depending on whether or not both games were zerg.
I'll give you a puzzle: I either ate a sandwich today, or I didn't.
What is the probability that I ate a sandwich?
With the information provided nobody is able to tell you what the probability that you eat a sandwich is, as eating a sandwich vs. not eating a sandwich is not a fixed probability event. Some people eat sandwiches every day, some people eat sandwiches once a week, some people never eat sandwiches. The only information that can be presented is that the probability that it is either of those two events is 100% because the probability of an event added to the compliment of that event is always 100%.
You could, however, give us a various sample of consecutive days stating whether you ate a sandwich on that day or not and an approximate probability to a certain confidence level based on sample size could be predicted provided the event of eating a sandwich is not pattern based.
You should specify the question that you are playing random. As for the Nestea vs TLO example, isn't that just not applicable? Because you know that Nestea will play zerg. So one side of the matchup will allways be zerg, hence it would be 1/3 in that case. But maybe you just added a non related fact to confuse people?
Also just because people understand what you mean with your question does not mean that the wording is good. In fact it could be better.
This whole thread makes me feel dumb. Never was good at math.
Anyway,
Here's a real probability mind screw for you: It's a real pic a friend of mine uploaded to facebook during her a recent trip to Vegas. I call it: 34 Red (lol) - + Show Spoiler +
On June 10 2011 15:15 Chairman Ray wrote: I've never learned about conditional probability before, so I ran a few calculations, and interesting to find out, 1/5 was indeed the correct answer.
For the couple posts above me that still don't understand, 1/3 would be the answer if the two games are individual events, but they are not. They are linked with the condition that if one game is P or T, then the other cannot be. If you factory in this condition into your calculations, you end up with 1/5.
They ARE individual events. I said this few pages ago and I'll say it again:
If the Bnet server told you that in the next two games you play as random, it will ensure that you play at zerg at least once, then the two games are related, and this would in fact be true.
But this question says this: I played two games as random. I got zerg once. Whats the chance that I got zerg the other game?
Also look at it this way: Assume he got zerg first game. Whats the probability of getting zerg second game? Vice versa? Add them? Still 1/3.
On June 10 2011 15:36 Tektos wrote: As someone who did quite an extensive amount of probability in my education it utterly offends me when I hear: "As someone who doesn't know anything about probability... I reread the question multiple times and after thinking about it the answer is definitely 1/5"
On June 10 2011 15:36 Tektos wrote: As someone who did quite an extensive amount of probability in my education it utterly offends me when I hear: "As someone who doesn't know anything about probability... I reread the question multiple times and after thinking about it the answer is definitely 1/5"
Haha wow there seems to be a lot of debate over a simple conditional probability problem. Yes, the wording was a little bit unclear, but his edits clarified what he was talking about. Some people seem to just not understand the concept of a conditional probability, or are still misreading the problem.
"I played two games as random today. In at least one of the games I was Zerg. (NOT the first game, but just in one of the two - this is an important distinction that the OP was a little bit unclear on at first but wasn't necessarily wrong either). Given this, what is the probability that I played both games as Zerg? (Not what the probability of getting Zerg in the second game was)"
Conditional probability P (A l B) = P(A and B) / P(A) (Probability of having gotten Zerg in two games given Zerg in at least one game) / (Probability of getting Zerg in one game)
The possible combinations are ZZ, ZP, ZT, TZ, PZ, PP, TT, TP, PT. Probability of getting Zerg in one game = 5/9 (since PP, TT, TP and PT are eliminated) Given this condition, the probability of getting another Zerg = 1/9 (only ZZ remains out of the 9)
Therefore, using the equation, we arrive at 1/9 divided by 5/9 = 9/45, or 1/5.
On June 10 2011 15:47 OneOther wrote: Haha wow there seems to be a lot of debate over a simple conditional probability problem. Yes, the wording was a little bit unclear, but his edits clarified what he was talking about. Some people seem to just not understand the concept of a conditional probability, or are still misreading the problem.
"I played two games as random today. In at least one of the games I was Zerg. (NOT the first game, but just in one of the two - this is an important distinction that the OP was a little bit unclear on at first but wasn't necessarily wrong either). Given this, what is the probability that I played both games as Zerg? (Not what the probability of getting Zerg in the second game was)"
Conditional probability P (A l B) = P(A and B) / P(A) (Probability of getting Zerg in two games given Zerg in at least one game) / (Probability of getting Zerg in one game)
The possible combinations are ZZ, ZP, ZT, TZ, PZ, PP, TT, TP, PT. Probability of getting Zerg in one game = 5/9 (since PP, TT, TP and PT are eliminated) Given this probability, the probability of getting another Zerg = 1/9 (only ZZ remains out of the 9)
Therefore, using the equation, we arrive at 1/9 divided by 5/9 = 9/40, or 1/5.
ZP and PZ are the same outcome (playing once as zerg and once as protoss) TZ and ZT are the same outcome (playing once as zerg and once as terran)
That consolidates your possible outcomes down to 3 results, giving 1/3 chance that the other game will be zerg.
The possible combinations are ZZ, ZP, ZT, TZ, PZ, PP, TT, TP, PT. Probability of getting Zerg in one game = 5/9 (since PP, TT, TP and PT are eliminated) Given this condition, the probability of getting another Zerg = 1/9 (only ZZ remains out of the 9)
Therefore, using the equation, we arrive at 1/9 divided by 5/9 = 9/45, or 1/5.
I just don't understand how ZP and PZ, or TZ and ZT are different. The OP can only be one of 3 races, why does the opponent's race have any bearing at all? Doesn't that invalidate this calculation?
Since the probability of getting zerg in 1 game is 1/3, getting zerg in 2 games = 2/6, which simplifies to 1/3 obviously, but is needed for the next step.
The unsimplified probability for a single result out of 6 games looks something like this: 1.Zerg------I 2.Zerg------I 3.Terran----I==========(2 opportunities to get zerg out of 6 games or 1/3 probability) 4.Terran----I 5.Protoss---I 6.Protoss---I
Once you already get zerg as the first game though, the fraction looks more like this:
1.XZergXXI==(this option has already been chosen and is not considered) 2.Zerg------I 3.Terran----I==========(1 opportunity to get zerg out of 5 remaining games or 1/5 probability) 4.Terran----I 5.Protoss---I 6.Protoss---I
So the probability for getting zerg in 2 consecutive games is 1/5 Do i win?
On June 10 2011 15:36 Tektos wrote: To everyone who says 1/5 is the correct answer:
Are you assuming getting Z P is a different event as getting P Z?
Well, yes, because they are different events, aren't they?
I play 2 games as random, at least one is zerg. The possible outcomes given this constraint are:
I play 2 games as zerg I play 1 game as zerg and 1 as protoss I play 1 game as zerg and 1 as terran
3 possible outcomes, playing zerg twice is one of the possible outcomes: 1 / 3
I play 2 games as Zerg I play game 1 as Zerg and game 2 as Protoss. I play game 1 as Protoss and game 2 as Zerg. I play game 1 as Zerg and game 2 as Terran. I play game 1 as Terran and game 2 as Zerg.
5 possible outcomes; playing Zerg twice is one of the possible outcomes.
Here's a similar problem that my stats teacher gave me some years ago. It's something along these lines.
You are at a party chatting with a guest and they share that they have 2 children. Right after they tell you this a girl walks up to the person you are talking to and the person says "oh, here is my daughter now." The question is what is the probability that the other child is also a girl. The answer I was given was 1/3 for the same reason, 3 possibilities, BG, GB, and GG so 1 out of 3, right?
Here's my problem.. The fact that the girl walked up is a completely random occurrence. It could have just as easily been a boy. If it was a boy, you would also say that the chance of the person having 2 boys is 1/3. Now you must be saying that the probability the unknown offspring is the opposite gender of the first offspring you see is 2/3. We know that the probability of having 2 different gender children vs 2 same-sex children is 50-50. What am I missing?
This is very simple as many others have said. Given that I played either game 1 as zerg or game 2 as zerg, what are the odds that i played zerg two times in a row.
First off, there are 9 possibile combinations pp pz pt tp tz tt zp zz zt and only in 1 scenario is zz. Get rid of any scenarios in which you don't play zerg and there are 5 left, thus 1/5.
There are many ways to do this problem and they all result in 1/5.
On June 10 2011 15:47 OneOther wrote: Haha wow there seems to be a lot of debate over a simple conditional probability problem. Yes, the wording was a little bit unclear, but his edits clarified what he was talking about. Some people seem to just not understand the concept of a conditional probability, or are still misreading the problem.
"I played two games as random today. In at least one of the games I was Zerg. (NOT the first game, but just in one of the two - this is an important distinction that the OP was a little bit unclear on at first but wasn't necessarily wrong either). Given this, what is the probability that I played both games as Zerg? (Not what the probability of getting Zerg in the second game was)"
Conditional probability P (A l B) = P(A and B) / P(A) (Probability of getting Zerg in two games given Zerg in at least one game) / (Probability of getting Zerg in one game)
The possible combinations are ZZ, ZP, ZT, TZ, PZ, PP, TT, TP, PT. Probability of getting Zerg in one game = 5/9 (since PP, TT, TP and PT are eliminated) Given this probability, the probability of getting another Zerg = 1/9 (only ZZ remains out of the 9)
Therefore, using the equation, we arrive at 1/9 divided by 5/9 = 9/40, or 1/5.
ZP and PZ are the same outcome (playing once as zerg and once as protoss) TZ and ZT are the same outcome (playing once as zerg and once as terran)
That consolidates your possible outcomes down to 3 results, giving 1/3 chance that the other game will be zerg.
By your reasoning, I can prove to you that you have a fifty percent chance of rolling a 1, using a fair die.
There are two possibilities: you roll 1 or you don't roll 1. 2, 3, 4, 5, 6 are the same result: not 1. That consolidates your possible outcomes down to 2 results, giving 1/2 chance that you roll a 1.
On June 10 2011 15:52 Stropheum wrote: Since the probability of getting zerg in 1 game is 1/3, getting zerg in 2 games = 2/6, which simplifies to 1/3 obviously, but is needed for the next step. Since you played 1 game already and got zerg 1 time, your probability for 2 zerg games remains 2/6, yet you factor in that you've already played one game, reducing it to 2/5, and you factor in the fact that you've gotten zerg, making the probability 1/5 that you will get zerg a second time. Do i win?
No the probability of getting zerg in two consecutive games is 1/3 * 1/3 = 1/9
On June 10 2011 10:11 theDreamStick wrote: It is correct that I am asking a conditional probability question: Given that at least one of my games was Zerg, what is the probability that both of my games are zerg.
It's either 100%, or 0%, depending on whether or not both games were zerg.
I'll give you a puzzle: I either ate a sandwich today, or I didn't.
What is the probability that I ate a sandwich?
With the information provided nobody is able to tell you what the probability that you eat a sandwich is, as eating a sandwich vs. not eating a sandwich is not a fixed probability event. Some people eat sandwiches every day, some people eat sandwiches once a week, some people never eat sandwiches. The only information that can be presented is that the probability that it is either of those two events is 100% because the probability of an event added to the compliment of that event is always 100%.
You could, however, give us a various sample of consecutive days stating whether you ate a sandwich on that day or not and an approximate probability to a certain confidence level based on sample size could be predicted provided the event of eating a sandwich is not pattern based.
So it's based on a huge set of assumptions and extra information.
Let's say that it happened like this:
Your friend sits down to watch TLO play a game. He randoms.
He only manages to watch one of the games.
After the game, he tells you, "holy shit, TLO's zerg is fucking awesome."
Or "TLO's protoss is fucking awesome."
Or "TLO's terran is fucking awesome."
Then, he asks you, because he's a clever sadistic bastard,
"What's the chance that TLO was zerg twice?"
Or "What's the chance that TLO was protoss twice?"
Or "What's the chance that TLO was terran twice?"
The answer to the first question is one third.
The answer to the second question is one third.
The answer to the third question is one third.
So the answer is one third if the backstory happened like this. But it's 1/5th if the backstory happened another way, say that your friend is this disgustingly patriotic zerg fanboy.and would only tell you that TLO played a zerg game and he doesn't give a shit about terran or protoss games.
But since you don't know the backstory, you can't answer the question.
The same way you can't answer my question without knowing if I ate a sandwich today.
On June 10 2011 15:47 OneOther wrote: Haha wow there seems to be a lot of debate over a simple conditional probability problem. Yes, the wording was a little bit unclear, but his edits clarified what he was talking about. Some people seem to just not understand the concept of a conditional probability, or are still misreading the problem.
"I played two games as random today. In at least one of the games I was Zerg. (NOT the first game, but just in one of the two - this is an important distinction that the OP was a little bit unclear on at first but wasn't necessarily wrong either). Given this, what is the probability that I played both games as Zerg? (Not what the probability of getting Zerg in the second game was)"
Conditional probability P (A l B) = P(A and B) / P(A) (Probability of getting Zerg in two games given Zerg in at least one game) / (Probability of getting Zerg in one game)
The possible combinations are ZZ, ZP, ZT, TZ, PZ, PP, TT, TP, PT. Probability of getting Zerg in one game = 5/9 (since PP, TT, TP and PT are eliminated) Given this probability, the probability of getting another Zerg = 1/9 (only ZZ remains out of the 9)
Therefore, using the equation, we arrive at 1/9 divided by 5/9 = 9/40, or 1/5.
ZP and PZ are the same outcome (playing once as zerg and once as protoss) TZ and ZT are the same outcome (playing once as zerg and once as terran)
That consolidates your possible outcomes down to 3 results, giving 1/3 chance that the other game will be zerg.
ZP and PZ are different outcomes. TZ and ZT are different outcomes. Getting a Zerg in the first game and Protoss in the second game is different from getting Protoss in the first game and then Zerg in the second. Same for ZT/TZ.
The possible combinations are ZZ, ZP, ZT, TZ, PZ, PP, TT, TP, PT. Probability of getting Zerg in one game = 5/9 (since PP, TT, TP and PT are eliminated) Given this condition, the probability of getting another Zerg = 1/9 (only ZZ remains out of the 9)
Therefore, using the equation, we arrive at 1/9 divided by 5/9 = 9/45, or 1/5.
I just don't understand how ZP and PZ, or TZ and ZT are different. The OP can only be one of 3 races, why does the opponent's race have any bearing at all? Doesn't that invalidate this calculation?
What do you mean? The opponent's races don't have to do anything with the problem.
On June 10 2011 15:47 OneOther wrote: Haha wow there seems to be a lot of debate over a simple conditional probability problem. Yes, the wording was a little bit unclear, but his edits clarified what he was talking about. Some people seem to just not understand the concept of a conditional probability, or are still misreading the problem.
"I played two games as random today. In at least one of the games I was Zerg. (NOT the first game, but just in one of the two - this is an important distinction that the OP was a little bit unclear on at first but wasn't necessarily wrong either). Given this, what is the probability that I played both games as Zerg? (Not what the probability of getting Zerg in the second game was)"
Conditional probability P (A l B) = P(A and B) / P(A) (Probability of getting Zerg in two games given Zerg in at least one game) / (Probability of getting Zerg in one game)
The possible combinations are ZZ, ZP, ZT, TZ, PZ, PP, TT, TP, PT. Probability of getting Zerg in one game = 5/9 (since PP, TT, TP and PT are eliminated) Given this probability, the probability of getting another Zerg = 1/9 (only ZZ remains out of the 9)
Therefore, using the equation, we arrive at 1/9 divided by 5/9 = 9/40, or 1/5.
ZP and PZ are the same outcome (playing once as zerg and once as protoss) TZ and ZT are the same outcome (playing once as zerg and once as terran)
That consolidates your possible outcomes down to 3 results, giving 1/3 chance that the other game will be zerg.
By your reasoning, I can prove to you that you have a fifty percent chance of rolling a 1, using a fair die.
There are two possibilities: you roll 1 or you don't roll 1. 2, 3, 4, 5, 6 are the same result: not 1. That consolidates your possible outcomes down to 2 results, giving 1/2 chance that you roll a 1.
You see the flaw in this, don't you?
ZP vs. PZ is comepletely different from P(A) vs. P(not A) you seem to have no clue what you're doing.
On June 10 2011 15:47 OneOther wrote: Haha wow there seems to be a lot of debate over a simple conditional probability problem. Yes, the wording was a little bit unclear, but his edits clarified what he was talking about. Some people seem to just not understand the concept of a conditional probability, or are still misreading the problem.
"I played two games as random today. In at least one of the games I was Zerg. (NOT the first game, but just in one of the two - this is an important distinction that the OP was a little bit unclear on at first but wasn't necessarily wrong either). Given this, what is the probability that I played both games as Zerg? (Not what the probability of getting Zerg in the second game was)"
Conditional probability P (A l B) = P(A and B) / P(A) (Probability of getting Zerg in two games given Zerg in at least one game) / (Probability of getting Zerg in one game)
The possible combinations are ZZ, ZP, ZT, TZ, PZ, PP, TT, TP, PT. Probability of getting Zerg in one game = 5/9 (since PP, TT, TP and PT are eliminated) Given this probability, the probability of getting another Zerg = 1/9 (only ZZ remains out of the 9)
Therefore, using the equation, we arrive at 1/9 divided by 5/9 = 9/40, or 1/5.
ZP and PZ are the same outcome (playing once as zerg and once as protoss) TZ and ZT are the same outcome (playing once as zerg and once as terran)
That consolidates your possible outcomes down to 3 results, giving 1/3 chance that the other game will be zerg.
ZP and PZ are different outcomes. TZ and ZT are different outcomes. Getting a Zerg in the first game and Protoss in the second game is different from getting Protoss in the first game and then Zerg in the second. Same for ZT/TZ.
No they are the same event because the games are independent. Everyone is confusing dependent event conditional probability with independent event conditional probability.
On June 10 2011 15:47 OneOther wrote: Haha wow there seems to be a lot of debate over a simple conditional probability problem. Yes, the wording was a little bit unclear, but his edits clarified what he was talking about. Some people seem to just not understand the concept of a conditional probability, or are still misreading the problem.
"I played two games as random today. In at least one of the games I was Zerg. (NOT the first game, but just in one of the two - this is an important distinction that the OP was a little bit unclear on at first but wasn't necessarily wrong either). Given this, what is the probability that I played both games as Zerg? (Not what the probability of getting Zerg in the second game was)"
Conditional probability P (A l B) = P(A and B) / P(A) (Probability of getting Zerg in two games given Zerg in at least one game) / (Probability of getting Zerg in one game)
The possible combinations are ZZ, ZP, ZT, TZ, PZ, PP, TT, TP, PT. Probability of getting Zerg in one game = 5/9 (since PP, TT, TP and PT are eliminated) Given this probability, the probability of getting another Zerg = 1/9 (only ZZ remains out of the 9)
Therefore, using the equation, we arrive at 1/9 divided by 5/9 = 9/40, or 1/5.
ZP and PZ are the same outcome (playing once as zerg and once as protoss) TZ and ZT are the same outcome (playing once as zerg and once as terran)
That consolidates your possible outcomes down to 3 results, giving 1/3 chance that the other game will be zerg.
By your reasoning, I can prove to you that you have a fifty percent chance of rolling a 1, using a fair die.
There are two possibilities: you roll 1 or you don't roll 1. 2, 3, 4, 5, 6 are the same result: not 1. That consolidates your possible outcomes down to 2 results, giving 1/2 chance that you roll a 1.
You see the flaw in this, don't you?
ZP vs. PZ is comepletely different from P(A) vs. P(not A) you seem to have no clue what you're doing.
What I'm doing is showing how your "consolidation process" can be abused 6 ways till Sunday. You're the one who doesn't seem to know what you're doing.
Playing a Zerg in game 1 and playing a Zerg in game 2 are different, mutually exclusive events. Saying that every event that includes a Zerg is the same event, is precisely as (in)valid as saying that every event that is non-1 is the same event.
On June 10 2011 15:47 OneOther wrote: Haha wow there seems to be a lot of debate over a simple conditional probability problem. Yes, the wording was a little bit unclear, but his edits clarified what he was talking about. Some people seem to just not understand the concept of a conditional probability, or are still misreading the problem.
"I played two games as random today. In at least one of the games I was Zerg. (NOT the first game, but just in one of the two - this is an important distinction that the OP was a little bit unclear on at first but wasn't necessarily wrong either). Given this, what is the probability that I played both games as Zerg? (Not what the probability of getting Zerg in the second game was)"
Conditional probability P (A l B) = P(A and B) / P(A) (Probability of getting Zerg in two games given Zerg in at least one game) / (Probability of getting Zerg in one game)
The possible combinations are ZZ, ZP, ZT, TZ, PZ, PP, TT, TP, PT. Probability of getting Zerg in one game = 5/9 (since PP, TT, TP and PT are eliminated) Given this probability, the probability of getting another Zerg = 1/9 (only ZZ remains out of the 9)
Therefore, using the equation, we arrive at 1/9 divided by 5/9 = 9/40, or 1/5.
ZP and PZ are the same outcome (playing once as zerg and once as protoss) TZ and ZT are the same outcome (playing once as zerg and once as terran)
That consolidates your possible outcomes down to 3 results, giving 1/3 chance that the other game will be zerg.
By your reasoning, I can prove to you that you have a fifty percent chance of rolling a 1, using a fair die.
There are two possibilities: you roll 1 or you don't roll 1. 2, 3, 4, 5, 6 are the same result: not 1. That consolidates your possible outcomes down to 2 results, giving 1/2 chance that you roll a 1.
You see the flaw in this, don't you?
ZP vs. PZ is comepletely different from P(A) vs. P(not A) you seem to have no clue what you're doing.
What I'm doing is showing how your "consolidation process" can be abused 6 ways till Sunday. You're the one who doesn't seem to know what you're doing.
Playing a Zerg in game 1 and playing a Zerg in game 2 are different, mutually exclusive events. Saying that every event that includes a Zerg is the same event, is precisely as (in)valid as saying that every event that is non-1 is the same event.
No it can't be abused 6 ways till Sunday unless you have no clue about a thing to do with probability. Order of occurrence does not play a factor in independent event conditional probability.
ZZ - Your friend reports, "TLO played at least one game as zerg." ZP - Your friend reports, "TLO played at least one game as zerg." ZT - Your friend reports, "TLO played at least one game as zerg." PZ - Your friend reports, "TLO played at least one game as zerg." TZ - Your friend reports, "TLO played at least one game as zerg." TT - Your friend goes to the pub. TP - Your friend goes to the pub. PT - Your friend goes to the pub. PP - Your friend goes to the pub.
Your friend asks you, "what is the chance that TLO played both games with the same race."
The answer to the question in this case is 1/5.
ZZ - Your friend reports, "TLO played at least one game as zerg." ZP - Your friend reports with 50% chance "TLO played at least one game as zerg" and 50% chance "TLO played at least one game as protoss." ZT - Your friend reports with 50% chance "TLO played at least one game as zerg" and 50% chance "TLO played at least one game as terran." PZ - Your friend reports with 50% chance "TLO played at least one game as zerg" and 50% chance "TLO played at least one game as protoss." TZ - Your friend reports with 50% chance "TLO played at least one game as zerg" and 50% chance "TLO played at least one game as terran." TT - Your friend reports, "TLO played at least one game as terran." TP - Your friend reports with 50% chance "TLO played at least one game as protoss" and 50% chance "TLO played at least one game as terran." PT - Your friend reports with 50% chance "TLO played at least one game as protoss" and 50% chance "TLO played at least one game as terran." PP - Your friend reports, "TLO played at least one game as protoss."
Your friend asks you, "what is the chance that TLO played both games with the same race."
The answer to the question in this case is 1/3.
You have no idea of knowing your friend's method of coming up with the question.
You cannot answer the question without further information.
On June 10 2011 16:01 Tektos wrote: No they are the same event because the games are independent. Everyone is confusing dependent event conditional probability with independent event conditional probability.
OK, let's do it this way. Imagine that a bunch of random games are played.
1/9 of them will be ZZ. 2/9 of them will be ZP/PZ 2/9 of them will be ZT/TZ
2/9 of them will be TP/PT 1/9 of them will be PP 1/9 of them will be TT.
Do you agree with the above?
If so, then consider the set of sets of games where at least one Z is played: 1/5 of these are ZZ.
On June 10 2011 15:47 OneOther wrote: Haha wow there seems to be a lot of debate over a simple conditional probability problem. Yes, the wording was a little bit unclear, but his edits clarified what he was talking about. Some people seem to just not understand the concept of a conditional probability, or are still misreading the problem.
