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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.
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by the way ZZ doesn't mean ZvZ. It means the first game is Z and the second game is Z.
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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.
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On June 10 2011 10:31 theDreamStick wrote:Show nested quote +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''?
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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.
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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?
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On June 10 2011 10:32 Hamster1800 wrote:Show nested quote +On June 10 2011 10:31 theDreamStick wrote: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.
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i understand how it's 1/5... but the way you worded it was VERY VERY fucking confusing...
when you said "what are the chances my other game was zerg" makes people think of the chances of getting zerg as a random player (which is 1/3)
what you SHOULD have said was:
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?
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Fuuuck. I answered before I had fully read the question. D'oh. Q_Q
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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...
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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."
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Canada13407 Posts
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.
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On June 10 2011 10:36 Zocat wrote:Show nested quote +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."
there we go... even better
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As most "brainteasers", it's just a badly formulated question.
Probability for the other game to be Z is 1/3
Probability for the second game to be the same race of your first is 1/5
"Probability that my other game was Zerg?" to me means the former.
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On June 10 2011 10:34 theDreamStick wrote:Show nested quote +On June 10 2011 10:32 Hamster1800 wrote:On June 10 2011 10:31 theDreamStick wrote: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.
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"D is for Diamond, E is for Everything Else"
even grand master? =O
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On June 10 2011 10:32 theDreamStick wrote:Show nested quote +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