"I played two games as random today. In at least one of the games I was Zerg. (NOT the first game, but just in one of the two - this is an important distinction that the OP was a little bit unclear on at first but wasn't necessarily wrong either). Given this, what is the probability that I played both games as Zerg? (Not what the probability of getting Zerg in the second game was)"
Conditional probability P (A l B) = P(A and B) / P(A) (Probability of getting Zerg in two games given Zerg in at least one game) / (Probability of getting Zerg in one game)
The possible combinations are ZZ, ZP, ZT, TZ, PZ, PP, TT, TP, PT. Probability of getting Zerg in one game = 5/9 (since PP, TT, TP and PT are eliminated) Given this probability, the probability of getting another Zerg = 1/9 (only ZZ remains out of the 9)
Therefore, using the equation, we arrive at 1/9 divided by 5/9 = 9/40, or 1/5.
ZP and PZ are the same outcome (playing once as zerg and once as protoss) TZ and ZT are the same outcome (playing once as zerg and once as terran)
That consolidates your possible outcomes down to 3 results, giving 1/3 chance that the other game will be zerg.
By your reasoning, I can prove to you that you have a fifty percent chance of rolling a 1, using a fair die.
There are two possibilities: you roll 1 or you don't roll 1. 2, 3, 4, 5, 6 are the same result: not 1. That consolidates your possible outcomes down to 2 results, giving 1/2 chance that you roll a 1.
You see the flaw in this, don't you?
ZP vs. PZ is comepletely different from P(A) vs. P(not A) you seem to have no clue what you're doing.
What I'm doing is showing how your "consolidation process" can be abused 6 ways till Sunday. You're the one who doesn't seem to know what you're doing.
Playing a Zerg in game 1 and playing a Zerg in game 2 are different, mutually exclusive events. Saying that every event that includes a Zerg is the same event, is precisely as (in)valid as saying that every event that is non-1 is the same event.
No it can't be abused 6 ways till Sunday unless you have no clue about a thing to do with probability. Order of occurrence does not play a factor in independent event conditional probability.
By your logic, the chance of flipping two coins and get two heads in a row is a 1/3, because the possible outcomes are HH, TT, and HT (which is the same as TH).
On June 10 2011 15:47 OneOther wrote: Haha wow there seems to be a lot of debate over a simple conditional probability problem. Yes, the wording was a little bit unclear, but his edits clarified what he was talking about. Some people seem to just not understand the concept of a conditional probability, or are still misreading the problem.
"I played two games as random today. In at least one of the games I was Zerg. (NOT the first game, but just in one of the two - this is an important distinction that the OP was a little bit unclear on at first but wasn't necessarily wrong either). Given this, what is the probability that I played both games as Zerg? (Not what the probability of getting Zerg in the second game was)"
Conditional probability P (A l B) = P(A and B) / P(A) (Probability of getting Zerg in two games given Zerg in at least one game) / (Probability of getting Zerg in one game)
The possible combinations are ZZ, ZP, ZT, TZ, PZ, PP, TT, TP, PT. Probability of getting Zerg in one game = 5/9 (since PP, TT, TP and PT are eliminated) Given this probability, the probability of getting another Zerg = 1/9 (only ZZ remains out of the 9)
Therefore, using the equation, we arrive at 1/9 divided by 5/9 = 9/40, or 1/5.
ZP and PZ are the same outcome (playing once as zerg and once as protoss) TZ and ZT are the same outcome (playing once as zerg and once as terran)
That consolidates your possible outcomes down to 3 results, giving 1/3 chance that the other game will be zerg.
ZP and PZ are different outcomes. TZ and ZT are different outcomes. Getting a Zerg in the first game and Protoss in the second game is different from getting Protoss in the first game and then Zerg in the second. Same for ZT/TZ.
No they are the same event because the games are independent. Everyone is confusing dependent event conditional probability with independent event conditional probability.
The problem you are having is that you are observing each event independently when in fact it's a combined event.
"Given that a player plays as zerg a game, what are the odds that he will play zerg in the next game?" This is 1/3. In this there are 3 scenarios zp zt zz
"Given that a player has played 2 games and at least 1 of them was as zerg, what are the odds both were as zerg". In this there are 5 possibilities all equally possible zp zt zz tz pz, only 1 has both as zerg.
In the first statement, only the first event has occurred, so the next indepedant event is calculated
In the second statement, both events have already occurred, so they are now calculated as single event.
On June 10 2011 15:47 OneOther wrote: Haha wow there seems to be a lot of debate over a simple conditional probability problem. Yes, the wording was a little bit unclear, but his edits clarified what he was talking about. Some people seem to just not understand the concept of a conditional probability, or are still misreading the problem.
"I played two games as random today. In at least one of the games I was Zerg. (NOT the first game, but just in one of the two - this is an important distinction that the OP was a little bit unclear on at first but wasn't necessarily wrong either). Given this, what is the probability that I played both games as Zerg? (Not what the probability of getting Zerg in the second game was)"
Conditional probability P (A l B) = P(A and B) / P(A) (Probability of getting Zerg in two games given Zerg in at least one game) / (Probability of getting Zerg in one game)
The possible combinations are ZZ, ZP, ZT, TZ, PZ, PP, TT, TP, PT. Probability of getting Zerg in one game = 5/9 (since PP, TT, TP and PT are eliminated) Given this probability, the probability of getting another Zerg = 1/9 (only ZZ remains out of the 9)
Therefore, using the equation, we arrive at 1/9 divided by 5/9 = 9/40, or 1/5.
ZP and PZ are the same outcome (playing once as zerg and once as protoss) TZ and ZT are the same outcome (playing once as zerg and once as terran)
That consolidates your possible outcomes down to 3 results, giving 1/3 chance that the other game will be zerg.
By your reasoning, I can prove to you that you have a fifty percent chance of rolling a 1, using a fair die.
There are two possibilities: you roll 1 or you don't roll 1. 2, 3, 4, 5, 6 are the same result: not 1. That consolidates your possible outcomes down to 2 results, giving 1/2 chance that you roll a 1.
You see the flaw in this, don't you?
ZP vs. PZ is comepletely different from P(A) vs. P(not A) you seem to have no clue what you're doing.
What I'm doing is showing how your "consolidation process" can be abused 6 ways till Sunday. You're the one who doesn't seem to know what you're doing.
Playing a Zerg in game 1 and playing a Zerg in game 2 are different, mutually exclusive events. Saying that every event that includes a Zerg is the same event, is precisely as (in)valid as saying that every event that is non-1 is the same event.
No it can't be abused 6 ways till Sunday unless you have no clue about a thing to do with probability. Order of occurrence does not play a factor in independent event conditional probability.
Please stop appealing to your education. You're just embarrassing yourself.
Would it help if you imagined a big label across the screen saying game 1 or game 2? What you don't seem to be getting is that order is one of the distinguishing factors of different events here, same as Protoss vs. Terran.
Suppose for argument's sake that we had some extra races: game 1-Protoss and game 2-Protoss being different things, as are game 1-Terran and game 2-Terran, game 1-Zerg and game 2-Zerg. All we know about the guy is that he chose one "game 1" race and one "game 2" race [edit: and that he chose at least one of the "Zerg" races]. The problem would boil down to exactly the same thing.
This kind of makes the problem sound more complicated, but maybe it helps you see why order matters here.
On June 10 2011 15:47 OneOther wrote: Haha wow there seems to be a lot of debate over a simple conditional probability problem. Yes, the wording was a little bit unclear, but his edits clarified what he was talking about. Some people seem to just not understand the concept of a conditional probability, or are still misreading the problem.
"I played two games as random today. In at least one of the games I was Zerg. (NOT the first game, but just in one of the two - this is an important distinction that the OP was a little bit unclear on at first but wasn't necessarily wrong either). Given this, what is the probability that I played both games as Zerg? (Not what the probability of getting Zerg in the second game was)"
Conditional probability P (A l B) = P(A and B) / P(A) (Probability of getting Zerg in two games given Zerg in at least one game) / (Probability of getting Zerg in one game)
The possible combinations are ZZ, ZP, ZT, TZ, PZ, PP, TT, TP, PT. Probability of getting Zerg in one game = 5/9 (since PP, TT, TP and PT are eliminated) Given this probability, the probability of getting another Zerg = 1/9 (only ZZ remains out of the 9)
Therefore, using the equation, we arrive at 1/9 divided by 5/9 = 9/40, or 1/5.
ZP and PZ are the same outcome (playing once as zerg and once as protoss) TZ and ZT are the same outcome (playing once as zerg and once as terran)
That consolidates your possible outcomes down to 3 results, giving 1/3 chance that the other game will be zerg.
By your reasoning, I can prove to you that you have a fifty percent chance of rolling a 1, using a fair die.
There are two possibilities: you roll 1 or you don't roll 1. 2, 3, 4, 5, 6 are the same result: not 1. That consolidates your possible outcomes down to 2 results, giving 1/2 chance that you roll a 1.
You see the flaw in this, don't you?
ZP vs. PZ is comepletely different from P(A) vs. P(not A) you seem to have no clue what you're doing.
What I'm doing is showing how your "consolidation process" can be abused 6 ways till Sunday. You're the one who doesn't seem to know what you're doing.
Playing a Zerg in game 1 and playing a Zerg in game 2 are different, mutually exclusive events. Saying that every event that includes a Zerg is the same event, is precisely as (in)valid as saying that every event that is non-1 is the same event.
No it can't be abused 6 ways till Sunday unless you have no clue about a thing to do with probability. Order of occurrence does not play a factor in independent event conditional probability.
By your logic, the chance of flipping two coins and get two heads in a row is a 1/3, because the possible outcomes are HH, TT, and HT (which is the same as TH).
The possible combinations are ZZ, ZP, ZT, TZ, PZ, PP, TT, TP, PT. Probability of getting Zerg in one game = 5/9 (since PP, TT, TP and PT are eliminated) Given this condition, the probability of getting another Zerg = 1/9 (only ZZ remains out of the 9)
Therefore, using the equation, we arrive at 1/9 divided by 5/9 = 9/45, or 1/5.
I just don't understand how ZP and PZ, or TZ and ZT are different. The OP can only be one of 3 races, why does the opponent's race have any bearing at all? Doesn't that invalidate this calculation?
What do you mean? The opponent's races don't have to do anything with the problem.
I mean that when you mark the player on the left as the opponent, and the OP as the player on the right you'd get these outcomes for ZP, ZT, PZ, TZ, and ZZ:
Opp Z v P OP Opp Z v T OP Opp P v Z OP Opp T v Z OP Opp Z v Z OP
If you just label the Opponent as race "X" in all five scenarios, because it doesn't matter for the problem, you'd get: X v P, X v T, X v Z, X v Z, X v Z.
Eliminating duplicate scenarios for the OP, you get: X v P, X v T, and X v Z. Isn't that only three options? Wouldn't getting X v Z twice in a row be 1/9?
I really just don't understand because I'm bad at probability. Help? ^^
On June 10 2011 15:47 OneOther wrote: Haha wow there seems to be a lot of debate over a simple conditional probability problem. Yes, the wording was a little bit unclear, but his edits clarified what he was talking about. Some people seem to just not understand the concept of a conditional probability, or are still misreading the problem.
"I played two games as random today. In at least one of the games I was Zerg. (NOT the first game, but just in one of the two - this is an important distinction that the OP was a little bit unclear on at first but wasn't necessarily wrong either). Given this, what is the probability that I played both games as Zerg? (Not what the probability of getting Zerg in the second game was)"
Conditional probability P (A l B) = P(A and B) / P(A) (Probability of getting Zerg in two games given Zerg in at least one game) / (Probability of getting Zerg in one game)
The possible combinations are ZZ, ZP, ZT, TZ, PZ, PP, TT, TP, PT. Probability of getting Zerg in one game = 5/9 (since PP, TT, TP and PT are eliminated) Given this probability, the probability of getting another Zerg = 1/9 (only ZZ remains out of the 9)
Therefore, using the equation, we arrive at 1/9 divided by 5/9 = 9/40, or 1/5.
ZP and PZ are the same outcome (playing once as zerg and once as protoss) TZ and ZT are the same outcome (playing once as zerg and once as terran)
That consolidates your possible outcomes down to 3 results, giving 1/3 chance that the other game will be zerg.
By your reasoning, I can prove to you that you have a fifty percent chance of rolling a 1, using a fair die.
There are two possibilities: you roll 1 or you don't roll 1. 2, 3, 4, 5, 6 are the same result: not 1. That consolidates your possible outcomes down to 2 results, giving 1/2 chance that you roll a 1.
You see the flaw in this, don't you?
ZP vs. PZ is comepletely different from P(A) vs. P(not A) you seem to have no clue what you're doing.
What I'm doing is showing how your "consolidation process" can be abused 6 ways till Sunday. You're the one who doesn't seem to know what you're doing.
Playing a Zerg in game 1 and playing a Zerg in game 2 are different, mutually exclusive events. Saying that every event that includes a Zerg is the same event, is precisely as (in)valid as saying that every event that is non-1 is the same event.
No it can't be abused 6 ways till Sunday unless you have no clue about a thing to do with probability. Order of occurrence does not play a factor in independent event conditional probability.
Please stop appealing to your education. You're just embarrassing yourself.
Would it help if you imagined a big label across the screen saying game 1 or game 2? What you don't seem to be getting is that order is one of the distinguishing factors of different events here, same as Protoss vs. Terran.
Suppose for argument's sake that we had some extra races: game 1-Protoss and game 2-Protoss being different things, as are game 1-Terran and game 2-Terran, game 1-Zerg and game 2-Zerg. All we know about the guy is that he chose one "game 1" race and one "game 2" race. The problem would boil down to exactly the same thing.
This kind of makes the problem sound more complicated, but maybe it helps you see why order matters here.
Hahaha ouch, completely embarrassing a guy trying to throw subtle personal jabs. I guess this is what athletes mean when they say "let your game speak." Anyways, thanks for clarifying that, I was away momentarily and it's nice to see that I don't have to explain everything.
The thing that is confusing people is the fact that you're not being asked between games what the chance of the next game is going to be. You're forced to make a call on the probability of two ZvZ occurring before any clarification regarding outcomes is provided. That chance is 1 in 9. You then learn that one of the games turned out to be a ZvZ, but that is not known when making the original probability. That occurrence eliminates the possibility of not getting at least one Z in the two games, bringing the total number of options to 5. Therefore, it is 1 in 5.
However, not knowing this does not make you stupid, or bad at maths. This is another perfect example of theoretical maths knowledge that has no practical application. Much like those ambiguous order of operations questions, it's not possible to know the answer until confronted with a question that requires learning the answer. Simple logic doesn't work. You need to understand exact syntax.
On June 10 2011 16:17 Clog wrote: this has nothing to do with the OP
But that's wrong.
No it isn't
The OP said "Given that at least one of my games was Zerg, what is the probability that both of my games are zerg."
You said something like "Given that at least one of my games was Zerg, what is the probability that the other game I am zerg."
The OP is discussing 2 games. You are discussing one game.
In all the situations that the answer to one of the questions is yes the other question is also yes, and in all the situations that the answer to one of the questions is no the other question is also no. Since the results concur in all situations they are precisely identical.
Just a note, threads like these are just purposefully divisive and humiliating. They're using logical ambiguity to attempt to embarrass those who don't already know the exact answer. Posts like this:
On June 10 2011 16:21 Eknoid4 wrote: This does not deserve a thread.
this is middle school math.
are just assholes trying to make others look stupid.
On June 10 2011 16:20 naggerNZ wrote: The thing that is confusing people is the fact that you're not being asked between games what the chance of the next game is going to be. You're forced to make a call on the probability of two ZvZ occurring before any clarification regarding outcomes is provided. That chance is 1 in 9. You then learn that one of the games turned out to be a ZvZ, but that is not known when making the original probability. That occurrence eliminates the possibility of not getting at least one Z in the two games, bringing the total number of options to 5. Therefore, it is 1 in 5.
However, not knowing this does not make you stupid, or bad at maths. This is another perfect example of theoretical maths knowledge that has no practical application. Much like those ambiguous order of operations questions, it's not possible to know the answer until confronted with a question that requires learning the answer. Simple logic doesn't work. You need to understand exact syntax.
One's opponent has no effect on this problem. You are making it very confusing. The rest of the post is great though ^^
On June 10 2011 16:17 Clog wrote: this has nothing to do with the OP
But that's wrong.
No it isn't
The OP said "Given that at least one of my games was Zerg, what is the probability that both of my games are zerg."
You said something like "Given that at least one of my games was Zerg, what is the probability that the other game I am zerg."
The OP is discussing 2 games. You are discussing one game.
In all the situations that the answer to one of the questions is yes the other question is also yes, and in all the situations that the answer to one of the questions is no the other question is also no. Since the results concur in all situations they are precisely identical.
You seem to be having trouble understanding conditional probability then. Since you are explicitly separating the games, basically into game 1 and game 2, the fact that you are zerg game 1 does not affect what race you are in game 2.
Since the op is making no such distinction, there are a greater number of instances in which you do *not* get zerg both games - hence the conditional probability.
It goes back to simply jotting down possibilities. When you refer to "the other game", the only possiblities are P, T, and Z. When the OP says that he was zerg one game (we do not know which), possbilities are PZ, ZP, TZ, ZT, ZZ. These are all possible and different scenarios.
If you still do not understand, I'm afraid I can't help you. But do understand that you are arguing against centuries of mathematics and statistics.
On June 10 2011 16:24 naggerNZ wrote: Just a note, threads like these are just purposefully divisive and humiliating. They're using logical ambiguity to attempt to embarrass those who don't already know the exact answer. Posts like this:
On June 10 2011 16:21 Eknoid4 wrote: This does not deserve a thread.
this is middle school math.
are just assholes trying to make others look stupid.
No, I'm serious. This doesn't deserve a thread. Ask a friend who is decent at math. Don't start a 10+ page argument. I don't have to try to make anybody i nthis thread look stupid. People arguing with basic math concepts when the OP has more to do with reading comprehension (which, in the context of math, you also should have learned in middle school) are making themselves look stupid enough.
On June 10 2011 16:24 naggerNZ wrote: Just a note, threads like these are just purposefully divisive and humiliating. They're using logical ambiguity to attempt to embarrass those who don't already know the exact answer. Posts like this:
On June 10 2011 16:21 Eknoid4 wrote: This does not deserve a thread.
this is middle school math.
are just assholes trying to make others look stupid.
No, I'm serious. This doesn't deserve a thread. Ask a friend who is decent at math. Don't start a 10+ page argument. I don't have to try to make anybody i nthis thread look stupid. People arguing with basic math concepts when the OP has more to do with reading comprehension (which, in the context of math, you also should have learned in middle school) are making themselves look stupid enough.
On June 10 2011 16:17 Clog wrote: this has nothing to do with the OP
But that's wrong.
No it isn't
The OP said "Given that at least one of my games was Zerg, what is the probability that both of my games are zerg."
You said something like "Given that at least one of my games was Zerg, what is the probability that the other game I am zerg."
The OP is discussing 2 games. You are discussing one game.
In all the situations that the answer to one of the questions is yes the other question is also yes, and in all the situations that the answer to one of the questions is no the other question is also no. Since the results concur in all situations they are precisely identical.
You seem to be having trouble understanding conditional probability then. Since you are explicitly separating the games, basically into game 1 and game 2, the fact that you are zerg game 1 does not affect what race you are in game 2.
I understand conditional probability just fine. This has nothing to do with the fact that the two questions are identical. If you asked those two questions to two different people who gave you the same answer they would be right or wrong precisely the same percentage of the time.
On June 10 2011 16:29 Clog wrote: Since the op is making no such distinction, there are a greater number of instances in which you do *not* get zerg both games - hence the conditional probability.
It goes back to simply jotting down possibilities. When you refer to "the other game", the only possiblities are P, T, and Z. When the OP says that he was zerg one game (we do not know which), possbilities are PZ, ZP, TZ, ZT, ZZ. These are all possible and different scenarios.
But you cannot claim that after your friend tells you "one player game was played as zerg," that PZ ZP TZ ZT ZZ occurred with equal probability because you have not considered the probability that your friend's report may have been based on the results of the random race selection in both games.
On June 10 2011 16:29 Clog wrote: If you still do not understand, I'm afraid I can't help you. But do understand that you are arguing against centuries of mathematics and statistics.
On June 10 2011 16:24 naggerNZ wrote: Just a note, threads like these are just purposefully divisive and humiliating. They're using logical ambiguity to attempt to embarrass those who don't already know the exact answer. Posts like this:
On June 10 2011 16:21 Eknoid4 wrote: This does not deserve a thread.
this is middle school math.
are just assholes trying to make others look stupid.
No, I'm serious. This doesn't deserve a thread. Ask a friend who is decent at math. Don't start a 10+ page argument. I don't have to try to make anybody i nthis thread look stupid. People arguing with basic math concepts when the OP has more to do with reading comprehension (which, in the context of math, you also should have learned in middle school) are making themselves look stupid enough.
Limiting your options to ZP, ZZ, ZT, PZ, TZ (order preserved).
That means there is only a 1/5 chance you got the ZZ, so that the other match was zerg.
When you said, ZP, ZZ, ZT, PZ, and TZ, you stated the number of possible combinations incorrectly. The order of the games doesn't actually matter, the only thing that does matter is how many times you played as a certain race. So now, ZP and PZ should be one outcome, and ZT and TZ should also be one outcome.
With those outcomes combined, there should only be three outcomes: ZZ, ZP, and ZT, so the chance of playing as zerg twice given that you played as zerg at least once, is 1/3.
On June 10 2011 16:24 naggerNZ wrote: Just a note, threads like these are just purposefully divisive and humiliating. They're using logical ambiguity to attempt to embarrass those who don't already know the exact answer. Posts like this:
On June 10 2011 16:21 Eknoid4 wrote: This does not deserve a thread.
this is middle school math.
are just assholes trying to make others look stupid.
No, I'm serious. This doesn't deserve a thread. Ask a friend who is decent at math. Don't start a 10+ page argument. I don't have to try to make anybody i nthis thread look stupid. People arguing with basic math concepts when the OP has more to do with reading comprehension (which, in the context of math, you also should have learned in middle school) are making themselves look stupid enough.
Limiting your options to ZP, ZZ, ZT, PZ, TZ (order preserved).
That means there is only a 1/5 chance you got the ZZ, so that the other match was zerg.
When you said, ZP, ZZ, ZT, PZ, and TZ, you stated the number of possible combinations incorrectly. The order of the games doesn't actually matter, the only thing that does matter is how many times you played as a certain race. So now, ZP and PZ should be one outcome, and ZT and TZ should also be one outcome.
With those outcomes combined, there should only be three outcomes: ZZ, ZP, and ZT, so the chance of playing as zerg twice given that you played as zerg at least once, is 1/3.
The problem with this thread isn't just the quarreling over the wording of the question, but it seems not everyone can even agree on the definition and axioms of probability. If you flip a coin and put a book over it before it's seen, is the probability that it is heads 0.5 or 1? Even for this simple question, you'll get different answers and debate because of this. I can only imagine the shitstorm if someone had posted the Monty Hall problem before it ever became well-known.
Limiting your options to ZP, ZZ, ZT, PZ, TZ (order preserved).
That means there is only a 1/5 chance you got the ZZ, so that the other match was zerg.
When you said, ZP, ZZ, ZT, PZ, and TZ, you stated the number of possible combinations incorrectly. The order of the games doesn't actually matter, the only thing that does matter is how many times you played as a certain race. So now, ZP and PZ should be one outcome, and ZT and TZ should also be one outcome.
With those outcomes combined, there should only be three outcomes: ZZ, ZP, and ZT, so the chance of playing as zerg twice given that you played as zerg at least once, is 1/3.
I understand the reasoning behind it, but I'd still argue it's incorrect. The order of the game does matter when considering its independent probability of happening. If you play two games as random, you are twice as likely to get 1 Zerg 1 Protoss than you are getting Zerg both games (which can be shown by splitting the Zerg + Protoss result into two based on order, and given them equal probabilities).
It's the same reason that after flipping a coin twice you have a 50% chance to get 1 Heads 1 Tails, and then 25% each for both Heads and both Tails. And this, as well, could be split up based on order saying that its just 25% for all the following: HH, HT, TH, TT
Basically, if you do something to identify a specific game or result (such as using the words "the other"), then the order does not matter, and the SC2 example answer would in fact be 1/3. But if you do not (such as the OP currently has in the non-spoiler chunk of his post), it does, for purposes as shown in the coin flipping scenario, and the SC2 example answer would be 1/5.
*Sigh I'm very surprised that this thread is still going on. For those who think the answer is 1/3 still after the rewording of the problem, you need to stop being stubborn and listen to the facts.
Independent Events:
The answer is 1/3 if they are independent events. This is the case in which the games are independent:
Total Possible Outcomes: PP PT PZ TP TT TZ ZP ZT ZZ
However "given that at least one of my games was Zerg, what is the probability that both of my games are zerg." So now we must eliminate to meet our conditions (hence conditional probability) We are now left with:
Total Outcomes that meet the conditions stated: PZ TZ ZP ZT ZZ
Now we simply count which meet condition one (one game played was zerg) and condition two (the other game is zerg AS WELL). One out of Five = 1/5.
Yes the wording was ambiguous at first, which led to confusion. But the "as well," makes it conditional. Now the OP changed so there is no ambiguity at all simply asking straight out that both games are played as zerg after one of them was already played as zerg. This is just a purely simple proof.
Limiting your options to ZP, ZZ, ZT, PZ, TZ (order preserved).
That means there is only a 1/5 chance you got the ZZ, so that the other match was zerg.
When you said, ZP, ZZ, ZT, PZ, and TZ, you stated the number of possible combinations incorrectly. The order of the games doesn't actually matter, the only thing that does matter is how many times you played as a certain race. So now, ZP and PZ should be one outcome, and ZT and TZ should also be one outcome.
With those outcomes combined, there should only be three outcomes: ZZ, ZP, and ZT, so the chance of playing as zerg twice given that you played as zerg at least once, is 1/3.
WRONG. Look at above all possible outcomes to see why order of games does matter.
Well it seems like the best way to settle this is for everybody to just go out and test this. Its simple, just go into a random number generator and set the range from 1 to 3.
1 = zerg, 2 = toss, 3 = terran
Now, generate a random number twice to create a set of data. Next, do this 9999 more times.
Done? Good. Now, decide which question you want to ask.
If you want to look only at the series that have at least one zerg in them, then eliminate all the data sets that don't have a 1 in them. Next, count how many of these sets have two zergs. The answer will approximately be equal to the chances of having two zergs in that situation.
If you want to look at all of the series, choose one set at random. Does it have a 1 in it? If so, record the other number in that set. Next, do this 9999 more times. The number of times that you record a double zerg will be approximately equal to the probability of finding a second zerg in a set if you randomly choose one that already contains a zerg.
There you go, there should be two different answers, both derived empirically. No more arguing should be allowed until everybody has done this.
I just fucking used brute force to prove it. If any idiot beyond this points wants to refute you deserve a permaban for inability to read and inability to do math.
OP: your updated question is much more ambiguous than the original with modifications
Let's say IMNestea (always Z) is playing against, say, TLO, and TLO is playing random. They're going to play exactly two games (for simplicity). Your friend spoils the fun by saying "OMG Nestea's ZvZ is absolutely sick!" (implying he saw a ZvZ game).
This leaves a lot up for interpretation. Your friend could have said that to you after the first game, with the 2nd game yet to be played, in this case, the answer is 1/3.
What you really meant was
Let's say IMNestea (always Z) is playing against, say, TLO, and TLO is playing random. They played exactly two games (for simplicity). Your friend spoils the fun by saying "OMG Nestea's ZvZ is absolutely sick!" (implying he saw at least one ZvZ game).
This is a simple application of conditional probability, and it's hardly a brainteaser. Those who say 1/5 are correct:
P(A | B) = P(A \cap B) / P(B) (\cap is the intersection sign)
In this case: P(A) = twice Z P(B) = at least Z once
P(A \cap B): Since P(A) is a subset of P(B), P(A \cap B) = P(A) = 1/3 * 1/3 = 1/9
P(B) = P(at least Z once) = 1 - P(no Z) = 1 - (2/3)^2 = 5/9
On June 10 2011 16:24 naggerNZ wrote: Just a note, threads like these are just purposefully divisive and humiliating. They're using logical ambiguity to attempt to embarrass those who don't already know the exact answer. Posts like this:
On June 10 2011 16:21 Eknoid4 wrote: This does not deserve a thread.
this is middle school math.
are just assholes trying to make others look stupid.
No, I'm serious. This doesn't deserve a thread. Ask a friend who is decent at math. Don't start a 10+ page argument. I don't have to try to make anybody i nthis thread look stupid. People arguing with basic math concepts when the OP has more to do with reading comprehension (which, in the context of math, you also should have learned in middle school) are making themselves look stupid enough.
On June 10 2011 12:29 Count9 wrote: I don't understand how people can know the paradox, see that it's worded in the exact same way (i.e. purposely ambiguous) and still argue that it's for sure one or the other.
It's because It's worded fine, he just misunderstands what he wrote. If I ask you what are the chances I flip a coin and it'll be heads, you'd say 1/2. if I told you I flipped it twice and at least 1 was heads, the chance that the other was heads is still 1 in 2.
by the OPs logic, it would be
HH TH HT
therefore 1 in 3 to get heads from 1 flip of a coin. That's wrong.
I don't know if the OP has changed the wording since I read it hours ago, but if he has, I maintain this about the original wording.
Limiting your options to ZP, ZZ, ZT, PZ, TZ (order preserved).
That means there is only a 1/5 chance you got the ZZ, so that the other match was zerg.
When you said, ZP, ZZ, ZT, PZ, and TZ, you stated the number of possible combinations incorrectly. The order of the games doesn't actually matter, the only thing that does matter is how many times you played as a certain race. So now, ZP and PZ should be one outcome, and ZT and TZ should also be one outcome.
With those outcomes combined, there should only be three outcomes: ZZ, ZP, and ZT, so the chance of playing as zerg twice given that you played as zerg at least once, is 1/3.
that make me think that it's a bad idea to teach probability as P(A) = (# ways A can happen)/(# ways anything can happen), simply because this assumes that every ``way'' is equally likely, which would lead to the conclusion that if we say the ``way'' A can happen is just that A happens and the ``ways'' anything can happen is that either A happens or A doesn't happen, then we get that there is 1 ``way'' for A to happen and 2 ``ways'' for anything to happen, so the probability of A happening is 1/2.
I'm sure everyone here agrees that is absurd (except for that poster many pages back who made a joke about exactly that).
The problem is that it can be very unclear when two different ``ways'' of something happening are actually different or the same, and I haven't come up with a good way to resolve it. Something about the ``size'' of a ``way'' is sort of necessary, but this seems to bring in probability circularly.
Here's another example of how that way of thinking can screw you up. Let's say I play a game with you. I have three cards, one with two red sides, one with two blue sides, and one with one red side and one blue side. I have a machine that shuffles them and randomly flips them over, then deals one out such that neither of us can see what is on the other side. If it comes up red, I'll bet you (at 1:1 odds) that the other side is red, and if it comes up blue, then I'll bet you that the other side is blue.
Now from your perspective, you see a red card and someone saying ``I bet you that the other side of the card is red''. You might think, well the other side is either red (which can happen in one way, with the red/red card), or the other side is blue (which can happen in one way also, with the red/blue card). So therefore I don't lose anything by taking this bet.
After a while, you lose all your money and wonder why. It's because the ``way'' of a red side coming up with a red/red card is twice as likely as the ``way'' of a red side coming up with a red/blue card.
Does anyone know of a good way to explain probability without using this idea of counting ``ways''?
On June 10 2011 17:23 Hamster1800 wrote: Does anyone know of a good way to explain probability without using this idea of counting ``ways''?
You mean permutations versus combinations?
No, I mean more basic than that. Like someone wants to understand what I mean when I say ``the probability of a die landing on 3 is 1/6''. Then the standard answer is to say ``Well, the die has 6 ways it can come up (1, 2, 3, 4, 5, or 6), and only one of those ways is a 3, so the probability is 1/6.''
But this leads to the misconceptions that I pointed out earlier, since this doesn't give any notion of relative size, which may well be important. So I want a better method of defining probability, which I haven't really seen.
On June 10 2011 12:29 Count9 wrote: I don't understand how people can know the paradox, see that it's worded in the exact same way (i.e. purposely ambiguous) and still argue that it's for sure one or the other.
It's because It's worded fine, he just misunderstands what he wrote. If I ask you what are the chances I flip a coin and it'll be heads, you'd say 1/2. if I told you I flipped it twice and at least 1 was heads, the chance that the other was heads is still 1 in 2.
by the OPs logic, it would be
HH TH HT
therefore 1 in 3 to get heads from 1 flip of a coin. That's wrong.
I don't know if the OP has changed the wording since I read it hours ago, but if he has, I maintain this about the original wording.
No. By the OPs logic, the question would be worded (essentially) "In 2 coin flips what is the probability that one flip yields two heads?" In which case the answer is 1/3
The more popular form of this question is about a family that has two children, and it asks the probability that both are boys if you know either A) at least one of them is a boy or B) there are not two girls. It's meant to show how people make assumptions about a starting condition, and don't consider order.
You will see answers for 1/2, 1/3, and 1/4. Yes, all three. Their reasoning:
1/2 - You know one child is a boy so there is a 1/2 chance the other child is a boy too 1/3 - You know both children can't be girls, so that narrows the possible scenarios to 3, only one of which is two boys. 1/4 - Knowing information about a particular family's children after both have been born does not actually change the probability that both children would be boys.
On June 10 2011 12:29 Count9 wrote: I don't understand how people can know the paradox, see that it's worded in the exact same way (i.e. purposely ambiguous) and still argue that it's for sure one or the other.
It's because It's worded fine, he just misunderstands what he wrote. If I ask you what are the chances I flip a coin and it'll be heads, you'd say 1/2. if I told you I flipped it twice and at least 1 was heads, the chance that the other was heads is still 1 in 2.
by the OPs logic, it would be
HH TH HT
therefore 1 in 3 to get heads from 1 flip of a coin. That's wrong.
I don't know if the OP has changed the wording since I read it hours ago, but if he has, I maintain this about the original wording.
No. By the OPs logic, the question would be worded (essentially) "In 2 coin flips what is the probability that one flip yields two heads?" In which case the answer is 1/3
Isn't it 1/4 ? p(event)=p(flip1)*p(flip2)=1/4
afaik the easiest way to handle chain events and probabilities is, if you know the probability of each smaller event and the relationship is "and" you can just multiply them. As long as order doesn't matter.
Elaborating: It makes sense to be 1/4 because you have 4 posible versions of the large event: HT TH HH TT
On June 10 2011 12:29 Count9 wrote: I don't understand how people can know the paradox, see that it's worded in the exact same way (i.e. purposely ambiguous) and still argue that it's for sure one or the other.
It's because It's worded fine, he just misunderstands what he wrote. If I ask you what are the chances I flip a coin and it'll be heads, you'd say 1/2. if I told you I flipped it twice and at least 1 was heads, the chance that the other was heads is still 1 in 2.
by the OPs logic, it would be
HH TH HT
therefore 1 in 3 to get heads from 1 flip of a coin. That's wrong.
I don't know if the OP has changed the wording since I read it hours ago, but if he has, I maintain this about the original wording.
No. By the OPs logic, the question would be worded (essentially) "In 2 coin flips what is the probability that one flip yields two heads?" In which case the answer is 1/3
Isn't it 1/4 ? p(event)=p(flip1)*p(flip2)=1/4
afaik the easiest way to handle chain events and probabilities is, if you know the probability of each smaller event and the relationship is "and" you can just multiply them. As long as order doesn't matter.
Elaborating: It makes sense to be 1/4 because you have 4 posible versions of the large event: HT TH HH TT
Yes but then he gives you the detail you are looking for at least 1 head, so TT is irrelevent since any time 2 tails pops up you trash that result entirely as for the purpose of the problem, it doesnt exist. In the example you are talking about he left out that detail but you have it right, he was just thinking in terms of the 1 of them is a heads situation which the problem this thread is based on.
On June 10 2011 17:23 Hamster1800 wrote:Does anyone know of a good way to explain probability without using this idea of counting ``ways''?
Coin flips is kind of classic but it's still about assigning correct probability to possible outcomes, or "counting ways" as you put it. Example:
Flipping two coins there are 3 possible outcomes: Head and head. Tail and tail. Head and tail.
Then we can illustrate the probabilities:
Have 2 people flip one coin each at the same time. Possible outcomes are: Person 1 Head, person 2 Tail Person 1 Tail, person 2 Head Person 1 Head, person 2 Head Person 1 Tail, person 2 Tail
Therefore if a person flips 2 coins the probabilities are: Head and head 25%. Tail and tail 25%. Head and tail 50%. This is because coin 1 head/coin 2 tail and coin 1 tail/coin 2 head both produce this result.
Another way to illustrate could be: If you bet on a normal double coin flip there are 3 possible outcomes, but if you flip one coin then another, what is the probability that the second coin will land on the same side as the first? Answers obviously depend on the questions asked and a lot of funny "problems" are based on trying to make our brains answer the wrong question. So if you for example use the answer to my last question in other 2 coin flip scenarios you need to understand that the question doesn't separate head/head and tail/tail just like a normal double coin flip does not separate head/tail and tail/head.
I'm probably a total noob now, but I still don't understand. Read the boy/girl paradox on wikipedia, still don't get it.
In the boy/girl paradox there's a difference between the two children outside of their gender, namely their age. One is older than the other, which gives us indeed three options. boy girl, where boy is older boy girl, where girl is older boy boy. then yes, it's 1/3.
but with your race, you're explicitly stating that *you* are the one playing zerg. Which makes the only possible matchups zvz, zvp and zvt. If you cound tvz and pvz as well, that would mean that you're not the one playing zerg. That is only when I assume you're naming yourself first in the xvx format.
On June 10 2011 12:29 Count9 wrote: I don't understand how people can know the paradox, see that it's worded in the exact same way (i.e. purposely ambiguous) and still argue that it's for sure one or the other.
It's because It's worded fine, he just misunderstands what he wrote. If I ask you what are the chances I flip a coin and it'll be heads, you'd say 1/2. if I told you I flipped it twice and at least 1 was heads, the chance that the other was heads is still 1 in 2.
by the OPs logic, it would be
HH TH HT
therefore 1 in 3 to get heads from 1 flip of a coin. That's wrong.
I don't know if the OP has changed the wording since I read it hours ago, but if he has, I maintain this about the original wording.
No. By the OPs logic, the question would be worded (essentially) "In 2 coin flips what is the probability that one flip yields two heads?" In which case the answer is 1/3
Did you read before he edited it (before Nestea, TLO and watching games got involved)? If so and that's what you're talking about, you're wrong. If you're not talking about that, I said I was talking about it's original iteration, so we're not really discussing the same thing.
I'm arguing against the people who say they agree with 1/5 with how it was originally phrased.
On June 10 2011 17:53 Dimagus wrote: The more popular form of this question is about a family that has two children, and it asks the probability that both are boys if you know either A) at least one of them is a boy or B) there are not two girls. It's meant to show how people make assumptions about a starting condition, and don't consider order.
You will see answers for 1/2, 1/3, and 1/4. Yes, all three. Their reasoning:
1/2 - You know one child is a boy so there is a 1/2 chance the other child is a boy too 1/3 - You know both children can't be girls, so that narrows the possible scenarios to 3, only one of which is two boys. 1/4 - Knowing information about a particular family's children after both have been born does not actually change the probability that both children would be boys.
An interesting variation on this question is: What is the probability of a family having two girls if one of the children is a girl named Kerrigan?
For any two games that a random player plays, given that in one of the games he was Zerg, what is the probability that he was also Zerg in the other? The answer here is clearly not 1/3.
On June 10 2011 16:11 Abenson wrote: erp. 1/5 - i don't see how hard this can be lol All you have to do, if you're confused, is to list out all the possibilities.
As I'm posting right now, 61% have answered 1/3 lol
That's because the OP has been edited. His wording was bad initially and it looked like he was asking the probability of getting zerg in a single game.
On June 10 2011 20:14 Dlok wrote: Answer is 1/3 that the other is zerg if you know witch one is being revealed and 2/6 if you do not know, ZT, ZP, (Z revealed)Z, PZ,TZ, Z(Z revealed)
This is a common error when dealing with set operations and the "at least" function. For instance, if you wanted all the two digit numbers that have at least one 5 in it, you would search for:
It's 1/3. The poll question is: "Probability that my other game was Zerg?" There are 3 options. T Z and P. You didn't ask about the specific order of games. Reword the poll question and it'd be fixed.
This shows that it at least works on the question in the OP. Let's try with 4 races, such as Human, Orc, Undead and Night Elf. You play random for 4 matches and at least one of them you play Undead. What are the odds you played Undead in all 4 games?
Chance = 1 / (4^3 + 3*(4^3 - 3^3)) = 1 / (64 + 3*(64 - 27) = 1 / (64 + 3*37) = 1 / 175 Had we not known you played Undead in at least one game, there would have been 4^4 = 256 possibilities, however, there are only 175 possibilities where Undead has been played (or in other words 81 where Undead has not been played), so the chances of playing Undead 4 times in a row KNOWING it has been played at least once is 1/175.
Independent Events The events A and B are independent if any one of the following three equivalent conditions hold.
P(AB) = P(A)P(B) P(A|B) = P(A) - - - - - B has no effect on A P(B|A) = P(B) - - - - - A has no effect on B Intuitively, two events are independent if the occurrence of one has no effect on the probability of the other.
If two events E and F are not independent, then they are dependent.
Independent Events The events A and B are independent if any one of the following three equivalent conditions hold.
P(AB) = P(A)P(B) P(A|B) = P(A) - - - - - B has no effect on A P(B|A) = P(B) - - - - - A has no effect on B Intuitively, two events are independent if the occurrence of one has no effect on the probability of the other.
If two events E and F are not independent, then they are dependent.
Everybody saying 1/5 is assuming this is a dependent event problem when it isn't.
On June 10 2011 21:13 Tektos wrote: Everybody saying 1/5 is assuming this is a dependent event problem when it isn't.
Can you explain how the diagram I drew is wrong?
We're answering different questions. The poll says probability that the other game is zerg (being 1/3), the diagram shows the answer to ZZ given at least 1 Z (1/5).
I jumped back into the thread after a few hours break and didn't see OP had edited everything to ask a completely different question to what he was asking earlier in the day.
We're both right just answering different questions. The OP sucks.
Independent Events The events A and B are independent if any one of the following three equivalent conditions hold.
P(AB) = P(A)P(B) P(A|B) = P(A) - - - - - B has no effect on A P(B|A) = P(B) - - - - - A has no effect on B Intuitively, two events are independent if the occurrence of one has no effect on the probability of the other.
If two events E and F are not independent, then they are dependent.
They are dependent given the original statement of the problem. He says that one of the two games he played zerg. That mean that if game 1 was protoss, game 2 HAD to be zerg. That is dependence.
I don't see how OTHER can mean SECOND unless it was specifically said that the other (see what I did there?) game was the FIRST game and not just ONE of the games.
"Probability of the other game being zerg" and "Probability of the other game being zerg given at least one game had zerg" are different questions.
I was answering the first, you're answering the second.
Or, there is the other question: "I played as Zerg at least once. What is the probability that my other game was as Zerg as well?"
Played zerg at least once, so the possible outcomes are: ZZ ZT ZP TZ PZ
Then "other game was zerg" if we take it as game 1 is the "other" game then the probability is 3/5. If you take it as game 2 is the "other" game then it is also 3/5.
All different questions, similar wording = confusion between everyone in the thread.
On June 10 2011 21:33 Tektos wrote: "Probability of the other game being zerg" and "Probability of the other game being zerg given at least one game had zerg" are different questions.
I was answering the first, you're answering the second.
The first question was never actually asked though. Both wordings imply the same thing.
On June 10 2011 21:33 Tektos wrote: "Probability of the other game being zerg" and "Probability of the other game being zerg given at least one game had zerg" are different questions.
I was answering the first, you're answering the second.
The first question was never actually asked though. Both wordings imply the same thing.
If you look at ONLY the poll without reading the rest of OP then yes that is the question. If you read the thread it should be 1/5, if you read the old question it should be 3/5. See my previous post's edit.
The answer isnt that obvious. Depending on the exact meaning it can have 2 answers. If were specificly looking at a game where always one of the games was zvz the answer would be 1/5. If we happen to stumble upon a game of zvz and then look at the other game the answer would be 1/3. Either way, not really that usefull to discuss over and over again.
@Tektos: No, you're simply failing to understand the wording of the original question.
On June 10 2011 21:33 Tektos wrote: Or, there is the other question: "I played as Zerg at least once. What is the probability that my other game was as Zerg as well?"
Played zerg at least once, so the possible outcomes are: ZZ ZT ZP TZ PZ
Then "other game was zerg" if we take it as game 1 is the "other" game then the probability is 3/5. If you take it as game 2 is the "other" game then it is also 3/5.
There is no 3/5 in here. The "other" refers to the game that is not already confirmed as zerg, not a random pick out of the two games. This is a simple failing of English comprehension on your part.
On June 10 2011 21:33 Tektos wrote: Or, there is the other question: "I played as Zerg at least once. What is the probability that my other game was as Zerg as well?"
Played zerg at least once, so the possible outcomes are: ZZ ZT ZP TZ PZ
Then "other game was zerg" if we take it as game 1 is the "other" game then the probability is 3/5. If you take it as game 2 is the "other" game then it is also 3/5.
There is no 3/5 in here. The "other" refers to the game that is not already confirmed as zerg, not a random pick out of the two games. This is a simple failing of English comprehension on your part.
"The "other" refers to the game that is not already confirmed as zerg" So you are not CONFIRMING that one game is already zerg? Then it is 1/3
It is not a failing of English comprehension on my part, it is ambiguity and interpretation of the meaning of "OTHER".
Hence why you wont ever see a real math problem worded using this use of "OTHER" due to the ambiguity.
On June 10 2011 21:41 Llama wrote: No, you're simply failing to understand the wording of the original question.
On June 10 2011 21:33 Tektos wrote: Or, there is the other question: "I played as Zerg at least once. What is the probability that my other game was as Zerg as well?"
Played zerg at least once, so the possible outcomes are: ZZ ZT ZP TZ PZ
Then "other game was zerg" if we take it as game 1 is the "other" game then the probability is 3/5. If you take it as game 2 is the "other" game then it is also 3/5.
There is no 3/5 in here. The "other" refers to the game that is not already confirmed as zerg, not a random pick out of the two games. This is a simple failing of English comprehension on your part.
"The "other" refers to the game that is not already confirmed as zerg" So you are not CONFIRMING that one game is already zerg? Then it is 1/3
It is not a failing of English comprehension on my part, it is ambiguity and interpretation of the meaning of "OTHER".
One game has already been confirmed as zerg. "I played as Zerg at least once". This confirms one game as zerg. There is no real ambiguity here.
it's 1/5! explanation: there are 5 scenarios in witch at least one game was as Zerg: (first T, second Z) , (first P, second Z) , (first Z, second Z) , (first Z, second T) , (first Z, second P) in only one of those the other game is as Zerg as well, so the odds are 1 in 5 Q.E.D
On June 10 2011 21:41 Llama wrote: No, you're simply failing to understand the wording of the original question.
On June 10 2011 21:33 Tektos wrote: Or, there is the other question: "I played as Zerg at least once. What is the probability that my other game was as Zerg as well?"
Played zerg at least once, so the possible outcomes are: ZZ ZT ZP TZ PZ
Then "other game was zerg" if we take it as game 1 is the "other" game then the probability is 3/5. If you take it as game 2 is the "other" game then it is also 3/5.
There is no 3/5 in here. The "other" refers to the game that is not already confirmed as zerg, not a random pick out of the two games. This is a simple failing of English comprehension on your part.
"The "other" refers to the game that is not already confirmed as zerg" So you are not CONFIRMING that one game is already zerg? Then it is 1/3
It is not a failing of English comprehension on my part, it is ambiguity and interpretation of the meaning of "OTHER".
One game has already been confirmed as zerg. "I played as Zerg at least once". This confirms one game as zerg. There is no real ambiguity here.
"OTHER" Other than what? Other than the game already confirmed as zerg? IT IS INCREDIBLY AMBIGUOUS. Tell me please what the "OTHER" refers to and I will give you the probability.
OP has given no indication of the starting point as to where the "OTHER" is based.
On June 10 2011 21:47 tomnov wrote: it's 1/5! explanation: there are 5 scenarios in witch at least one game was as Zerg: (first T, second Z) , (first P, second Z) , (first Z, second Z) , (first Z, second T) , (first Z, second P) in only one of those the other game is as Zerg as well, so the odds are 1 in 5 Q.E.D
So you're implying that the original game to what the "OTHER" is based off is indeed confirmed to be a zerg sample. That is your interpretation of the word other. Given that interpretation, yes 1/5 is correct.
On June 10 2011 21:41 Llama wrote: No, you're simply failing to understand the wording of the original question.
On June 10 2011 21:33 Tektos wrote: Or, there is the other question: "I played as Zerg at least once. What is the probability that my other game was as Zerg as well?"
Played zerg at least once, so the possible outcomes are: ZZ ZT ZP TZ PZ
Then "other game was zerg" if we take it as game 1 is the "other" game then the probability is 3/5. If you take it as game 2 is the "other" game then it is also 3/5.
There is no 3/5 in here. The "other" refers to the game that is not already confirmed as zerg, not a random pick out of the two games. This is a simple failing of English comprehension on your part.
"The "other" refers to the game that is not already confirmed as zerg" So you are not CONFIRMING that one game is already zerg? Then it is 1/3
It is not a failing of English comprehension on my part, it is ambiguity and interpretation of the meaning of "OTHER".
One game has already been confirmed as zerg. "I played as Zerg at least once". This confirms one game as zerg. There is no real ambiguity here.
"OTHER" Other than what? Other than the game already confirmed as zerg? IT IS INCREDIBLY AMBIGUOUS. Tell me please what the "OTHER" refers to and I will give you the probability.
OP has given no indication of the starting point as to where the "OTHER" is based.
How can it refer to anything else? "At least one game is zerg, what is the other?" How can you interpret other as anything else? I don't understand. There are only two elements in this question. One has been identified and the asker is now asking for the other. Where is the ambiguity?
On June 10 2011 21:41 Llama wrote: No, you're simply failing to understand the wording of the original question.
On June 10 2011 21:33 Tektos wrote: Or, there is the other question: "I played as Zerg at least once. What is the probability that my other game was as Zerg as well?"
Played zerg at least once, so the possible outcomes are: ZZ ZT ZP TZ PZ
Then "other game was zerg" if we take it as game 1 is the "other" game then the probability is 3/5. If you take it as game 2 is the "other" game then it is also 3/5.
There is no 3/5 in here. The "other" refers to the game that is not already confirmed as zerg, not a random pick out of the two games. This is a simple failing of English comprehension on your part.
"The "other" refers to the game that is not already confirmed as zerg" So you are not CONFIRMING that one game is already zerg? Then it is 1/3
It is not a failing of English comprehension on my part, it is ambiguity and interpretation of the meaning of "OTHER".
One game has already been confirmed as zerg. "I played as Zerg at least once". This confirms one game as zerg. There is no real ambiguity here.
"OTHER" Other than what? Other than the game already confirmed as zerg? IT IS INCREDIBLY AMBIGUOUS. Tell me please what the "OTHER" refers to and I will give you the probability.
OP has given no indication of the starting point as to where the "OTHER" is based.
How can it refer to anything else? "At least one game is zerg, what is the other?" How can you interpret other as anything else? I don't understand. There are only two elements in this question. One has been identified and the asker is now asking for the other. Where is the ambiguity?
"at least one game" is not a starting position for a reference of the word OTHER. If you're implying that at least one game is zerg, given that you choose that game which is zerg what is the probability of the other game being zerg. THEN IT WOULD BE 1/3. It is incredibly ambiguous.
At least one game is zerg, you're choose that game and then looking at the other game to that one zerg game? The "at least one zerg" condition is already fulfilled so there is no dependence on that constraint for the other game that you choose hence it being 1/3.
Answer this: Both games are zerg. What is the "OTHER" game.
It's the only starting position available and "other" can only be defined relatively. It's not strictly correct grammatically but the author's intent and meaning is obvious since other in this context can be defined no other way.
vv Damn I forgot about that haha, pretty unambiguous
On June 10 2011 21:59 aRRoSC2 wrote: I went to OP, opened the spoiler with original problem, and found this:
I played as Zerg at least once. What is the probability that my other game was as Zerg as well?
You see that "as well" at the end? Or do you just randomly pick what words of the OP you want to read and completely disregard those you don't like?
By that logic it is 1/3 because you've chosen the zerg game, from that starting point you're asking what the OTHER game will be. You've already got at least 1 zerg game so that condition is fulfilled so your possibilities for the OTHER game are Z, T, P =
On June 10 2011 21:59 aRRoSC2 wrote: I went to OP, opened the spoiler with original problem, and found this:
I played as Zerg at least once. What is the probability that my other game was as Zerg as well?
You see that "as well" at the end? Or do you just randomly pick what words of the OP you want to read and completely disregard those you don't like?
By that logic it is 1/3 because you've chosen the zerg game, from that starting point you're asking what the OTHER game will be. You've already got at least 1 zerg game so that condition is fulfilled so your possibilities for the OTHER game are Z, T, P =
0.33333333333333333333333333333333333
This is "setting" the first game as the zerg and only looking at Z_
I'm pretty sure that most of the people who picked 1/3 now realize their mistake (if they read through the thread) and wish they could change their poll response to 1/5.
On June 10 2011 21:59 aRRoSC2 wrote: I went to OP, opened the spoiler with original problem, and found this:
I played as Zerg at least once. What is the probability that my other game was as Zerg as well?
You see that "as well" at the end? Or do you just randomly pick what words of the OP you want to read and completely disregard those you don't like?
By that logic it is 1/3 because you've chosen the zerg game, from that starting point you're asking what the OTHER game will be. You've already got at least 1 zerg game so that condition is fulfilled so your possibilities for the OTHER game are Z, T, P =
0.33333333333333333333333333333333333
This is "setting" the first game as the zerg and only looking at Z_
You completely ignore the possibility of _Z
No I'm not setting the FIRST game as zerg, I'm setting the game chosen at random as to which the "OTHER" positioning statement is based off.
We're arguing semantics on an ambiguous statement.
On June 10 2011 21:59 aRRoSC2 wrote: I went to OP, opened the spoiler with original problem, and found this:
I played as Zerg at least once. What is the probability that my other game was as Zerg as well?
You see that "as well" at the end? Or do you just randomly pick what words of the OP you want to read and completely disregard those you don't like?
By that logic it is 1/3 because you've chosen the zerg game, from that starting point you're asking what the OTHER game will be. You've already got at least 1 zerg game so that condition is fulfilled so your possibilities for the OTHER game are Z, T, P =
0.33333333333333333333333333333333333
This is "setting" the first game as the zerg and only looking at Z_
You completely ignore the possibility of _Z
This.
In other words, "What's the probability of me ending up getting Zerg twice in two games if I'm a Random player, given the knowledge that I'll get Zerg at least once (doesn't have to be the first game)."
ZT ZP ZZ TZ PZ
1 out of 5 Zerg choices give me double Zerg. Therefore, 1/5. QED ^^
On June 10 2011 21:59 aRRoSC2 wrote: I went to OP, opened the spoiler with original problem, and found this:
I played as Zerg at least once. What is the probability that my other game was as Zerg as well?
You see that "as well" at the end? Or do you just randomly pick what words of the OP you want to read and completely disregard those you don't like?
By that logic it is 1/3 because you've chosen the zerg game, from that starting point you're asking what the OTHER game will be. You've already got at least 1 zerg game so that condition is fulfilled so your possibilities for the OTHER game are Z, T, P =
0.33333333333333333333333333333333333
This is "setting" the first game as the zerg and only looking at Z_
You completely ignore the possibility of _Z
No I'm not setting the FIRST game as zerg, I'm setting the game chosen at random as to which the "OTHER" positioning statement is based off.
We're arguing semantics on an ambiguous statement.
Both games have already occurred. You are arguing the position that only the first game has been determined and only look at the probability for the 2nd game.
He could even show you the replay of a zerg game, but you don't know whether it was the first or the second that he played.
On June 10 2011 22:07 DarkPlasmaBall wrote: In other words, "What's the probability of me ending up getting Zerg twice in two games if I'm a Random player, given the knowledge that I'll get Zerg at least once (doesn't have to be the first game)."
That is a different question to "What is the probability of the other game being zerg" though.
Answering the question you stated, then yes you are correct with your math.
On June 10 2011 21:59 aRRoSC2 wrote: I went to OP, opened the spoiler with original problem, and found this:
I played as Zerg at least once. What is the probability that my other game was as Zerg as well?
You see that "as well" at the end? Or do you just randomly pick what words of the OP you want to read and completely disregard those you don't like?
By that logic it is 1/3 because you've chosen the zerg game, from that starting point you're asking what the OTHER game will be. You've already got at least 1 zerg game so that condition is fulfilled so your possibilities for the OTHER game are Z, T, P =
0.33333333333333333333333333333333333
This is "setting" the first game as the zerg and only looking at Z_
You completely ignore the possibility of _Z
No I'm not setting the FIRST game as zerg, I'm setting the game chosen at random as to which the "OTHER" positioning statement is based off.
We're arguing semantics on an ambiguous statement.
Both games have already occurred. You are arguing the position that only the first game has been determined and only look at the probability for the 2nd game.
He could even show you the replay of a zerg game, but you don't know whether it was the first or the second that he played.
Z Z Z T Z P T Z P Z
Look at each of those zerg games that is 6 games, then look at the "OTHER" games to those zerg games you selected. The possible outcomes to that are, if you choose each zerg from top left to bottom right sequentially: Z, Z, T, P, T, P 2 of those games are Z, 2 successes out of 6 trials = 2/6 = 1/3.
On June 10 2011 22:07 DarkPlasmaBall wrote: In other words, "What's the probability of me ending up getting Zerg twice in two games if I'm a Random player, given the knowledge that I'll get Zerg at least once (doesn't have to be the first game)."
That is a different question to "What is the probability of the other game being zerg" though.
Answering the question you stated, then yes you are correct with your math.
It's conditional probability though, not independent trials, because there is the explicit statement made that we know that at least one game you must play as Zerg.
Therefore, if during the first game you're Protoss, then the probability of you being Zerg in game two is 100%. It's not 1/3, because of the initial condition. It's as if you have omniscience, despite you being a Random player.
On June 10 2011 22:13 Tektos wrote: Z Z Z T Z P T Z P Z
Look at each of those zerg games that is 6 games, then look at the "OTHER" games to those zerg games you selected. The possible outcomes to that are, if you choose each zerg from top left to bottom right sequentially: Z, Z, T, P, T, P 2 of those games are Z, 2 successes out of 6 trials = 2/6 = 1/3.
NOT 1/5
There are five trials there, not six. The "other" in the first instance is arbitrary but it's irrelevant since both games are zerg.
On June 10 2011 21:59 aRRoSC2 wrote: I went to OP, opened the spoiler with original problem, and found this:
I played as Zerg at least once. What is the probability that my other game was as Zerg as well?
You see that "as well" at the end? Or do you just randomly pick what words of the OP you want to read and completely disregard those you don't like?
By that logic it is 1/3 because you've chosen the zerg game, from that starting point you're asking what the OTHER game will be. You've already got at least 1 zerg game so that condition is fulfilled so your possibilities for the OTHER game are Z, T, P =
0.33333333333333333333333333333333333
This is "setting" the first game as the zerg and only looking at Z_
You completely ignore the possibility of _Z
No I'm not setting the FIRST game as zerg, I'm setting the game chosen at random as to which the "OTHER" positioning statement is based off.
We're arguing semantics on an ambiguous statement.
Both games have already occurred. You are arguing the position that only the first game has been determined and only look at the probability for the 2nd game.
He could even show you the replay of a zerg game, but you don't know whether it was the first or the second that he played.
1) Z Z 2) Z T 3) Z P 4) T Z 5) P Z
Look at each of those zerg games that is 6 games ...
On June 10 2011 12:51 Beez wrote: the answer doesnt rely on knowing the race of your opponent so you can be T, P, or Z. theres no reason to even think about matchups.
indeed you are correct, but everybody knows this and nobody is arguing about matchups it is simply "i play two games as random, i spawn as zerg atleast once, what's the probability of me spawning as zerg in both games"
my point is that for the games since you cannot tell which game is played as zerg you think of it as you can be 2 of 6 races Z,Z,P,P,T,T and since one game is garrenteed to be zerg you can get rid of one of the Zs so there is a 1/5 chance that both will be Z
On June 10 2011 21:59 aRRoSC2 wrote: I went to OP, opened the spoiler with original problem, and found this:
I played as Zerg at least once. What is the probability that my other game was as Zerg as well?
You see that "as well" at the end? Or do you just randomly pick what words of the OP you want to read and completely disregard those you don't like?
By that logic it is 1/3 because you've chosen the zerg game, from that starting point you're asking what the OTHER game will be. You've already got at least 1 zerg game so that condition is fulfilled so your possibilities for the OTHER game are Z, T, P =
0.33333333333333333333333333333333333
This is "setting" the first game as the zerg and only looking at Z_
You completely ignore the possibility of _Z
No I'm not setting the FIRST game as zerg, I'm setting the game chosen at random as to which the "OTHER" positioning statement is based off.
We're arguing semantics on an ambiguous statement.
Both games have already occurred. You are arguing the position that only the first game has been determined and only look at the probability for the 2nd game.
He could even show you the replay of a zerg game, but you don't know whether it was the first or the second that he played.
1) Z Z 2) Z T 3) Z P 4) T Z 5) P Z
Look at each of those zerg games that is 6 games ...
Ummmmm....
There are two zerg games in row one = 2 one zerg game in each row after that = 4 2 + 4 = 6
If you can't even get basic arithmetic correct why are you arguing probability?
On June 10 2011 21:59 aRRoSC2 wrote: I went to OP, opened the spoiler with original problem, and found this:
I played as Zerg at least once. What is the probability that my other game was as Zerg as well?
You see that "as well" at the end? Or do you just randomly pick what words of the OP you want to read and completely disregard those you don't like?
By that logic it is 1/3 because you've chosen the zerg game, from that starting point you're asking what the OTHER game will be. You've already got at least 1 zerg game so that condition is fulfilled so your possibilities for the OTHER game are Z, T, P =
0.33333333333333333333333333333333333
This is "setting" the first game as the zerg and only looking at Z_
You completely ignore the possibility of _Z
No I'm not setting the FIRST game as zerg, I'm setting the game chosen at random as to which the "OTHER" positioning statement is based off.
We're arguing semantics on an ambiguous statement.
Both games have already occurred. You are arguing the position that only the first game has been determined and only look at the probability for the 2nd game.
He could even show you the replay of a zerg game, but you don't know whether it was the first or the second that he played.
1) Z Z 2) Z T 3) Z P 4) T Z 5) P Z
Look at each of those zerg games that is 6 games ...
Ummmmm....
There are two zerg games in row one = 2 one zerg game in each row after that = 4 2 + 4 = 6
I love these kinds of brainteasers, especially when you test TL. Usually the majority is correct but apparently not this time. It will soon have the majority correct though because people will cheat now that the correct answer is out. :/
On June 10 2011 21:59 aRRoSC2 wrote: I went to OP, opened the spoiler with original problem, and found this:
I played as Zerg at least once. What is the probability that my other game was as Zerg as well?
You see that "as well" at the end? Or do you just randomly pick what words of the OP you want to read and completely disregard those you don't like?
By that logic it is 1/3 because you've chosen the zerg game, from that starting point you're asking what the OTHER game will be. You've already got at least 1 zerg game so that condition is fulfilled so your possibilities for the OTHER game are Z, T, P =
0.33333333333333333333333333333333333
This is "setting" the first game as the zerg and only looking at Z_
You completely ignore the possibility of _Z
No I'm not setting the FIRST game as zerg, I'm setting the game chosen at random as to which the "OTHER" positioning statement is based off.
We're arguing semantics on an ambiguous statement.
Both games have already occurred. You are arguing the position that only the first game has been determined and only look at the probability for the 2nd game.
He could even show you the replay of a zerg game, but you don't know whether it was the first or the second that he played.
1) Z Z 2) Z T 3) Z P 4) T Z 5) P Z
Look at each of those zerg games that is 6 games ...
Ummmmm....
There are two zerg games in row one = 2 one zerg game in each row after that = 4 2 + 4 = 6
Why and how are you trying to dupe ZZ somehow...?
He played game 1 as zerg, he also played game 2 as zerg, that is two games played as zerg.
On June 10 2011 21:59 aRRoSC2 wrote: I went to OP, opened the spoiler with original problem, and found this:
I played as Zerg at least once. What is the probability that my other game was as Zerg as well?
You see that "as well" at the end? Or do you just randomly pick what words of the OP you want to read and completely disregard those you don't like?
By that logic it is 1/3 because you've chosen the zerg game, from that starting point you're asking what the OTHER game will be. You've already got at least 1 zerg game so that condition is fulfilled so your possibilities for the OTHER game are Z, T, P =
0.33333333333333333333333333333333333
This is "setting" the first game as the zerg and only looking at Z_
You completely ignore the possibility of _Z
No I'm not setting the FIRST game as zerg, I'm setting the game chosen at random as to which the "OTHER" positioning statement is based off.
We're arguing semantics on an ambiguous statement.
Both games have already occurred. You are arguing the position that only the first game has been determined and only look at the probability for the 2nd game.
He could even show you the replay of a zerg game, but you don't know whether it was the first or the second that he played.
1) Z Z 2) Z T 3) Z P 4) T Z 5) P Z
Look at each of those zerg games that is 6 games ...
Ummmmm....
There are two zerg games in row one = 2 one zerg game in each row after that = 4 2 + 4 = 6
the important part is that there are 5 such possible matches, and each has _equal_ probability (the ZvZ is not weighted more, lol.)
Or think about it this way: you have a 5/9 chance of having at least one opponent be Zerg (the condition), and a 1/9 chance of both being Zergs. Hence the probability of having both opponents being Zerg given at least one of them is Zerg is (1/9) / (5/9) = 1/5.
On June 10 2011 21:59 aRRoSC2 wrote: I went to OP, opened the spoiler with original problem, and found this:
I played as Zerg at least once. What is the probability that my other game was as Zerg as well?
You see that "as well" at the end? Or do you just randomly pick what words of the OP you want to read and completely disregard those you don't like?
By that logic it is 1/3 because you've chosen the zerg game, from that starting point you're asking what the OTHER game will be. You've already got at least 1 zerg game so that condition is fulfilled so your possibilities for the OTHER game are Z, T, P =
0.33333333333333333333333333333333333
This is "setting" the first game as the zerg and only looking at Z_
You completely ignore the possibility of _Z
No I'm not setting the FIRST game as zerg, I'm setting the game chosen at random as to which the "OTHER" positioning statement is based off.
We're arguing semantics on an ambiguous statement.
Both games have already occurred. You are arguing the position that only the first game has been determined and only look at the probability for the 2nd game.
He could even show you the replay of a zerg game, but you don't know whether it was the first or the second that he played.
1) Z Z 2) Z T 3) Z P 4) T Z 5) P Z
Look at each of those zerg games that is 6 games ...
Ummmmm....
There are two zerg games in row one = 2 one zerg game in each row after that = 4 2 + 4 = 6
Why and how are you trying to dupe ZZ somehow...?
He played game 1 as zerg, he also played game 2 as zerg, that is two games played as zerg.
So the *1* scenario where he plays zerg twice, you are therefore duping as 2 scenarios ZZ and ZZ?
On June 10 2011 21:59 aRRoSC2 wrote: I went to OP, opened the spoiler with original problem, and found this:
I played as Zerg at least once. What is the probability that my other game was as Zerg as well?
You see that "as well" at the end? Or do you just randomly pick what words of the OP you want to read and completely disregard those you don't like?
By that logic it is 1/3 because you've chosen the zerg game, from that starting point you're asking what the OTHER game will be. You've already got at least 1 zerg game so that condition is fulfilled so your possibilities for the OTHER game are Z, T, P =
0.33333333333333333333333333333333333
This is "setting" the first game as the zerg and only looking at Z_
You completely ignore the possibility of _Z
No I'm not setting the FIRST game as zerg, I'm setting the game chosen at random as to which the "OTHER" positioning statement is based off.
We're arguing semantics on an ambiguous statement.
Both games have already occurred. You are arguing the position that only the first game has been determined and only look at the probability for the 2nd game.
He could even show you the replay of a zerg game, but you don't know whether it was the first or the second that he played.
1) Z Z 2) Z T 3) Z P 4) T Z 5) P Z
Look at each of those zerg games that is 6 games ...
Ummmmm....
There are two zerg games in row one = 2 one zerg game in each row after that = 4 2 + 4 = 6
Why and how are you trying to dupe ZZ somehow...?
He played game 1 as zerg, he also played game 2 as zerg, that is two games played as zerg.
But it's only one case.
We're counting cases, not games won as Zerg.
For instance, a case that's irrelevant is PT because neither is Zerg. but that's not two examples.
On June 10 2011 21:59 aRRoSC2 wrote: I went to OP, opened the spoiler with original problem, and found this:
I played as Zerg at least once. What is the probability that my other game was as Zerg as well?
You see that "as well" at the end? Or do you just randomly pick what words of the OP you want to read and completely disregard those you don't like?
By that logic it is 1/3 because you've chosen the zerg game, from that starting point you're asking what the OTHER game will be. You've already got at least 1 zerg game so that condition is fulfilled so your possibilities for the OTHER game are Z, T, P =
0.33333333333333333333333333333333333
This is "setting" the first game as the zerg and only looking at Z_
You completely ignore the possibility of _Z
No I'm not setting the FIRST game as zerg, I'm setting the game chosen at random as to which the "OTHER" positioning statement is based off.
We're arguing semantics on an ambiguous statement.
Both games have already occurred. You are arguing the position that only the first game has been determined and only look at the probability for the 2nd game.
He could even show you the replay of a zerg game, but you don't know whether it was the first or the second that he played.
1) Z Z 2) Z T 3) Z P 4) T Z 5) P Z
Look at each of those zerg games that is 6 games ...
Ummmmm....
There are two zerg games in row one = 2 one zerg game in each row after that = 4 2 + 4 = 6
the important part is that there are 5 such possible matches, and each has _equal_ probability (the ZvZ is not weighted more, lol.)
Or think about it this way: you have a 5/9 chance of having at least one opponent be Zerg (the condition), and a 1/9 chance of both being Zergs. Hence the probability of having both opponents being Zerg given at least one of them is Zerg is (1/9) / (5/9) = 1/5.
Each match has equal probability, but he is TWICE as likely to play zerg in a one game selection of a sample size of 2 in Z Z as he is in T Z. We were talking about:
"Here is a replay where I played zerg, it is part of a 2 game series" hence if you pick 1 game of a 2 game series:
100% chance vs. 50% chance thus that game gets double weighting when arguing the "OTHER" game question.
On June 10 2011 21:59 aRRoSC2 wrote: I went to OP, opened the spoiler with original problem, and found this:
I played as Zerg at least once. What is the probability that my other game was as Zerg as well?
You see that "as well" at the end? Or do you just randomly pick what words of the OP you want to read and completely disregard those you don't like?
By that logic it is 1/3 because you've chosen the zerg game, from that starting point you're asking what the OTHER game will be. You've already got at least 1 zerg game so that condition is fulfilled so your possibilities for the OTHER game are Z, T, P =
0.33333333333333333333333333333333333
This is "setting" the first game as the zerg and only looking at Z_
You completely ignore the possibility of _Z
No I'm not setting the FIRST game as zerg, I'm setting the game chosen at random as to which the "OTHER" positioning statement is based off.
We're arguing semantics on an ambiguous statement.
Both games have already occurred. You are arguing the position that only the first game has been determined and only look at the probability for the 2nd game.
He could even show you the replay of a zerg game, but you don't know whether it was the first or the second that he played.
1) Z Z 2) Z T 3) Z P 4) T Z 5) P Z
Look at each of those zerg games that is 6 games ...
Ummmmm....
There are two zerg games in row one = 2 one zerg game in each row after that = 4 2 + 4 = 6
the important part is that there are 5 such possible matches, and each has _equal_ probability (the ZvZ is not weighted more, lol.)
Or think about it this way: you have a 5/9 chance of having at least one opponent be Zerg (the condition), and a 1/9 chance of both being Zergs. Hence the probability of having both opponents being Zerg given at least one of them is Zerg is (1/9) / (5/9) = 1/5.
Each match has equal probability, but he is TWICE as likely to play zerg in Z Z as he is in T Z 100% chance vs. 50% chance thus that game gets double weighting when arguing the "OTHER" game question.
By that logic it is 1/3 because you've chosen the zerg game, from that starting point you're asking what the OTHER game will be. You've already got at least 1 zerg game so that condition is fulfilled so your possibilities for the OTHER game are Z, T, P =
0.33333333333333333333333333333333333
This is "setting" the first game as the zerg and only looking at Z_
You completely ignore the possibility of _Z
No I'm not setting the FIRST game as zerg, I'm setting the game chosen at random as to which the "OTHER" positioning statement is based off.
We're arguing semantics on an ambiguous statement.
Both games have already occurred. You are arguing the position that only the first game has been determined and only look at the probability for the 2nd game.
He could even show you the replay of a zerg game, but you don't know whether it was the first or the second that he played.
1) Z Z 2) Z T 3) Z P 4) T Z 5) P Z
Look at each of those zerg games that is 6 games ...
Ummmmm....
There are two zerg games in row one = 2 one zerg game in each row after that = 4 2 + 4 = 6
the important part is that there are 5 such possible matches, and each has _equal_ probability (the ZvZ is not weighted more, lol.)
Or think about it this way: you have a 5/9 chance of having at least one opponent be Zerg (the condition), and a 1/9 chance of both being Zergs. Hence the probability of having both opponents being Zerg given at least one of them is Zerg is (1/9) / (5/9) = 1/5.
Each match has equal probability, but he is TWICE as likely to play zerg in Z Z as he is in T Z 100% chance vs. 50% chance thus that game gets double weighting when arguing the "OTHER" game question.
And.... there's the flaw.
You were talking single game selection not paired game selection.
YES if you choose the games as a pair then it has equal probability, but if you hand me a replay of a zerg playing it is twice as likely to be from that series as it is from any of the others.
By that logic it is 1/3 because you've chosen the zerg game, from that starting point you're asking what the OTHER game will be. You've already got at least 1 zerg game so that condition is fulfilled so your possibilities for the OTHER game are Z, T, P =
0.33333333333333333333333333333333333
This is "setting" the first game as the zerg and only looking at Z_
You completely ignore the possibility of _Z
No I'm not setting the FIRST game as zerg, I'm setting the game chosen at random as to which the "OTHER" positioning statement is based off.
We're arguing semantics on an ambiguous statement.
Both games have already occurred. You are arguing the position that only the first game has been determined and only look at the probability for the 2nd game.
He could even show you the replay of a zerg game, but you don't know whether it was the first or the second that he played.
1) Z Z 2) Z T 3) Z P 4) T Z 5) P Z
Look at each of those zerg games that is 6 games ...
Ummmmm....
There are two zerg games in row one = 2 one zerg game in each row after that = 4 2 + 4 = 6
the important part is that there are 5 such possible matches, and each has _equal_ probability (the ZvZ is not weighted more, lol.)
Or think about it this way: you have a 5/9 chance of having at least one opponent be Zerg (the condition), and a 1/9 chance of both being Zergs. Hence the probability of having both opponents being Zerg given at least one of them is Zerg is (1/9) / (5/9) = 1/5.
Each match has equal probability, but he is TWICE as likely to play zerg in Z Z as he is in T Z 100% chance vs. 50% chance thus that game gets double weighting when arguing the "OTHER" game question.
And.... there's the flaw.
Agreed. You're not twice as likely to play Zerg in ZZ as you are in TZ. You just happen to have twice as many Zs. Big difference.
If you get T first game, then you absolutely must be Z next game, given the condition that at least one of your games is Zerg. 100% chance of getting Zerg.
If you're Zerg the first game... then you can be any of the three races next game. 1/3 chance you'll get Zerg again.
It's a sort of omniscience you have of both games you're going to play. You being Random just means you have the option to get any of the three races, which opens up all scenarios for you. It's conditional probability.
Twice as likely to play Zerg if my opponent is Zerg. It's kinda like saying you're twice as likely to be in a traffic accident if the idiot who hit you also was in a traffic accident. I think. Oh fuck it. I'm done here, some people just really cannot be convinced they are wrong even though it's as obvious as the day. It's kinda like Flat Earth Society.
I will repost my formula because it's kind of a shame if I spent that time on this thread and no one sees it :/
Here are 10 games played. Answer the following question:
1) I hand you a replay of 1 game: What is the probability that I am playing zerg in that replay?
2) I hand you the second game in that series, what is the probability of the second replay I hand you that I am playing a zerg?
On June 10 2011 22:30 DarkPlasmaBall wrote: Agreed. You're not twice as likely to play Zerg in ZZ as you are in TZ. You just happen to have twice as many Zs. Big difference.
I flip a standard coin, what is the probability of heads?
I flip a coin that has heads on both sides, what is the probability of heads?
If you have twice as many of something you're twice as likely to pick said thing.
The part most people don't understand is that the condition doesn't actually affect the independent probability of anything happening, but it does remove the possibility of some SETS of outcomes. - normally, the chance of random rolling zerg in both games would be 1/9. However, we know that 4 out of those 9 "possibilities" are impossible, leaving 5 scenarios left.
Normally, these scenarios are all equally likely: 1 2 Z Z Z T Z P T Z T T T P P Z P T P P
However, we know that 4 of these 9 equally likely scenarios couldn't have been the final outcome. We don't know whether it was game 1 or game 2 (or both) were zerg, but it wasn't neither. The remaining 5 are all still equally likely. Leaving us with 1/5 chance.
This is "setting" the first game as the zerg and only looking at Z_
You completely ignore the possibility of _Z
No I'm not setting the FIRST game as zerg, I'm setting the game chosen at random as to which the "OTHER" positioning statement is based off.
We're arguing semantics on an ambiguous statement.
Both games have already occurred. You are arguing the position that only the first game has been determined and only look at the probability for the 2nd game.
He could even show you the replay of a zerg game, but you don't know whether it was the first or the second that he played.
1) Z Z 2) Z T 3) Z P 4) T Z 5) P Z
Look at each of those zerg games that is 6 games ...
Ummmmm....
There are two zerg games in row one = 2 one zerg game in each row after that = 4 2 + 4 = 6
the important part is that there are 5 such possible matches, and each has _equal_ probability (the ZvZ is not weighted more, lol.)
Or think about it this way: you have a 5/9 chance of having at least one opponent be Zerg (the condition), and a 1/9 chance of both being Zergs. Hence the probability of having both opponents being Zerg given at least one of them is Zerg is (1/9) / (5/9) = 1/5.
Each match has equal probability, but he is TWICE as likely to play zerg in Z Z as he is in T Z 100% chance vs. 50% chance thus that game gets double weighting when arguing the "OTHER" game question.
And.... there's the flaw.
You were talking single game selection not paired game selection.
YES if you choose the games as a pair then it has equal probability, but if you hand me a replay of a zerg playing it is twice as likely to be from that series as it is from any of the others.
Wait, so you're not answering any of the questions actually being asked, you zigzagged off on a tangent and are doing the math for something completely unrelated?
Like if he played all 5 match ups ZZ PZ TZ ZP ZT, you're arguing that there's double the chance the replay came from the ZZ? Okay...
I am always amazed how even in the face of reason, people remain convinced that the wrong answer is correct. I really don't understand how people can't get that MORE information changes the probability. Maybe they'd rather just believe that they were "tricked" than to be shown to have a poor grasp of probability.
On June 10 2011 22:33 Tektos wrote: Z Z Z P Z T T Z P Z
Here are 10 games played. Answer the following question:
1) I hand you a replay of 1 game: What is the probability that I am playing zerg in that replay?
Um, 100% given the premise of the original question.
What?
I can be playing the following: Z, Z, Z, P, Z, T, T, Z, P, Z Do those all look like Zs to you? Wow omg.
Just saying that in the original question you are essentially being given the zerg replay 100% of the time. When you ask for the race of the second replay in each pair isn't it obvious that the chance it also being zerg is one in five?
Given that at least one of my games was Zerg, what is the probability that both of my games are zerg.
Is a different question to this:
I played as Zerg at least once. What is the probability that my other game was as Zerg as well?
But I did learn about conditional probability. So everyone's a winner! Learning is fun!
These two are the same question. If one game is zerg and the other game is also zerg then both games are zerg. This is what both means.
Not quite.
Given that at least one of my games was Zerg, what is the probability that both of my games are zerg
This asks specifically for the probability of both games.
I played as Zerg at least once. What is the probability that my other game was as Zerg as well?
This asks specifically for the probability of one game.The word "other" means the game that is not already specified as being zerg. The order of the games doesn't matter, when you say "What is the probability that my other game was as Zerg as well?" you are referring to the probability of one game.
You need to refer to the probability of both games for the answer to be 1/5.
Having said that, I answered 1/3 to the updated OP question right away without realising I was wrong.
On June 10 2011 22:33 Tektos wrote: Z Z Z P Z T T Z P Z
Here are 10 games played. Answer the following question:
1) I hand you a replay of 1 game: What is the probability that I am playing zerg in that replay?
Um, 100% given the premise of the original question.
What?
I can be playing the following: Z, Z, Z, P, Z, T, T, Z, P, Z Do those all look like Zs to you? Wow omg.
Just saying that in the original question you are essentially being given the zerg replay 100% of the time. When you ask for the race of the second replay in each pair isn't it obvious that the chance it also being zerg is one in five?
So look at all the zerg replays, there are 6 of them. Now look at the other replay in each of those pairs.
Protip: If I played zerg twice in the first combination, I'll be handing you two replays from that series.
On June 10 2011 22:33 Tektos wrote: Z Z Z P Z T T Z P Z
Here are 10 games played. Answer the following question:
1) I hand you a replay of 1 game: What is the probability that I am playing zerg in that replay?
Um, 100% given the premise of the original question.
What?
I can be playing the following: Z, Z, Z, P, Z, T, T, Z, P, Z Do those all look like Zs to you? Wow omg.
Just saying that in the original question you are essentially being given the zerg replay 100% of the time. When you ask for the race of the second replay in each pair isn't it obvious that the chance it also being zerg is one in five?
So look at all the zerg replays, there are 6 of them. Now look at the other replay in each of those pairs.
Protip: If I played zerg twice in the first combination, I'll be handing you two replays from that series.
Each "pair" here has an equal likelyhood. So I send you one zerg replay from the pairs [ZZ] [ZP] [ZT] [TZ] [PZ]. There isn't a double chance that I send you a Z from pair one, but I do have to pick one of the two at random. Then I send you the second half of the pair which is non-Z in four cases, Z in the remaining one.
On June 10 2011 22:30 DarkPlasmaBall wrote: Agreed. You're not twice as likely to play Zerg in ZZ as you are in TZ. You just happen to have twice as many Zs. Big difference.
I flip a standard coin, what is the probability of heads?
I flip a coin that has heads on both sides, what is the probability of heads?
If you have twice as many of something you're twice as likely to pick said thing.
Two different games are two different coin flips. Not the same coinflip. It's not even close to the same thing. And you're still forgetting about conditional probability.
Tektos suddenly stopped trying to do the probability for match-ups, but instead starting calculating # of replays. So that resulted in the last couple of pages...
On June 10 2011 22:33 Tektos wrote: Z Z Z P Z T T Z P Z
Here are 10 games played. Answer the following question:
1) I hand you a replay of 1 game: What is the probability that I am playing zerg in that replay?
Um, 100% given the premise of the original question.
What?
I can be playing the following: Z, Z, Z, P, Z, T, T, Z, P, Z Do those all look like Zs to you? Wow omg.
Just saying that in the original question you are essentially being given the zerg replay 100% of the time. When you ask for the race of the second replay in each pair isn't it obvious that the chance it also being zerg is one in five?
So look at all the zerg replays, there are 6 of them. Now look at the other replay in each of those pairs.
Protip: If I played zerg twice in the first combination, I'll be handing you two replays from that series.
Each "pair" here has an equal likelyhood. So I send you one zerg replay from the pairs [ZZ] [ZP] [ZT] [TZ] [PZ]. There isn't a double chance that I send you a Z from pair one, but I do have to pick one of the two at random. Then I send you the second half of the pair which is non-Z in four cases, Z in the remaining one.
If I play two games as zerg, and I send you the replay of one of those games, what is the probability that the replay has be playing zerg in it? 100% If I play one game as zerg and one game as terran, what is the probability that if I send you one of those replays I'll be playing zerg? 50%
We're talking about sending one replay, not choosing a series then sending you one of those two replays. Your ignorance is astounding.
On June 10 2011 22:30 DarkPlasmaBall wrote: Agreed. You're not twice as likely to play Zerg in ZZ as you are in TZ. You just happen to have twice as many Zs. Big difference.
I flip a standard coin, what is the probability of heads?
I flip a coin that has heads on both sides, what is the probability of heads?
If you have twice as many of something you're twice as likely to pick said thing.
Two different games are two different coin flips. Not the same coinflip. It's not even close to the same thing. And you're still forgetting about conditional probability.
Anyways, I gotta go. Enjoy your day
You're utterly confused.
If I have two replays and I play the same race in each replay, and you select one of those replays at random, it is twice as likely to be that race than if you have two replays where you play a different race in each replay.
The coin flip was to represent which of the two replays in the series gets sent to you.
I'm upset that the education system hasn't worked for so many people.
On June 10 2011 22:33 Tektos wrote: Z Z Z P Z T T Z P Z
Here are 10 games played. Answer the following question:
1) I hand you a replay of 1 game: What is the probability that I am playing zerg in that replay?
Um, 100% given the premise of the original question.
What?
I can be playing the following: Z, Z, Z, P, Z, T, T, Z, P, Z Do those all look like Zs to you? Wow omg.
Just saying that in the original question you are essentially being given the zerg replay 100% of the time. When you ask for the race of the second replay in each pair isn't it obvious that the chance it also being zerg is one in five?
So look at all the zerg replays, there are 6 of them. Now look at the other replay in each of those pairs.
Protip: If I played zerg twice in the first combination, I'll be handing you two replays from that series.
Each "pair" here has an equal likelyhood. So I send you one zerg replay from the pairs [ZZ] [ZP] [ZT] [TZ] [PZ]. There isn't a double chance that I send you a Z from pair one, but I do have to pick one of the two at random. Then I send you the second half of the pair which is non-Z in four cases, Z in the remaining one.
If I play two games as zerg, and I send you the replay of one of those games, what is the probability that the replay has be playing zerg in it? 100% If I play one game as zerg and one game as terran, what is the probability that if I send you one of those replays I'll be playing zerg? 50%
We're talking about sending one replay, not choosing a series then sending you one of those two replays. Your ignorance is astounding.
Sorry, I was just trying to help you understand this in the context of the original question in which one game is guaranteed to be zerg.
You seem to have lost track of any semblance of a point.
On June 10 2011 22:33 Tektos wrote: Z Z Z P Z T T Z P Z
Here are 10 games played. Answer the following question:
1) I hand you a replay of 1 game: What is the probability that I am playing zerg in that replay?
Um, 100% given the premise of the original question.
What?
I can be playing the following: Z, Z, Z, P, Z, T, T, Z, P, Z Do those all look like Zs to you? Wow omg.
Just saying that in the original question you are essentially being given the zerg replay 100% of the time. When you ask for the race of the second replay in each pair isn't it obvious that the chance it also being zerg is one in five?
So look at all the zerg replays, there are 6 of them. Now look at the other replay in each of those pairs.
Protip: If I played zerg twice in the first combination, I'll be handing you two replays from that series.
Each "pair" here has an equal likelyhood. So I send you one zerg replay from the pairs [ZZ] [ZP] [ZT] [TZ] [PZ]. There isn't a double chance that I send you a Z from pair one, but I do have to pick one of the two at random. Then I send you the second half of the pair which is non-Z in four cases, Z in the remaining one.
If I play two games as zerg, and I send you the replay of one of those games, what is the probability that the replay has be playing zerg in it? 100% If I play one game as zerg and one game as terran, what is the probability that if I send you one of those replays I'll be playing zerg? 50%
We're talking about sending one replay, not choosing a series then sending you one of those two replays. Your ignorance is astounding.
Sorry, I was just trying to help you understand this in the context of the original question in which one game is guaranteed to be zerg.
You seem to have lost track of any semblance of a point.
So lets take it back to the original point. Forget everything in the OP and read this as a standalone:
If I play two games, I hand you a replay where I played zerg, what is the probability of the other replay I have also being a zerg game?
(If you answer anything other than 1/3 I don't think I can help you to understand even the most basic of probability)
I've played two games. Then the possible combinations are: ZZ, ZP, ZT, PZ, PP, PT, TZ, TP, TT.
However, I've said I played Zerg. Then that eliminates PP, PT, TP, TT.
Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices, and that is the only which corresponds to "The other game is Zerg."
Then the correct answer is 1/5.
The elimination only occurs on the first game.
You said you're talking about your race in the second game, so I don't get why you exclude PP, PT, TP, TT. You made no assumptions about the second game, therefore your chance of getting Zerg is 3/9, or 1/3
On June 10 2011 22:57 Tektos wrote: So lets take it back to the original point. Forget everything in the OP and read this as a standalone:
If I play two games, I hand you a replay where I played zerg, what is the probability of the other replay I have also being a zerg game?
(If you answer anything other than 1/3 I don't think I can help you to understand even the most basic of probability)
If you are asking "what are the chances that I get zerg if I play random" then the answer is 1/3. I don't think that was ever in dispute.
But the original question is: "You play two games, I ask you to give me a zerg replay if you have one. What race is the other replay?"
Yes, the probability of getting zerg if you play random is 1/3, and because there are no constraints and the second replay is independent of the first replay it is still 1/3.
Now add this to the equation: Between that first zerg replay I've handed you, and the replay I still have, at least one of them is of zerg. I've already handed you one zerg replay, so the condition of "At least one replay being zerg" is fulfilled. Therefore the probability of the other replay that I still hold being zerg is still 1/3.
But the original question is: "You play two games, I ask you to give me a zerg replay if you have one. What race is the other replay?"
That is not the original question, as saying "IF YOU HAVE ONE" implies that TT, TP, PT, PP are all still options.
But the original question is: "You play two games, I ask you to give me a zerg replay if you have one. What race is the other replay?"
That is not the original question, as saying "IF YOU HAVE ONE" implies that TT, TP, PT, PP are all still options.
They are options if you random twice--that's the nature of random. Obviously they get eliminated if we get handed a zerg replay, but not until then.
The point is that the asking for one zerg replay out of two (if there is at least one) is different from simply receiving a zerg replay. The first applies the condition of at least one being zerg before the question is asked, the second is a statement of fact after gaining the relevant information.
Please understand that as a question of probability the two are distinct.
But the original question is: "You play two games, I ask you to give me a zerg replay if you have one. What race is the other replay?"
That is not the original question, as saying "IF YOU HAVE ONE" implies that TT, TP, PT, PP are all still options.
They are options if you random twice--that's the nature of random. Obviously they get eliminated if we get handed a zerg replay, but not until then.
The point is that the asking for one zerg replay out of two (if there is at least one) is different from simply receiving a zerg replay. The first applies the condition of at least one being zerg before the question is asked, the second is a statement of fact after gaining the relevant information.
Please understand that as a question of probability the two are distinct.
Okay so everything is an option. Now before handing you a replay I tell you that at least one of them is of a zerg, that eliminates the possibilities of the games down to just being: Z Z Z T Z P T Z P Z
Now I hand you one zerg replay. There are 6 zerg replays in those 5 possible outcomes (this is where your error comes in).
I hand you one of those 6 zerg replays. Of those 6 possible replays that I could have handed you, can you tell me what the 6 possible other race outcomes that I played in that pair are?
But the original question is: "You play two games, I ask you to give me a zerg replay if you have one. What race is the other replay?"
That is not the original question, as saying "IF YOU HAVE ONE" implies that TT, TP, PT, PP are all still options.
They are options if you random twice--that's the nature of random. Obviously they get eliminated if we get handed a zerg replay, but not until then.
The point is that the asking for one zerg replay out of two (if there is at least one) is different from simply receiving a zerg replay. The first applies the condition of at least one being zerg before the question is asked, the second is a statement of fact after gaining the relevant information.
Please understand that as a question of probability the two are distinct.
Okay so everything is an option. Now before handing you a replay I tell you that at least one of them is of a zerg, that eliminates the possibilities of the games down to just being: Z Z Z T Z P T Z P Z
Now I hand you one zerg replay. There are 6 zerg replays in those 5 possible outcomes (this is where your error comes in).
I hand you one of those 6 zerg replays. Of those 6 possible replays that I could have handed you, can you tell me what the other race I played in that pair is?
There are six replays but that doesn't mean that each one has the same probability. If each of the pairs has equal possibility then each "Z" in the first pair is halved since you can only pick one of them. Each "Z" does not represent a unit of probability. If you reframe the question in terms of, say, Z and NOTZ then you get:
Z Z Z NOTZ NOTZ Z
but that doesn't imply anything about the respective probabilities of each Z or even each pairing.
edit: to clarify further, there aren't six replays at all, there are one or two. Each letter does not represent an actual replay but the possibility of a replay.
lol if he plays random, there are three races that he could possibly be.. however he can only land as one of those three. if the question is simply asking what the probability that his next game (worded as if the first game has any relevance what so ever) then the answer is one of three or 1/3
edit : busted i didnt read it right.. what are the odds that he is zerg in BOTH games.. well he had a 1/3 chance to be zerg in the first game and a 1/3 chance in the second game.. 1/9?
to those who say 1/5: flip two coins, then tell me one of the results in any way that does not exclude the possibility of a pair. you would give me 1/3 odds on a pair, while the odds clearly havent changed and really are 1/2. now we do this 100 times and I make alot of money.
There's 9 possibilities total for TLO: TT, TZ, TP, ZT, ZZ, ZP, PT, PZ, PP Out out those, we already know that something with zerg has happened, which is TZ ZT ZZ ZP PZ. Out of those, only ZZ means two mirror matches, so it's 1/5 to get two ZvZ when we already know there's at least one ZvZ.
On June 10 2011 23:41 Dlok wrote: to those who say 1/5: flip two coins, then tell me one of the results in any way that does not exclude the possibility of a pair. you would give me 1/3 odds on a pair, while the odds clearly havent changed and really are 1/2. now we do this 100 times and I make alot of money.
If we do the test 100s of times the result will in fact be 1/3rd and I would gladly take you up on such odds at something less than 1/2 but greater than 1/3rd as it'd be free money. You could even write a very simple program to test it.
Not understanding probability (which is very counter-intuitive) doesn't change reality.
On June 10 2011 23:41 Dlok wrote: to those who say 1/5: flip two coins, then tell me one of the results in any way that does not exclude the possibility of a pair. you would give me 1/3 odds on a pair, while the odds clearly havent changed and really are 1/2. now we do this 100 times and I make alot of money.
Uhm, if you get told one outcome, it's 1/2 for a pair? Possible outcomes are HH, HT and TT. Being told that one coin came up heads would exclude TT and thus it's 1/2 for a pair.
On June 10 2011 23:41 Dlok wrote: to those who say 1/5: flip two coins, then tell me one of the results in any way that does not exclude the possibility of a pair. you would give me 1/3 odds on a pair, while the odds clearly havent changed and really are 1/2. now we do this 100 times and I make alot of money.
Uhm, if you get told one outcome, it's 1/2 for a pair? Possible outcomes are HH, HT and TT. Being told that one coin came up heads would exclude TT and thus it's 1/2 for a pair.
No, HT and TH aren't the same.
Or they are the same, but have 2x the chance of appearing.
Your wrong. The orders are considered different match ups because you always put the person your referring to race first. Jinro doesn't play ZvT, but he does play TvZ.
Since NesTea is always Z that means it is XvZ so you eliminate either ZvP and ZvT or PvZ and TvZ (either set not a mix and match)
What you are calling a paradox is actually your failure to understand starcraft terminology.
Just read the original problem and you still fail. You don't make any mention of a match up only the race you play, the games are indeed independent events of one another. If order mattered then the odds would be different, but you never specify that. You just suck at trying to sate a question correctly.
On June 11 2011 00:01 wswordsmen wrote: Your wrong. The orders are considered different match ups because you always put the person your referring to race first. Jinro doesn't play ZvT, but he does play TvZ.
Since NesTea is always Z that means it is XvZ so you eliminate either ZvP and ZvT or PvZ and TvZ (either set not a mix and match)
What you are calling a paradox is actually your failure to understand starcraft terminology.
They aren't talking about matchups. They're talking about the 2 games.
ZZ is TLO getting zerg twice (so 2 ZvZ's) ZP is TLO getting zerg then protoss (ZvZ ZvP)
We're given that he played two games and that he played Z at least once. The question is: what is the probability that in the other game he also played Z? If he played Z in the other game as well, then hes played Z twice overall. So the question is equivalent to: what is the probability he played Z twice.
Combining the question and our givens into the same sentence, we get: What is the probability he played Z twice, given that he played Z at least once out of two games.
First lets define some events: Event A = He played Z twice (out of two games) Event B = He played Z at least once (out of two games)
P(A|B) = Probability of event A occurring given event B has occurred. Translating this into a sentence with the events we've defined, we get: P(A|B) = Probability he played Z twice, given that he played Z at least once out of two games.
Notice the two bolded parts are the same. So now we just need to find P(A|B).
I'm not going to prove Bayes' Theorem, but it is taught in intro stats courses and the proof is pretty straightforward. It states:
P(B|A) = The probability he played Z at least once, given that he played twice..well if we know he played Z twice, hes obviously played it at least once, making this probability = 1
P(A) = The probability he played Z twice out of two games = (1/3)^2 = 1/9 P(B) = The probability he played Z at least once out of two games = (1/3)*1 + (2/3)*(1/3) = 5/9 Explanation: + Show Spoiler +
We can break this down into two independent events: Event C: He gets Z the first game (probability=1/3) and gets anything the second game (probability=1). So P(C) = 1/3*1 = 1/3 This covers ZZ, ZT, and ZP. Event D: He does not get Z in the first game(probability = 2/3) and gets Z the second game (probability=1/3). So P(D) = 2/3*1/3 = 2/9. This covers TZ and PZ.
Because events C and D encompass all the ways he can get Z at least once (B is the union of C and D) and have no overlaps (the intersection is zero), we can find P(B) by adding P(C) and P(D).
On June 11 2011 00:08 TastyMuffins wrote: We're given that he played two games and that he played Z at least once. The question is: what is the probability that in the other game he also played Z? If he played Z in the other game as well, then hes played Z twice overall. So the question is equivalent to: what is the probability he played Z twice.
Combining the question and our givens into the same sentence, we get: What is the probability he played Z twice, given that he played Z at least once out of two games.
First lets define some events: Event A = He played Z twice (out of two games) Event B = He played Z at least once (out of two games)
P(A|B) = Probability of event A occurring given event B has occurred. Translating this into a sentence with the events we've defined, we get: P(A|B) = Probability he played Z twice, given that he played Z at least once out of two games.
Notice the two bolded parts are the same. So now we just need to find P(A|B).
I'm not going to prove Bayes' Theorem, but it is taught in intro stats courses and the proof is pretty straightforward. It states:
P(B|A) = The probability he played Z at least once, given that he played twice..well if we know he played Z twice, hes obviously played it at least once, making this probability = 1
P(A) = The probability he played Z twice out of two games = (1/3)^2 = 1/9 P(B) = The probability he played Z at least once out of two games = (1/3)*1 + (2/3)*(1/3) = 5/9 Explanation: + Show Spoiler +
We can break this down into two independent events: Event C: He gets Z the first game (probability=1/3) and gets anything the second game (probability=1). So P(C) = 1/3*1 = 1/3 This covers ZZ, ZT, and ZP. Event D: He does not get Z in the first game(probability = 2/3) and gets Z the second game (probability=1/3). So P(D) = 2/3*1/3 = 2/9. This covers TZ and PZ.
Because events C and D encompass all the ways he can get Z at least once (B is the union of C and D) and have no overlaps (the intersection is zero), we can find P(B) by adding P(C) and P(D).
On June 11 2011 00:01 wswordsmen wrote: Your wrong. The orders are considered different match ups because you always put the person your referring to race first. Jinro doesn't play ZvT, but he does play TvZ.
Since NesTea is always Z that means it is XvZ so you eliminate either ZvP and ZvT or PvZ and TvZ (either set not a mix and match)
What you are calling a paradox is actually your failure to understand starcraft terminology.
They aren't talking about matchups. They're talking about the 2 games.
ZZ is TLO getting zerg twice (so 2 ZvZ's) ZP is TLO getting zerg then protoss (ZvZ ZvP)
I was wrong about that but it doesn't change the fact he never says order matters. TZ and ZT are still the same so he is double counting them.
They're not the same...
or more accurately if you ignore order you have the possibilities of:
TZ ZZ ZP
but they're not equal probability. If you random twice you are twice as likely to get TZ or ZP as you are ZZ.
The paradox has frequently stimulated a great deal of controversy. Many people argued strongly for both sides with a great deal of confidence, sometimes showing disdain for those who took the opposing view.
On June 10 2011 23:41 Dlok wrote: to those who say 1/5: flip two coins, then tell me one of the results in any way that does not exclude the possibility of a pair. you would give me 1/3 odds on a pair, while the odds clearly havent changed and really are 1/2. now we do this 100 times and I make alot of money.
If we do the test 100s of times the result will in fact be 1/3rd and I would gladly take you up on such odds at something less than 1/2 but greater than 1/3rd as it'd be free money. You could even write a very simple program to test it.
Not understanding probability (which is very counter-intuitive) doesn't change reality.
On June 10 2011 23:41 Dlok wrote: to those who say 1/5: flip two coins, then tell me one of the results in any way that does not exclude the possibility of a pair. you would give me 1/3 odds on a pair, while the odds clearly havent changed and really are 1/2. now we do this 100 times and I make alot of money.
If we do the test 100s of times the result will in fact be 1/3rd and I would gladly take you up on such odds at something less than 1/2 but greater than 1/3rd as it'd be free money. You could even write a very simple program to test it.
Not understanding probability (which is very counter-intuitive) doesn't change reality.
The paradox has frequently stimulated a great deal of controversy. Many people argued strongly for both sides with a great deal of confidence, sometimes showing disdain for those who took the opposing view.
So uh you linked us the answer your own question.
Haha these two sentences from wikipedia are also ambiguous. It's unclear that the many people who argued strongly for both sides knew whether it was a paradox or not. Since that link was already posted at the beginning of the thread, I assume people here knew it's was paradox when arguing, so why simply not admit that the wording has it's importance and that it's pointless to argue forever, since both sides are right depending on how you interpret the wording ?
On June 10 2011 23:41 Dlok wrote: to those who say 1/5: flip two coins, then tell me one of the results in any way that does not exclude the possibility of a pair. you would give me 1/3 odds on a pair, while the odds clearly havent changed and really are 1/2. now we do this 100 times and I make alot of money.
If we do the test 100s of times the result will in fact be 1/3rd and I would gladly take you up on such odds at something less than 1/2 but greater than 1/3rd as it'd be free money. You could even write a very simple program to test it.
Not understanding probability (which is very counter-intuitive) doesn't change reality.
I recommend you stay away from Vegas. Do you even realize what you'd be betting on?
If I win every time we get heads + heads then you'd win with heads + tails (or tails + heads) and every time we get tails + tails we'd call it a draw. You'd be a fool to give someone odds on heads + heads at > 1/3 when you're not winning on two of the 4 possible outcomes.
Haha these two sentences from wikipedia are also ambiguous. It's unclear that the many people who argued strongly for both sides knew whether it was a paradox or not. Since that link was already posted at the beginning of the thread, I assume people here knew it's was paradox when arguing, so why simply not admit that the wording has it's importance and that it's pointless to argue forever, since both sides are right depending on how you interpret the wording ?
The point is the paradox causes intense reactions by those who don't accept the answer as true while those who understand the right answer are... well right. I don't know why you'd think that people wouldn't argue over a paradox that's known to create controversy and argument.
On June 10 2011 23:41 Dlok wrote: to those who say 1/5: flip two coins, then tell me one of the results in any way that does not exclude the possibility of a pair. you would give me 1/3 odds on a pair, while the odds clearly havent changed and really are 1/2. now we do this 100 times and I make alot of money.
Wrong. It is the ambiguity of the paradox that causes us to say 1/5. I would not take your bet.
This is the bet I would make. Flip 2 coins. For each set, throw it out if there is no heads result. Only keep the results which have at least one heads. I give you that 1/3 the sets have a pair of heads, not 1/2.
...since we know that one of the games was zerg, all we need to do is find the probability that the other game was zerg. This gets affected in no way by the game we KNOW was zerg. Therefore, there are three possibilities for the 'other' game played:
He played as zerg. He played as terran. He played as protoss.
Zerg will occur 1/3 of the time. Therefore the answer is 1/3. There is absolutely no purpose to writing out the different combinations of games (ZZ PP TT PT TP... etc.), as we only need to find the probability he played one game as zerg, since we already know one of his games was zerg.
I will happily respond to any counter-arguments, but I would most likely be rewording / repeating myself. (^_^)
On June 10 2011 23:41 Dlok wrote: to those who say 1/5: flip two coins, then tell me one of the results in any way that does not exclude the possibility of a pair. you would give me 1/3 odds on a pair, while the odds clearly havent changed and really are 1/2. now we do this 100 times and I make alot of money.
If we do the test 100s of times the result will in fact be 1/3rd and I would gladly take you up on such odds at something less than 1/2 but greater than 1/3rd as it'd be free money. You could even write a very simple program to test it.
Not understanding probability (which is very counter-intuitive) doesn't change reality.
I recommend you stay away from Vegas. Do you even realize what you'd be betting on?
If I win every time we get heads + heads then you'd win with heads + tails (or tails + heads) and every time we get tails + tails we'd call it a draw. You'd be a fool to give someone odds on heads + heads at > 1/3 when you're not winning on two of the 4 possible outcomes.
Read his bet carefully. He does not throw out tails/tails, which is different from what we expect when we say 1/5. See my above post for details.
On June 10 2011 10:11 theDreamStick wrote: Solution: I've played two games. Then the possible combinations are: ZZ, ZP, ZT, PZ, PP, PT, TZ, TP, TT.
However, I've said I played Zerg. Then that eliminates PP, PT, TP, TT.
Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices, and that is the only which corresponds to "The other game is Zerg."
Then the correct answer is 1/5.
I don't get this: why do you say "i've played two games" and then enumerate the possible pairings for only one game?
And then, regaring only one game, i believe you are wrong. Why is the probability for ZZ 1/5?
In the solution spoiler, it states: "Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices"
But i think, this is wrong. Here's the possible stuff. Note that [Z] means "you" [Z]Z [Z]P [Z]T Z[Z] P[Z] T[Z]
if you don't mark yourself, you get ZZ and ZZ, which looks exactly the same and you would thus - when writing it into a set - resolve those to leave only one ZZ. But it's actually two separate events if you count the positions of the players.
So, either [Z]Z and Z[Z] are separate events, or ZP and PZ and ZT and TZ are equal as well.
Adding this up: there are 2/6 ZZ (namely [Z]Z and Z[Z] out of the six mentioned above), or there is 1/3 ZZ (namely ZZ out of ZZ, ZT, ZP). So the Probability for ZZ is actually 1/3, not 1/5?
This in mind - i think the whole discussion about conditional probability is completely pointless. Why on earth would you argue "this is conditional probability, so it's counter intuitive" and bla bla, but then post a solution where the word "conditional" is not mentioned once? Your answer is not based on conditional probability, and in addition i believe it's wrong because you miscounted the possible events.
On June 11 2011 00:29 Ivs wrote: People are still arguing because OP wanted to present the Boy/Girl paradox, but messed up the wording.
Now there are 3 camps of people
1. People interpreted the OP as the boy/girl paradox, even though OP failed. They say 1/5
2. People who don't really understand whats going on and go with the simplest reasoning. They say 1/3
3. People who are calling out OP's original wording and poor usage of "other", and also get the answer of 1/3.
Chill out guys, no need for name/credential calling. There is no argument here.
There is no error in the original wording. The statement, "Given that I played zerg at least once, I played zerg both times," and "I played zerg once and I played the other game zerg as well" are equivalent. People are just misconceiving how many choices they actually have, thinking that "the other game" is in reference to a specific game, when it can not be.
On June 10 2011 10:11 theDreamStick wrote: Solution: I've played two games. Then the possible combinations are: ZZ, ZP, ZT, PZ, PP, PT, TZ, TP, TT.
However, I've said I played Zerg. Then that eliminates PP, PT, TP, TT.
Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices, and that is the only which corresponds to "The other game is Zerg."
Then the correct answer is 1/5.
I don't get this: why do you say "i've played two games" and then enumerate the possible pairings for only one game?
And then, regaring only one game, i believe you are wrong. Why is the probability for ZZ 1/5?
In the solution spoiler, it states: "Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices"
But i think, this is wrong. Here's the possible stuff. Note that [Z] means "you" [Z]Z [Z]P [Z]T Z[Z] P[Z] T[Z]
if you don't mark yourself, you get ZZ and ZZ, which looks exactly the same and you would thus - when writing it into a set - resolve those to leave only one ZZ. But it's actually two separate events if you count the positions of the players.
So, either [Z]Z and Z[Z] are separate events, or ZP and PZ and ZT and TZ are equal as well.
Adding this up: there are 2/6 ZZ (namely [Z]Z and Z[Z] out of the six mentioned above), or there is 1/3 ZZ (namely ZZ out of ZZ, ZT, ZP). So the Probability for ZZ is actually 1/3, not 1/5?
This in mind - i think the whole discussion about conditional probability is completely pointless. Why on earth would you argue "this is conditional probability, so it's counter intuitive" and bla bla, but then post a solution where the word "conditional" is not mentioned once? Your answer is not based on conditional probability, and in addition i believe it's wrong because you miscounted the possible events.
/edit: clarified wording a little
you've misinterpreted.
ZZ means you were Zerg the first game, and Zerg the second game. it isn't a "matchup".
On June 10 2011 23:41 Dlok wrote: to those who say 1/5: flip two coins, then tell me one of the results in any way that does not exclude the possibility of a pair. you would give me 1/3 odds on a pair, while the odds clearly havent changed and really are 1/2. now we do this 100 times and I make alot of money.
If we do the test 100s of times the result will in fact be 1/3rd and I would gladly take you up on such odds at something less than 1/2 but greater than 1/3rd as it'd be free money. You could even write a very simple program to test it.
Not understanding probability (which is very counter-intuitive) doesn't change reality.
I recommend you stay away from Vegas. Do you even realize what you'd be betting on?
If I win every time we get heads + heads then you'd win with heads + tails (or tails + heads) and every time we get tails + tails we'd call it a draw. You'd be a fool to give someone odds on heads + heads at > 1/3 when you're not winning on two of the 4 possible outcomes.
Haha these two sentences from wikipedia are also ambiguous. It's unclear that the many people who argued strongly for both sides knew whether it was a paradox or not. Since that link was already posted at the beginning of the thread, I assume people here knew it's was paradox when arguing, so why simply not admit that the wording has it's importance and that it's pointless to argue forever, since both sides are right depending on how you interpret the wording ?
The point is the paradox causes intense reactions by those who don't accept the answer as true while those who understand the right answer are... well right. I don't know why you'd think that people wouldn't argue over a paradox that's known to create controversy and argument.
Every time you say Heads i say it will be pair of heads, if you say tailes i say pair of tails, now we know 50 out of a hundred are likely to be pairs so i will do fine.
If i say i take only pair of heads i will win 1/4 of said 100 but if im allowed to withdraw when atleast one is not heads i raise my odds to 1/3. This however was not how the problem was stated, and I cant se how it could be interpited that way.
On June 10 2011 10:11 theDreamStick wrote: Solution: I've played two games. Then the possible combinations are: ZZ, ZP, ZT, PZ, PP, PT, TZ, TP, TT.
However, I've said I played Zerg. Then that eliminates PP, PT, TP, TT.
Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices, and that is the only which corresponds to "The other game is Zerg."
Then the correct answer is 1/5.
I don't get this: why do you say "i've played two games" and then enumerate the possible pairings for only one game?
And then, regaring only one game, i believe you are wrong. Why is the probability for ZZ 1/5?
In the solution spoiler, it states: "Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices"
But i think, this is wrong. Here's the possible stuff. Note that [Z] means "you" [Z]Z [Z]P [Z]T Z[Z] P[Z] T[Z]
if you don't mark yourself, you get ZZ and ZZ, which looks exactly the same and you would thus - when writing it into a set - resolve those to leave only one ZZ. But it's actually two separate events if you count the positions of the players.
So, either [Z]Z and Z[Z] are separate events, or ZP and PZ and ZT and TZ are equal as well.
Adding this up: there are 2/6 ZZ (namely [Z]Z and Z[Z] out of the six mentioned above), or there is 1/3 ZZ (namely ZZ out of ZZ, ZT, ZP). So the Probability for ZZ is actually 1/3, not 1/5?
This in mind - i think the whole discussion about conditional probability is completely pointless. Why on earth would you argue "this is conditional probability, so it's counter intuitive" and bla bla, but then post a solution where the word "conditional" is not mentioned once? Your answer is not based on conditional probability, and in addition i believe it's wrong because you miscounted the possible events.
/edit: clarified wording a little
you've misinterpreted.
ZZ means you were Zerg the first game, and Zerg the second game. it isn't a "matchup".
ahhh okay, that's how it's meant! xD thanks. okay, then this sounds fine. Still, the point about discussing conditional probabilities stands - this answer does not use them, so don't argue they are needed ^^
On June 11 2011 00:29 Ivs wrote: People are still arguing because OP wanted to present the Boy/Girl paradox, but messed up the wording.
Now there are 3 camps of people
1. People interpreted the OP as the boy/girl paradox, even though OP failed. They say 1/5
2. People who don't really understand whats going on and go with the simplest reasoning. They say 1/3
3. People who are calling out OP's original wording and poor usage of "other", and also get the answer of 1/3.
Chill out guys, no need for name/credential calling. There is no argument here.
There is no error in the original wording. The statement, "Given that I played zerg at least once, I played zerg both times," and "I played zerg once and I played the other game zerg as well" are equivalent. People are just misconceiving how many choices they actually have, thinking that "the other game" is in reference to a specific game, when it can not be.
Exactly, the two statements are logically equivalent. The only difference is that the 2nd one is misleading, encouraging the reader to focus more on the "other game" and disregard the given information.
The 3 groups of people should be:
1. People who messed up by reading the question too quickly (even though they understand conditional probability). So, instead they to try to convince themselves and others that the OP was wrong, not them.
2. People who don't really understand whats going on and go with the simplest reasoning. They say 1/3
I'm going with the "I am no native speaker" excuse. I've read the sentence probably 10 times by now and the wording still makes my brain hurt. I would still choose 1/3, although it's obviously a trap.
Sometimes you have to embrace being stupid, that's what I'm doing here.
edit: yeah, as the poster below me said, the poll question still says "Probability that my other game was Zerg?". What kind of english is that?
On June 11 2011 00:29 Ivs wrote: People are still arguing because OP wanted to present the Boy/Girl paradox, but messed up the wording.
Now there are 3 camps of people
1. People interpreted the OP as the boy/girl paradox, even though OP failed. They say 1/5
2. People who don't really understand whats going on and go with the simplest reasoning. They say 1/3
3. People who are calling out OP's original wording and poor usage of "other", and also get the answer of 1/3.
Chill out guys, no need for name/credential calling. There is no argument here.
There is no error in the original wording. The statement, "Given that I played zerg at least once, I played zerg both times," and "I played zerg once and I played the other game zerg as well" are equivalent. People are just misconceiving how many choices they actually have, thinking that "the other game" is in reference to a specific game, when it can not be.
The unedited OP didn't say "other game as well". It just said "other game". The poll itself STILL says "other game". People are going to continue to say 1/3 because, regardless of the explanation at the beginning, the poll itself still asking the probability of drawing Zerg in a single game.
The unedited OP didn't say "other game as well". It just said "other game". The poll itself STILL says "other game". People are going to continue to say 1/3 because, regardless of the explanation at the beginning, the poll itself still asking the probability of drawing Zerg in a single game.
Adding "as well" doesn't change the fact that "other game" isn't in reference to a specific game. It can be either the first game, or the second game, and your probability should reflect this.
On June 11 2011 00:41 iStarKraft wrote: ...since we know that one of the games was zerg, all we need to do is find the probability that the other game was zerg. This gets affected in no way by the game we KNOW was zerg. Therefore, there are three possibilities for the 'other' game played:
He played as zerg. He played as terran. He played as protoss.
Zerg will occur 1/3 of the time. Therefore the answer is 1/3. There is absolutely no purpose to writing out the different combinations of games (ZZ PP TT PT TP... etc.), as we only need to find the probability he played one game as zerg, since we already know one of his games was zerg.
I will happily respond to any counter-arguments, but I would most likely be rewording / repeating myself. (^_^)
iSK
LOL this thread is hopeless, my friends. Even after countless explanations people are still wrong and not reading other posts to see why. I guess the three categorizations are accurate.
Please try to understand the difference between "the probability of having gotten Zerg in both games given Zerg in at least one" and "the probability of getting Zerg in a game." The OP's wordings are more than clear enough to make this distinction. The latter is 1/3 (no shit), the former 1/5. Good God
Here's an easier way to understand the problem. Say I flip two coins, we can all agree that the possible outcomes are:
H-T 1/4 H-H 1/4 T-T 1/4 T-H 1/4
Now I tell you that if, and only if, a heads comes up, if the other coin is tails, I win. If the other coin is heads, you win. Simply making this proposition does not change the probability of each result coming up.
H-T I win H-H I lose T-T Draw T-H I win
We already know each of the outcomes has an equal chance of occuring, so in 2/4 results, I win, 1/4 you win. Therefore the odds are 2:1, or 2/3 in my favour, and 1/3 in yours. NOT 1/2.
On June 11 2011 02:11 foxmeep wrote: Here's an easier way to understand the problem. Say I flip two coins, we can all agree that the possible outcomes are:
H-T 1/4 H-H 1/4 T-T 1/4 T-H 1/4
Now I tell you that if, and only if, a heads comes up, if the other coin is tails, I win. If the other coin is heads, you win. Simply making this proposition does not change the probability of each result coming up.
H-T I win H-H I lose T-T Draw T-H I win
We already know each of the outcomes has an equal chance of occuring, so in 2/4 results, I win, 1/4 you win. Therefore the odds are 2:1, or 2/3 in my favour, and 1/3 in yours. NOT 1/2.
That's because you represent the problem differently, the question is what are the odds for TT to drop if we flip it twice in a row. It's only understandable in the way you say it because the OP was a bit bad ;p
On June 11 2011 02:11 foxmeep wrote: Here's an easier way to understand the problem. Say I flip two coins, we can all agree that the possible outcomes are:
H-T 1/4 H-H 1/4 T-T 1/4 T-H 1/4
Now I tell you that if, and only if, a heads comes up, if the other coin is tails, I win. If the other coin is heads, you win. Simply making this proposition does not change the probability of each result coming up.
H-T I win H-H I lose T-T Draw T-H I win
We already know each of the outcomes has an equal chance of occuring, so in 2/4 results, I win, 1/4 you win. Therefore the odds are 2:1, or 2/3 in my favour, and 1/3 in yours. NOT 1/2.
That's because you represent the problem differently, the question is what are the odds for TT to drop if we flip it twice in a row. It's only understandable in the way you say it because the OP was a bit bad ;p
Incorrect. I represent the problem identical to the OP. If heads comes up, what are the chances that the other coin is heads. If I random Zerg in one game, what are the chances that I random Zerg the other game.
so your friend says, basically at least 1 of the games was ZvZ so the next game has a 1/3 chance of being zerg again, surely? so if definitely not 1/3?
aah i get it it might not be the first game was a ZVZ so then it will be 1/5. nice-kind-0f-a teaser
Z = Zerg Probability = 1/3 N = Not zerg Probability = 2/3
We know 1 match is zerg so the possibilities left are ZN, NZ, ZZ which happen with probability (2/3)*(1/3) = (2/9), again 2/9 and (1/3)*(1/3) = (1/9). Normalizing so that they sum up to 1 we need to multiply with a factor (1/(5/9)) = 9/5. Hence the probability is (1/9)*(9/5)=1/5.
On June 11 2011 02:49 JKira wrote: I learned conditional probability in grade 11, and I don't even think that's early or anything. I was 16 at the time.
Are the majority of people on TL really under 16 years old? I was under the impression that the majority is over 20 years old.
Condescending much? Even the great mathematician Erdos initially got the Monty Hall problem wrong, which is considered a basic conditional probability problem. And most of the fuss is over the wording ambiguity of the problem, not the problem solving itself.
i read the solution, then voted 1/3. youre making the mistake of counting ZT and TZ as two seperate cases, when they are actually exactly the same, given that ordering doesnt matter.
OP, In your solution, you claim that there are 5 possible games, ZZ, ZP, ZT, PZ, TZ. This is nonsense, because ZvT is the same as TvZ; meaning you count two of the matchups twice for no reason.
On June 11 2011 03:00 Millitron wrote: OP, In your solution, you claim that there are 5 possible games, ZZ, ZP, ZT, PZ, TZ. This is nonsense, because ZvT is the same as TvZ; meaning you count two of the matchups twice for no reason.
Its 1/3.
ZvT is same as TvZ, but playing Z and then playing T is not the same as playing T and then playing Z.
Though realistically
1) Why did you create this thread? Obvious flame... but yes, the answer is technically 1/5.
2) Realistically the answer is close to 1 since most people who play zerg just play zerg every game
On June 11 2011 02:11 foxmeep wrote: Here's an easier way to understand the problem. Say I flip two coins, we can all agree that the possible outcomes are:
H-T 1/4 H-H 1/4 T-T 1/4 T-H 1/4
Now I tell you that if, and only if, a heads comes up, if the other coin is tails, I win. If the other coin is heads, you win. Simply making this proposition does not change the probability of each result coming up.
H-T I win H-H I lose T-T Draw T-H I win
We already know each of the outcomes has an equal chance of occuring, so in 2/4 results, I win, 1/4 you win. Therefore the odds are 2:1, or 2/3 in my favour, and 1/3 in yours. NOT 1/2.
no, because you are making ordering a necessaity in this. the OP's problem does not take ordering into account.
On June 11 2011 02:59 L3g3nd_ wrote: i read the solution, then voted 1/3. youre making the mistake of counting ZT and TZ as two seperate cases, when they are actually exactly the same, given that ordering doesnt matter.
It technically doesn't matter but ZT (or ZT/TZ) is twice as likely as ZZ, so it does matter in a sense.
Because English isn't my first language; I had to reread the statement twice. That the chance 'me being zerg in a game' is a lot bigger than 'me being zerg in a game, one game prior or one game after being zerg in an other game' is obvious; but with the whole story with it you certainly "teased my brain" :D Very nice!
On the other hand; I haven't been drinking that much the first days after graduating because I wanted to remember all that stuff ^^
Edit: Now I'm ******** up the wording... Stupid English language
On June 11 2011 02:49 JKira wrote: I learned conditional probability in grade 11, and I don't even think that's early or anything. I was 16 at the time.
Are the majority of people on TL really under 16 years old? I was under the impression that the majority is over 20 years old.
Condescending much? Even the great mathematician Erdos initially got the Monty Hall problem wrong, which is considered a basic conditional probability problem. And most of the fuss is over the wording ambiguity of the problem, not the problem solving itself.
Sorry if I came off that way. I was genuinely surprised at the amount of people saying it's 1/3. I see nothing wrong with the wording of the problem now but I didn't read it pre-edit.
On June 10 2011 23:41 Dlok wrote: to those who say 1/5: flip two coins, then tell me one of the results in any way that does not exclude the possibility of a pair. you would give me 1/3 odds on a pair, while the odds clearly havent changed and really are 1/2. now we do this 100 times and I make alot of money.
If we do the test 100s of times the result will in fact be 1/3rd and I would gladly take you up on such odds at something less than 1/2 but greater than 1/3rd as it'd be free money. You could even write a very simple program to test it.
Not understanding probability (which is very counter-intuitive) doesn't change reality.
I recommend you stay away from Vegas. Do you even realize what you'd be betting on?
If I win every time we get heads + heads then you'd win with heads + tails (or tails + heads) and every time we get tails + tails we'd call it a draw. You'd be a fool to give someone odds on heads + heads at > 1/3 when you're not winning on two of the 4 possible outcomes.
Haha these two sentences from wikipedia are also ambiguous. It's unclear that the many people who argued strongly for both sides knew whether it was a paradox or not. Since that link was already posted at the beginning of the thread, I assume people here knew it's was paradox when arguing, so why simply not admit that the wording has it's importance and that it's pointless to argue forever, since both sides are right depending on how you interpret the wording ?
The point is the paradox causes intense reactions by those who don't accept the answer as true while those who understand the right answer are... well right. I don't know why you'd think that people wouldn't argue over a paradox that's known to create controversy and argument.
Every time you say Heads i say it will be pair of heads, if you say tailes i say pair of tails, now we know 50 out of a hundred are likely to be pairs so i will do fine.
If i say i take only pair of heads i will win 1/4 of said 100 but if im allowed to withdraw when atleast one is not heads i raise my odds to 1/3. This however was not how the problem was stated, and I cant se how it could be interpited that way.
Because that is how the problem was stated? Many people are arguing about the wording of the problem, so you can refer to those posts.
The problem as you say it is:
"What are the odds of getting a pair knowing one result"
when the question is...
"What are the odds of getting a pair if one result is heads"
On June 11 2011 02:59 L3g3nd_ wrote: i read the solution, then voted 1/3. youre making the mistake of counting ZT and TZ as two seperate cases, when they are actually exactly the same, given that ordering doesnt matter.
It technically doesn't matter but ZT (or ZT/TZ) is twice as likely as ZZ, so it does matter in a sense.
Alternatively you could look at it as
ZZ ZT x2 ZP x2
ZT x2 is the exact same as ZT and TZ
Its like saying instead of 4+4+4+4 do 3*4
Let me do some math,
So i think one issue is the T/P, but im not sure. anyway ill use Z for zerg and N for not zerg.
On June 11 2011 02:59 L3g3nd_ wrote: i read the solution, then voted 1/3. youre making the mistake of counting ZT and TZ as two seperate cases, when they are actually exactly the same, given that ordering doesnt matter.
It technically doesn't matter but ZT (or ZT/TZ) is twice as likely as ZZ, so it does matter in a sense.
Alternatively you could look at it as
ZZ ZT x2 ZP x2
ZT x2 is the exact same as ZT and TZ
Its like saying instead of 4+4+4+4 do 3*4
Let me do some math,
So i think one issue is the T/P, but im not sure. anyway ill use Z for zerg and N for not zerg.
Well balls now i have an answer of 1/9 and that wasnt even an option. what i do wrong
You aren't getting rid of all the possibilities.
1/9 is correct on the odds of getting ZvZ. However we know that N/N did not happen so those gets thrown out completely because it's impossible for that to be the case since we were told one of the games was ZvZ. However this does not change the relative odds of the other situations (Z/N is still x2 as likely as Z/Z)
So then you have
Z/N 2/X N/Z 2/X Z/Z 1/X
and the sum of the probabilities have to equal 1 so we know it's 2/5 2/5 1/5.
On June 11 2011 02:59 L3g3nd_ wrote: i read the solution, then voted 1/3. youre making the mistake of counting ZT and TZ as two seperate cases, when they are actually exactly the same, given that ordering doesnt matter.
It technically doesn't matter but ZT (or ZT/TZ) is twice as likely as ZZ, so it does matter in a sense.
Alternatively you could look at it as
ZZ ZT x2 ZP x2
ZT x2 is the exact same as ZT and TZ
Its like saying instead of 4+4+4+4 do 3*4
Let me do some math,
So i think one issue is the T/P, but im not sure. anyway ill use Z for zerg and N for not zerg.
Well balls now i have an answer of 1/9 and that wasnt even an option. what i do wrong
You aren't getting rid of all the possibilities.
1/9 is correct on the odds of getting ZvZ. However we know that N/N did not happen so those gets thrown out completely because it's impossible for that to be the case since we were told one of the games was ZvZ. However this does not change the relative odds of the other situations (Z/N is still x2 as likely as Z/Z)
So then you have
Z/N 2/X N/Z 2/X Z/Z 1/X
and the sum of the probabilities have to equal 1 so we know it's 2/5 2/5 1/5.
ah, thatll be why its called conditional probability!
Well, while i still think its 1/3 when it comes to words,
Math doesnt lie, so i understand why its 1/5, mathematically at least
Thanks to those people pointing out where us people who dont get it are going wrong! for a lot of us its awhile since we did math at school, and like for me i havent used it since i left school
incase you guys didn't know, on ladder you don't have a 33.3% chance of getting each race. Long detailed explanation that I don't feel like getting into and explaining, but it is not completely random.
On June 11 2011 02:11 foxmeep wrote: Here's an easier way to understand the problem. Say I flip two coins, we can all agree that the possible outcomes are:
H-T 1/4 H-H 1/4 T-T 1/4 T-H 1/4
Now I tell you that if, and only if, a heads comes up, if the other coin is tails, I win. If the other coin is heads, you win. Simply making this proposition does not change the probability of each result coming up.
H-T I win H-H I lose T-T Draw T-H I win
We already know each of the outcomes has an equal chance of occuring, so in 2/4 results, I win, 1/4 you win. Therefore the odds are 2:1, or 2/3 in my favour, and 1/3 in yours. NOT 1/2.
no, because you are making ordering a necessaity in this. the OP's problem does not take ordering into account.
On June 11 2011 02:59 L3g3nd_ wrote: i read the solution, then voted 1/3. youre making the mistake of counting ZT and TZ as two seperate cases, when they are actually exactly the same, given that ordering doesnt matter.
It technically doesn't matter but ZT (or ZT/TZ) is twice as likely as ZZ, so it does matter in a sense.
Alternatively you could look at it as
ZZ ZT x2 ZP x2
19 pages and the first thing that actually makes sense to me. I didn't realize the probability of ZT is twice the probability of ZZ. This helps, thanks
On June 11 2011 02:59 L3g3nd_ wrote: i read the solution, then voted 1/3. youre making the mistake of counting ZT and TZ as two seperate cases, when they are actually exactly the same, given that ordering doesnt matter.
It technically doesn't matter but ZT (or ZT/TZ) is twice as likely as ZZ, so it does matter in a sense.
Alternatively you could look at it as
ZZ ZT x2 ZP x2
19 pages and the first thing that actually makes sense to me. I didn't realize the probability of ZT is twice the probability of ZZ. This helps, thanks
Explain me why the chance to get ZT is twice of ZZ, as one plays zerg and one plays random.
On June 11 2011 02:59 L3g3nd_ wrote: i read the solution, then voted 1/3. youre making the mistake of counting ZT and TZ as two seperate cases, when they are actually exactly the same, given that ordering doesnt matter.
It technically doesn't matter but ZT (or ZT/TZ) is twice as likely as ZZ, so it does matter in a sense.
Alternatively you could look at it as
ZZ ZT x2 ZP x2
19 pages and the first thing that actually makes sense to me. I didn't realize the probability of ZT is twice the probability of ZZ. This helps, thanks
Explain me why the chance to get ZT is twice of ZZ, as one plays zerg and one plays random.
On June 11 2011 03:43 Kikimiki wrote: Probability of getting zerg two consecutive times is 1/9..Given that probability to get zerg is 1/3
the reason this is wrong is on this page, read a little
OK ill explain it to you,why you are wrong. This is right if the question was "whats the probability of getting zerg 2 consecutive times ?"
This situation could be described using joint probability rules.... If "A" was used to denote getting zerg at game 1, P(A)= 1/3 If "B" was used to denote getting zerg at game 2, P(B)= 1/3 The probability of occurrence of both of A and B is denoted (for independant variables or probabilities, which means the result of event A (getting zerg in game 1) in the 1st try doesn't alter event B) Which is also equal to P(A) x P(B) = 1/9
If the question was "Whats the probability of occurrence of event be given that A had occured ?" That would be denoted by: P(B|A) which is also equal to P(B)= 1/3, for independent event.. Thats because getting zerg in the 1st try doesn't change your chances of getting zerg the 2nd try..
On June 11 2011 03:43 Kikimiki wrote: Probability of getting zerg two consecutive times is 1/9..Given that probability to get zerg is 1/3
the reason this is wrong is on this page, read a little
OK ill explain it to you,why you are wrong. This is right if the question was "whats the probability of getting zerg 2 consecutive times ?"
This situation could be described using joint probability rules.... If "A" was used to denote getting zerg at game 1, P(A)= 1/3 If "B" was used to denote getting zerg at game 2, P(B)= 1/3 The probability of occurrence of both of A and B is denoted (for independant variables or probabilities, which means the result of event A (getting zerg in game 1) in the 1st try doesn't alter event B) Which is also equal to P(A) x P(B) = 1/9
If the question was "Whats the probability of occurrence of event be given that A had occured ?" That would be denoted by: P(B|A) which is also equal to P(B)= 1/3, for independent event.. Thats because getting zerg in the 1st try doesn't change your chances of getting zerg the 2nd try..
But A does not equal getting zerg on your first try. It's getting Zerg on one of the two tries. The events are not independent. B is also "Zerg on both of two games" not "Zerg on the 2nd game"
It's easy to illustrate. If you look at one of the two games and it's not zerg the probability of the other game being Zerg is 1 (100%) otherwise our known condition (at least one game is zerg) wouldn't be true.
The question is ambiguous for the same reason the Boy/Girl Paradox question is ambiguous. Draw an analogy from this quote about the Boy/Girl Paradox:
"While it is certainly true that every possible Mr. Smith has at least one boy - i.e., the condition is necessary - it is not clear that every Mr. Smith with at least one boy is intended. That is, the problem statement does not say that having a boy is a sufficient condition for Mr. Smith to be identified as having a boy this way."
i.e. - if your friend would not always say that he saw a ZvZ, but perhaps some of the times that TLO played zerg in one game, he said instead that Nestea has sick ZvT because he only watched the other game where TLO played Terran... then the conditional probability of your friend telling you that Nestea has a sick ZvZ when TLO played at least once as zerg is less than one. This provides a different answer than if TLO playing one game as zerg is sufficient for your friend to report that Nestea has sick ZvZ.
If your friend could possibly have told you that Nestea has sick ZvT when TLO played zerg in the other game, then the probability that TLO played zerg twice is much higher because the conditional probability of your friend telling you Nestea played a sick ZvZ when TLO rolled ZZ is 1... and 1/2 for all other combinations.
From what I gathered on the Wikipedia article previously posted, the problem is only paradoxical because of the ambiguous nature of the statement of the problem. The statistical analysis is correct when viewed from both sides, it is only the view which causes the inconsistency. By changing a few words in the statement the view can change from a statistical analysis of a singular event (e.g. I played 2 games. The first game was played as zerg. What is the probability the next game will be played as zerg? In this case the first and second statement are true, but unimportant, as the problem asks for the probability that the next game will be zerg. We can deduce that only 1 of 3 options fits the parameters of the question, therefore the answer is 1/3.) to an analysis of all possibilities involving at least one zerg (e.g. I played 2 games. One game was played as Zerg. What is the probability of playing zerg in both games? Given these parameters you weed out all non-zerg possibilities and come to the aforementioned 1/5 answer.) as well as a further possibility of analysis involving all possibilities over two games when what race is played either game is unknown (e.g. I played 2 games. What is the probability of playing zerg in both games? This time the statement of one game being played as zerg is removed and all possibilities must be included. Since there are 9 possibilities, and only 1 of them is ZZ, the answer is 1/9.).
For an additional example of the linguistic nature of this type of paradox I will quote a different one from Wikipedia:
For example, consider a situation in which a father and his son are driving down the road. The car crashes into a tree and the father is killed. The boy is rushed to the nearest hospital where he is prepared for emergency surgery. On entering the surgery suite, the surgeon says, "I can't operate on this boy. He's my son."
In this situation the reader may first assume that it is an impossibility and completely illogical. The father was killed in the crash, therefore it is impossible for him to be a surgeon at the hospital when his son arrives. While this conclusion is logically correct, it is founded on the assumption that the father is the only possible parent. If it is given that the mother is a surgeon at the hospital, then it can be concluded that the mother is the surgeon in question and as such the statement is entirely possible and logical. It is the specific way in which the information is presented that leads to false assumptions, providing the ambiguity of the problem.
Hopefully this will help shift the focus of the discussion from whether the answer is 1/3 or 1/5 to the actual problem of the inherent ambiguity of the original question and the erroneous assumptions and conclusions made from it. And for anyone who denies that the original question is ambiguous, the words used are specifically designed to create confusion leading to it being "open to more than one interpretation" which is the definition of ambiguous. If it wasn't designed in such a way we wouldn't be discussing it.
[An Aside] I've been up all night and I am terribly tired, so if I messed something up I apologize.
On June 10 2011 22:30 DarkPlasmaBall wrote: Agreed. You're not twice as likely to play Zerg in ZZ as you are in TZ. You just happen to have twice as many Zs. Big difference.
I flip a standard coin, what is the probability of heads?
I flip a coin that has heads on both sides, what is the probability of heads?
If you have twice as many of something you're twice as likely to pick said thing.
Two different games are two different coin flips. Not the same coinflip. It's not even close to the same thing. And you're still forgetting about conditional probability.
Anyways, I gotta go. Enjoy your day
You're utterly confused.
If I have two replays and I play the same race in each replay, and you select one of those replays at random, it is twice as likely to be that race than if you have two replays where you play a different race in each replay.
The coin flip was to represent which of the two replays in the series gets sent to you.
I'm upset that the education system hasn't worked for so many people.
Oh, the irony in this post. You're one of the few people who has been arguing for 1/3 throughout pages and pages of this thread, despite us attempting to explain conditional probability to you.
Me too. Me too.
I guess we math teachers need to do a little better in explaining these problems so that people don't misinterpret the question and end up with a wrong analogy and answer, huh?
On June 11 2011 03:43 Kikimiki wrote: Probability of getting zerg two consecutive times is 1/9..Given that probability to get zerg is 1/3
the reason this is wrong is on this page, read a little
OK ill explain it to you,why you are wrong. This is right if the question was "whats the probability of getting zerg 2 consecutive times ?"
This situation could be described using joint probability rules.... If "A" was used to denote getting zerg at game 1, P(A)= 1/3 If "B" was used to denote getting zerg at game 2, P(B)= 1/3 The probability of occurrence of both of A and B is denoted (for independant variables or probabilities, which means the result of event A (getting zerg in game 1) in the 1st try doesn't alter event B) Which is also equal to P(A) x P(B) = 1/9
If the question was "Whats the probability of occurrence of event be given that A had occured ?" That would be denoted by: P(B|A) which is also equal to P(B)= 1/3, for independent event.. Thats because getting zerg in the 1st try doesn't change your chances of getting zerg the 2nd try..
But B does not equal getting zerg on your first try. It's getting Zerg on one of the two tries. The events are not independent.
The original problem is : "So I was playing random today, and I played 2 games of Starcraft 2!
I played as Zerg at least once. What is the probability that my other game was as Zerg as well?" 2 Games played. A = getting zerg in 1st game B = getting zerg in 2nd game My translation of these words would be P(B|A), the probability of getting zerg in the 2nd try is independent of the 1st try.....In other words getting zerg the 1st game doesnt change the probability of getting zerg 2nd game, and thus events are independent.
On June 11 2011 03:43 Kikimiki wrote: Probability of getting zerg two consecutive times is 1/9..Given that probability to get zerg is 1/3
the reason this is wrong is on this page, read a little
OK ill explain it to you,why you are wrong. This is right if the question was "whats the probability of getting zerg 2 consecutive times ?"
This situation could be described using joint probability rules.... If "A" was used to denote getting zerg at game 1, P(A)= 1/3 If "B" was used to denote getting zerg at game 2, P(B)= 1/3 The probability of occurrence of both of A and B is denoted (for independant variables or probabilities, which means the result of event A (getting zerg in game 1) in the 1st try doesn't alter event B) Which is also equal to P(A) x P(B) = 1/9
If the question was "Whats the probability of occurrence of event be given that A had occured ?" That would be denoted by: P(B|A) which is also equal to P(B)= 1/3, for independent event.. Thats because getting zerg in the 1st try doesn't change your chances of getting zerg the 2nd try..
But B does not equal getting zerg on your first try. It's getting Zerg on one of the two tries. The events are not independent.
The original problem is : "So I was playing random today, and I played 2 games of Starcraft 2!
I played as Zerg at least once. What is the probability that my other game was as Zerg as well?" 2 Games played. A = getting zerg in 1st game B = getting zerg in 2nd game My translation of these words would be P(B|A), the probability of getting zerg in the 2nd try is independent of the 1st try.....In other words getting zerg the 1st game doesnt change the probability of getting zerg 2nd game, and thus events are independent.
but that's not reading English correctly because he never said the first game was played as Zerg.
If I said, "I played the lottery a million times and won at least once" you wouldn't assume I won the lottery on my very first try then played it 999,999 more times (in fact you're likely to assume the exact opposite). You're reading more into his statements than he's said because you consider playing Zerg a common event and he's only done 2 trials.
Your probability is correct if you assume he meant he got Zerg the first game, but that's not what he said if you ready precisely what he said and not what you think he's implying.
The order of the games doesn't matter though. One of them being Zerg is given and the other is not so you are only calculating the probability of the one that isn't given.
On June 11 2011 03:43 Kikimiki wrote: Probability of getting zerg two consecutive times is 1/9..Given that probability to get zerg is 1/3
the reason this is wrong is on this page, read a little
OK ill explain it to you,why you are wrong. This is right if the question was "whats the probability of getting zerg 2 consecutive times ?"
This situation could be described using joint probability rules.... If "A" was used to denote getting zerg at game 1, P(A)= 1/3 If "B" was used to denote getting zerg at game 2, P(B)= 1/3 The probability of occurrence of both of A and B is denoted (for independant variables or probabilities, which means the result of event A (getting zerg in game 1) in the 1st try doesn't alter event B) Which is also equal to P(A) x P(B) = 1/9
If the question was "Whats the probability of occurrence of event be given that A had occured ?" That would be denoted by: P(B|A) which is also equal to P(B)= 1/3, for independent event.. Thats because getting zerg in the 1st try doesn't change your chances of getting zerg the 2nd try..
But B does not equal getting zerg on your first try. It's getting Zerg on one of the two tries. The events are not independent.
The original problem is : "So I was playing random today, and I played 2 games of Starcraft 2!
I played as Zerg at least once. What is the probability that my other game was as Zerg as well?" 2 Games played. A = getting zerg in 1st game B = getting zerg in 2nd game My translation of these words would be P(B|A), the probability of getting zerg in the 2nd try is independent of the 1st try.....In other words getting zerg the 1st game doesnt change the probability of getting zerg 2nd game, and thus events are independent.
If your first game is Protoss or Terran, then you know your second game will be Zerg no matter what, since the axiom is that at least one of your games must be Zerg. Your second game is indeed dependent on the first in those cases. If your first game is Zerg, then your second game is truly random (can be Zerg, Terran, or Protoss). Therefore, there are 5 total outcomes which give you your result of at least one Zerg game. One of those cases has double Zerg. 1/5 is the answer.
Being a Random player only means you're able to roll each race unless restricted by axioms; you're actually omniscient in these mathematical situations (you'll know what you're going to get, based on certain restrictions).
I've played two games. Then the possible combinations are: ZZ, ZP, ZT, PZ, PP, PT, TZ, TP, TT.
However, I've said I played Zerg. Then that eliminates PP, PT, TP, TT.
Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices, and that is the only which corresponds to "The other game is Zerg."
Then the correct answer is 1/5.
I was going to flame and say "why didn't you also eliminate ZP and ZT, your opponent is only playing zerg, then I realised you made it all confusing with your wording.
Same quote, made clearer:
I've played two games. My opponents race doesn't matter. The possible combinations of races I got are: (Z,Z), (Z,P), (Z,T), (P,Z), (P,P), (P,T), (T,Z), (T,P) and (T,T)
However, I've said I played as Zerg at least once. Then that eliminates PP, PT, TP, TT.
Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices, and that is the only which corresponds to "I got zerg both games"
Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever.
Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever.
He listed 5 pairs of games (not 6), and they aren't matchups.
ZP =/= ZvP
ZT =/= TZ, ZP =/= PZ because of the fact that they're pairs of games (getting Zerg then Terran, as opposed to Terran then Zerg).
You clearly didn't even attempt to read the OP. You literally could not be more wrong with the brain teaser's information.
I've played thousands of games of random in SCBW, and I've gotten streaks of like 18 in a row as terran, maybe like 11 in a row as protoss, etc. As well as streaks of just not getting one race at all for like 30 games.
Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever.
He listed 5 pairs of games (not 6), and they aren't matchups.
ZP =/= ZvP
ZT =/= TZ, ZP =/= PZ because of the fact that they're pairs of games (getting Zerg then Terran, as opposed to Terran then Zerg).
You clearly didn't even attempt to read the OP. You literally could not be more wrong with the brain teaser's information.
Player 1 is Nestea (always playes zerg).
So player 1 is constant as Z
so the Zv? is the only question there is. IF one player is playing zerg and one is playing random, the chances of it being ZvZ are 1/3 not 1/5. Its not my fault this thread is full of condescending people bickering over a poorly written "brain teaser"
If the OP wanted people to understand WTF he was asking he would say.
2 players played 2 starcraft games as Random Vs. Random. The first game was ZVZ, AND THEN ask his question, it would make some sense. But he doesnt. He lead's you to believe that ONE player is ZERG and the OTHER is RANDOM. It's not a brain teaser if he can't even ask the question properly.
Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever.
He listed 5 pairs of games (not 6), and they aren't matchups.
ZP =/= ZvP
ZT =/= TZ, ZP =/= PZ because of the fact that they're pairs of games (getting Zerg then Terran, as opposed to Terran then Zerg).
You clearly didn't even attempt to read the OP. You literally could not be more wrong with the brain teaser's information.
Player 1 is Nestea (always playes zerg).
So player 1 is constant as Z
so the Zv? is the only question there is. IF one player is playing zerg and one is playing random, the chances of it being ZvZ are 1/3 not 1/5. Its not my fault this thread is full of condescending people bickering over a poorly written "brain teaser"
That's not the question though. The question is over a 2 game series what's the chance of there being 2 ZvZs played given that at least one game was a ZvZ.
To do it out in long form notation for you here are the possible realities....
NesTea (Z) v TLO (R) Game 1 Game 2 ZvZ ZvZ ZvZ ZvT ZvZ ZvP ZvT ZvZ ZvT ZvT ZvT ZvP ZvP ZvZ ZvP ZvT ZvP ZvP
Now we know that at least one game was ZvZ so we can remove all realities that don't have a ZvZ.
NesTea (Z) v TLO (R) Game 1 Game 2 ZvZ ZvZ ZvZ ZvT ZvZ ZvP ZvT ZvZ ZvP ZvZ
Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever.
He listed 5 pairs of games (not 6), and they aren't matchups.
ZP =/= ZvP
ZT =/= TZ, ZP =/= PZ because of the fact that they're pairs of games (getting Zerg then Terran, as opposed to Terran then Zerg).
You clearly didn't even attempt to read the OP. You literally could not be more wrong with the brain teaser's information.
Player 1 is Nestea (always playes zerg).
So player 1 is constant as Z
so the Zv? is the only question there is. IF one player is playing zerg and one is playing random, the chances of it being ZvZ are 1/3 not 1/5. Its not my fault this thread is full of condescending people bickering over a poorly written "brain teaser"
That's not the question though. The question is over a 2 game series what's the chance of there being 2 ZvZs played given that at least one game was a ZvZ.
Nope. The question is over a 2 game series what is the probability of getting zerg twice when playing as random, if at least one time you get zerg, and the order you play them in matters.
the question IS worded properly, and the ops examples just add to the confusion. stating in your post that it isnt badly worded doesnt make it so. but if you understand what the op meant, yes its 1/5. but its easy to see why people are saying 1/3
Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever.
He listed 5 pairs of games (not 6), and they aren't matchups.
ZP =/= ZvP
ZT =/= TZ, ZP =/= PZ because of the fact that they're pairs of games (getting Zerg then Terran, as opposed to Terran then Zerg).
You clearly didn't even attempt to read the OP. You literally could not be more wrong with the brain teaser's information.
Player 1 is Nestea (always playes zerg).
So player 1 is constant as Z
so the Zv? is the only question there is. IF one player is playing zerg and one is playing random, the chances of it being ZvZ are 1/3 not 1/5. Its not my fault this thread is full of condescending people bickering over a poorly written "brain teaser"
That's not the question though. The question is over a 2 game series what's the chance of there being 2 ZvZs played given that at least one game was a ZvZ.
Nope. The question is over a 2 game series what is the probability of getting zerg twice when playing as random, if at least one time you get zerg, and the order you play them in matters.
Uh that's the exact same thing as what I said except I worded it towards the example where he has NesTea playing TLO for 2 games.
The last part of your statement is not necessarily as well. It doesn't matter if you consider order important or not.
Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever.
He listed 5 pairs of games (not 6), and they aren't matchups.
ZP =/= ZvP
ZT =/= TZ, ZP =/= PZ because of the fact that they're pairs of games (getting Zerg then Terran, as opposed to Terran then Zerg).
You clearly didn't even attempt to read the OP. You literally could not be more wrong with the brain teaser's information.
Player 1 is Nestea (always playes zerg).
So player 1 is constant as Z
so the Zv? is the only question there is. IF one player is playing zerg and one is playing random, the chances of it being ZvZ are 1/3 not 1/5. Its not my fault this thread is full of condescending people bickering over a poorly written "brain teaser"
That's not the question though. The question is over a 2 game series what's the chance of there being 2 ZvZs played given that at least one game was a ZvZ.
Nope. The question is over a 2 game series what is the probability of getting zerg twice when playing as random, if at least one time you get zerg, and the order you play them in matters.
Uh that's the exact same thing as what I said.
Not really, you got the same result, but the question wasn't about playing a specific match like zerg vs zerg. It was about playing zerg as a race twice while playing random. It's not the same thing, despite the same answer.
Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever.
He listed 5 pairs of games (not 6), and they aren't matchups.
ZP =/= ZvP
ZT =/= TZ, ZP =/= PZ because of the fact that they're pairs of games (getting Zerg then Terran, as opposed to Terran then Zerg).
You clearly didn't even attempt to read the OP. You literally could not be more wrong with the brain teaser's information.
Player 1 is Nestea (always playes zerg).
So player 1 is constant as Z
so the Zv? is the only question there is. IF one player is playing zerg and one is playing random, the chances of it being ZvZ are 1/3 not 1/5. Its not my fault this thread is full of condescending people bickering over a poorly written "brain teaser"
That's not the question though. The question is over a 2 game series what's the chance of there being 2 ZvZs played given that at least one game was a ZvZ.
Nope. The question is over a 2 game series what is the probability of getting zerg twice when playing as random, if at least one time you get zerg, and the order you play them in matters.
Uh that's the exact same thing as what I said.
Not really, you got the same result, but the question wasn't about playing a specific match like zerg vs zerg. It was about playing zerg as a race twice while playing random. It's not the same thing, despite the same answer.
It breaks down to the same probability question of P(A|B) where A = at least one game as a zerg and B = both games as zerg since one side of the match-up is constant in the OP's example.
Let's say IMNestea (always Z) is playing against, say, TLO, and TLO is playing random. They're going to play exactly two games (for simplicity). Your friend spoils the fun by saying "OMG Nestea's ZvZ is absolutely sick!" (implying he saw a ZvZ game).
How is anyone supposed to read that and assume you are asking about a RANDOM vs RANDOM situation when you clearly state that one player is "always z".
Ridiculous how condescending people are being. "oh YOU just don't understand this question I'm posing even though I"M sabotaging it myself by making it obnoxiously confusing with examples that I gave.
None one is assuming that. NesTea is always Z AS IT SAYS. It's a 2 game series with 1 player (TLO) playing random both games. Honestly you aren't even trying to read the problem.
The most disturbing thing about this is that this thread is 21 pages long. This is not ambiguous. It is not a paradox. It is possible counterintuitive, but the answer is unquestionably right - get over it and get on with your lives.
Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever.
He listed 5 pairs of games (not 6), and they aren't matchups.
ZP =/= ZvP
ZT =/= TZ, ZP =/= PZ because of the fact that they're pairs of games (getting Zerg then Terran, as opposed to Terran then Zerg).
You clearly didn't even attempt to read the OP. You literally could not be more wrong with the brain teaser's information.
Player 1 is Nestea (always playes zerg).
So player 1 is constant as Z
so the Zv? is the only question there is. IF one player is playing zerg and one is playing random, the chances of it being ZvZ are 1/3 not 1/5. Its not my fault this thread is full of condescending people bickering over a poorly written "brain teaser"
That's not the question though. The question is over a 2 game series what's the chance of there being 2 ZvZs played given that at least one game was a ZvZ.
Nope. The question is over a 2 game series what is the probability of getting zerg twice when playing as random, if at least one time you get zerg, and the order you play them in matters.
Uh that's the exact same thing as what I said.
Not really, you got the same result, but the question wasn't about playing a specific match like zerg vs zerg. It was about playing zerg as a race twice while playing random. It's not the same thing, despite the same answer.
It breaks down to the same probability question of P(A|B) where A = at least one game as a zerg and B = both games as zerg since one side of the match-up is constant in the OP's example.
Yeah I realize that, and stated that. Was just correcting you as the original question wasn't about matchups.
Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever.
He listed 5 pairs of games (not 6), and they aren't matchups.
ZP =/= ZvP
ZT =/= TZ, ZP =/= PZ because of the fact that they're pairs of games (getting Zerg then Terran, as opposed to Terran then Zerg).
You clearly didn't even attempt to read the OP. You literally could not be more wrong with the brain teaser's information.
Player 1 is Nestea (always playes zerg).
So player 1 is constant as Z
so the Zv? is the only question there is. IF one player is playing zerg and one is playing random, the chances of it being ZvZ are 1/3 not 1/5. Its not my fault this thread is full of condescending people bickering over a poorly written "brain teaser"
That's not the question though. The question is over a 2 game series what's the chance of there being 2 ZvZs played given that at least one game was a ZvZ.
Nope. The question is over a 2 game series what is the probability of getting zerg twice when playing as random, if at least one time you get zerg, and the order you play them in matters.
Uh that's the exact same thing as what I said.
Not really, you got the same result, but the question wasn't about playing a specific match like zerg vs zerg. It was about playing zerg as a race twice while playing random. It's not the same thing, despite the same answer.
It breaks down to the same probability question of P(A|B) where A = at least one game as a zerg and B = both games as zerg since one side of the match-up is constant in the OP's example.
Yeah I realize that, and stated that. Was just correcting you as the original question wasn't about matchups.
Well yeah... sort of. The OP has posted 2 versions of the same question (after his edits) which is what's causing a lot of confusion.
Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever.
Check that punctuation, one of those guys is a period!
Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever.
He listed 5 pairs of games (not 6), and they aren't matchups.
ZP =/= ZvP
ZT =/= TZ, ZP =/= PZ because of the fact that they're pairs of games (getting Zerg then Terran, as opposed to Terran then Zerg).
You clearly didn't even attempt to read the OP. You literally could not be more wrong with the brain teaser's information.
Player 1 is Nestea (always playes zerg).
So player 1 is constant as Z
so the Zv? is the only question there is. IF one player is playing zerg and one is playing random, the chances of it being ZvZ are 1/3 not 1/5. Its not my fault this thread is full of condescending people bickering over a poorly written "brain teaser"
If the OP wanted people to understand WTF he was asking he would say.
2 players played 2 starcraft games as Random Vs. Random. The first game was ZVZ, AND THEN ask his question, it would make some sense. But he doesnt. He lead's you to believe that ONE player is ZERG and the OTHER is RANDOM. It's not a brain teaser if he can't even ask the question properly.
Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever.
He listed 5 pairs of games (not 6), and they aren't matchups.
ZP =/= ZvP
ZT =/= TZ, ZP =/= PZ because of the fact that they're pairs of games (getting Zerg then Terran, as opposed to Terran then Zerg).
You clearly didn't even attempt to read the OP. You literally could not be more wrong with the brain teaser's information.
Player 1 is Nestea (always playes zerg).
So player 1 is constant as Z
so the Zv? is the only question there is. IF one player is playing zerg and one is playing random, the chances of it being ZvZ are 1/3 not 1/5. Its not my fault this thread is full of condescending people bickering over a poorly written "brain teaser"
That's not the question though. The question is over a 2 game series what's the chance of there being 2 ZvZs played given that at least one game was a ZvZ.
Nope. The question is over a 2 game series what is the probability of getting zerg twice when playing as random, if at least one time you get zerg, and the order you play them in matters.
Uh that's the exact same thing as what I said.
Not really, you got the same result, but the question wasn't about playing a specific match like zerg vs zerg. It was about playing zerg as a race twice while playing random. It's not the same thing, despite the same answer.
It breaks down to the same probability question of P(A|B) where A = at least one game as a zerg and B = both games as zerg since one side of the match-up is constant in the OP's example.
Yeah I realize that, and stated that. Was just correcting you as the original question wasn't about matchups.
Well yeah... sort of. The OP has posted 2 versions of the same question (after his edits) which is what's causing a lot of confusion.
On June 11 2011 05:40 eluv wrote: The most disturbing thing about this is that this thread is 21 pages long. This is not ambiguous. It is not a paradox. It is possible counterintuitive, but the answer is unquestionably right - get over it and get on with your lives.
It is ambiguous for reasons that are well documented. This is a rephrasing of the ambiguous question originally posed as part of the "Boy or Girl Paradox."
Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever.
He listed 5 pairs of games (not 6), and they aren't matchups.
ZP =/= ZvP
ZT =/= TZ, ZP =/= PZ because of the fact that they're pairs of games (getting Zerg then Terran, as opposed to Terran then Zerg).
You clearly didn't even attempt to read the OP. You literally could not be more wrong with the brain teaser's information.
Player 1 is Nestea (always playes zerg).
So player 1 is constant as Z
so the Zv? is the only question there is. IF one player is playing zerg and one is playing random, the chances of it being ZvZ are 1/3 not 1/5. Its not my fault this thread is full of condescending people bickering over a poorly written "brain teaser"
If the OP wanted people to understand WTF he was asking he would say.
2 players played 2 starcraft games as Random Vs. Random. The first game was ZVZ, AND THEN ask his question, it would make some sense. But he doesnt. He lead's you to believe that ONE player is ZERG and the OTHER is RANDOM. It's not a brain teaser if he can't even ask the question properly.
Yea this is my conclusion as well.
It's not random vs random. It's zerg vs random for 2 games. The question is how often does random roll zerg both games when your friend tells you that he saw an awesome ZvZ?
On June 11 2011 05:40 eluv wrote: The most disturbing thing about this is that this thread is 21 pages long. This is not ambiguous. It is not a paradox. It is possible counterintuitive, but the answer is unquestionably right - get over it and get on with your lives.
It is ambiguous for reasons that are well documented. This is a rephrasing of the ambiguous question originally posed as part of the "Boy or Girl Paradox."
It's mostly ambiguous for external factors to the intention of the problem, as in stuff like you are not equally likely to have a girl as you are a boy and selecting random isn't truly random.
Either way people should accept that the answer (in the OP's example) can range from 1/5 to 1/3 inclusive depending on all of the underlying factors, assumptions, and wording. Claiming it's 1/3rd absolutely and not acknowledging that the general idea of the problem can produce a 1/5 probability is not understanding the problem fully.
Nice brainteaser, almost got me! The easiest way I solved it with is just writing all the possible outcomes that the random player could get, if at least one of his game was as zerg: PZ, TZ, ZZ, ZP and ZT ^^
1 player play 2 games of starcraft as random. In at least one of the games he got zerg but it is unknown which one. What is the probability that the player played zerg in both games?
All of the possibilities would be TT, TP, TZ, PP, PT, PZ, ZZ, ZT, ZP. We know this to be a fact because without any conditions every race has 1/3 chance of being chosen, so every outcome here has 1/3*1/3=1/9 chance of being positive. There are 9 possible outcomes and 1/9*9=1 which ensures that we have every outcome listed.
Now we know that zerg was played in at least one of the games which would be the outcomes
TZ, PZ, ZZ, ZT, ZP
the chance of that happening is 1/9+1/9+1/9+1/9+1/9=5/9 but that is not what we want to solve.
Our condition is that zerg was played at least once with a probability of 1, which means that
Pr(TZ) + Pr(PZ) + Pr(ZZ) + Pr(ZT) + Pr(ZP) = 1, and all the probabilities are equal.
If you solve the above equation you get that Pr(ZZ) = 1/5
Hope this is helpful for some, who had problems understanding the op.
The part where you introduced Nestea and TLO to give some hints just made it more confusing for some people. DON'T think of his opponent at all, only focus on him, the player who's playing random.
2 games, 9 sets of possible outcomes (for his own race only, still ignore the opponent). 4 of these don't include zerg at all. Left are T+Z, P+Z, Z+T, Z+P and Z+Z. Only Z+Z fulfils the requirements of him getting zerg both games. ^^ 1/5.
On June 11 2011 05:40 eluv wrote: The most disturbing thing about this is that this thread is 21 pages long. This is not ambiguous. It is not a paradox. It is possible counterintuitive, but the answer is unquestionably right - get over it and get on with your lives.
It is ambiguous for reasons that are well documented. This is a rephrasing of the ambiguous question originally posed as part of the "Boy or Girl Paradox."
It's mostly ambiguous for external factors to the intention of the problem, as in stuff like you are not equally likely to have a girl as you are a boy and selecting random isn't truly random.
Those factors are not at all the reason that it is ambiguous. Those factors can be determined. However, the wording of the problem makes it impossible to determine whether having played at least one game of zerg is sufficient for your friend to report that there was an awesome ZvZ. This alters the Probability that your friend will tell you that nestea is amazing ZvZ. If it IS sufficient, the Probability is 5/9 (1/5 of which is ZZ), if it IS NOT, the probability is 1/3 (1/3 of which is ZZ).
If player 1 and player 2 played two random vs random games, and you know at least one of the games was a ZvZ, what is the probability that both games were ZvZs?
If player 1 and player 2 played two random vs random games, and you know player 1 played zerg in at least one of the games, what is the probability that both games were ZvZs?
If player 1 and player 2 played two random vs random games, and you know that at least one of the players played zerg in at least one of the games, what is the probability that both games were ZvZs?
On June 11 2011 06:34 LastPrime wrote: Extension problems (answers in spoilers):
If player 1 and player 2 played two random vs random games, and you know at least one of the games was a ZvZ, what is the probability that both games were ZvZs?
If player 1 and player 2 played two random vs random games, and you know player 1 played zerg in at least one of the games, what is the probability that both games were ZvZs?
If player 1 and player 2 played two random vs random games, and you know that at least one of the players played zerg in at least one of the games, what is the probability that both games were ZvZs?
and you know player 1 played zerg in at least one of the games doesn't exclude player 2 from also getting zerg, making it equivalent to "at least one of the players played zerg".
The 1/5th is correct. In the post you say the combinations are ZZ, ZP, ZT, PZ, PP, PT, TZ, TP, TT
Perhaps it will be easier for people to understand if you say there are 3*3=9 combinations listed here (ZvZ,ZvZ), (ZvZ,ZvP), (ZvZ,ZvT), (ZvP,ZvZ), (ZvP,ZvP), (ZvP,ZvT), (ZvT,ZvZ), (ZvT,ZvP), (ZvT,ZvT)
This is the same thing of course, but some might not understand what ZZ means or why some of those combinations are eliminated when you say a ZvZ was played.
I should also note that there is no confusion with the way the problem is stated originally or now. The confusion is because of people's inability to understand a complicated problem and so they ignorantly and naturally believe it is worded wrong.
On June 11 2011 06:34 LastPrime wrote: Extension problems (answers in spoilers):
If player 1 and player 2 played two random vs random games, and you know at least one of the games was a ZvZ, what is the probability that both games were ZvZs?
If player 1 and player 2 played two random vs random games, and you know player 1 played zerg in at least one of the games, what is the probability that both games were ZvZs?
If player 1 and player 2 played two random vs random games, and you know that at least one of the players played zerg in at least one of the games, what is the probability that both games were ZvZs?
and you know player 1 played zerg in at least one of the games doesn't exclude player 2 from also getting zerg, making it equivalent to "at least one of the players played zerg".
The latter case includes the former. You must consider the case in which player 2 player plays zerg but not player 1.
You played two games random of which at least one you were Z and ask what the probobility is that the other be Z aswell. there are five outcomes as many have claimed, but they are not as likly.
ZZ we can be sure ZN and NZ appear to be four outcomes giving ZZ 1/5 odds
what is forgotten is the possibility of op asking for the N in those cases, e.i ''of which atleast one was P'' that would happen half the time, when this is taken to account we get the real odds 1/3. Good brain teser!
what you want to ask is: I played two games with Z, atlest one was against Z whats the probability the other was aswell.
Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever.
He listed 5 pairs of games (not 6), and they aren't matchups.
ZP =/= ZvP
ZT =/= TZ, ZP =/= PZ because of the fact that they're pairs of games (getting Zerg then Terran, as opposed to Terran then Zerg).
You clearly didn't even attempt to read the OP. You literally could not be more wrong with the brain teaser's information.
Player 1 is Nestea (always playes zerg).
So player 1 is constant as Z
so the Zv? is the only question there is. IF one player is playing zerg and one is playing random, the chances of it being ZvZ are 1/3 not 1/5. Its not my fault this thread is full of condescending people bickering over a poorly written "brain teaser"
If the OP wanted people to understand WTF he was asking he would say.
2 players played 2 starcraft games as Random Vs. Random. The first game was ZVZ, AND THEN ask his question, it would make some sense. But he doesnt. He lead's you to believe that ONE player is ZERG and the OTHER is RANDOM. It's not a brain teaser if he can't even ask the question properly.
Yea this is my conclusion as well.
It's not random vs random. It's zerg vs random for 2 games. The question is how often does random roll zerg both games when your friend tells you that he saw an awesome ZvZ?
If it's zerg vs random then the only combinations for all the outcomes of the games are
Game 1: (I'll use a and b for players a and b a is nestea who always plays zerg)
(Za,Tb), (Za,Pb), (Za,Zb)
Game 2: (Za,Tb), (Za,Pb) (Za,Zb)
Because we know in one game there was a ZvZ we can infer from that that while there are still 5 possibilities remaining in our random series only the possibilities of the remaining game matter. Thus proving what the possibility of one out come in one game is to be 1/3rd.
If it's detecting the probability that both games are ZvZ (under no prior conditions) it's 1/9th. Under the assumption that 1 game has already been played and it was stated to be a ZvZ it would seemingly eliminate one of the possibilities out of the 6 but this isn't true. (see above paragraph)
Think of it this way: A player rolls 2 die. One of them is a 6, what is the possibility of the other being a 6 thus both of them being a 6. Well since there are 6 sides it's going to occur 1/6th of the time. Because one outcome is guaranteed to occur you can eliminate that from the experiment.
On June 11 2011 00:29 Ivs wrote: People are still arguing because OP wanted to present the Boy/Girl paradox, but messed up the wording.
Now there are 3 camps of people
1. People interpreted the OP as the boy/girl paradox, even though OP failed. They say 1/5
2. People who don't really understand whats going on and go with the simplest reasoning. They say 1/3
3. People who are calling out OP's original wording and poor usage of "other", and also get the answer of 1/3.
Chill out guys, no need for name/credential calling. There is no argument here.
There is no error in the original wording. The statement, "Given that I played zerg at least once, I played zerg both times," and "I played zerg once and I played the other game zerg as well" are equivalent. People are just misconceiving how many choices they actually have, thinking that "the other game" is in reference to a specific game, when it can not be.
Exactly, the two statements are logically equivalent. The only difference is that the 2nd one is misleading, encouraging the reader to focus more on the "other game" and disregard the given information.
The 3 groups of people should be:
1. People who messed up by reading the question too quickly (even though they understand conditional probability). So, instead they to try to convince themselves and others that the OP was wrong, not them.
2. People who don't really understand whats going on and go with the simplest reasoning. They say 1/3
3. People who got the correct answer 1/5.
Although they seem equivalent, I don't think they actually are. One is "at least 1 is zerg" The other is "this one is zerg"
"this one is zerg" implies "at least 1 is zerg" but "at least 1 is zerg" does NOT imply "this one is zerg"
So "this one is zerg" is giving strictly more information. Hence why the wording of "other" is important, and not the right choice for what the OP wanted to achieve.
EDIT: the OP used "at least once" which is good, but then the question becomes meaningless as it is addressing "the other", hence why it's misleading.
I think anyone who's still confused should take time to understand the different interpretations as well as reading up on the paradox. Like I said there is not much to argue here. Then again it IS the internet.
EPT
) so a1+1=a2, a2+1=a3 and so on and so forth
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Monty Hall![[image loading]](http://i.imgur.com/68oPu.png)
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(for independant variables or probabilities, which means the result of event A (getting zerg in game 1) in the 1st try doesn't alter event B)
DON'T think of his opponent at all, only focus on him, the player who's playing random.
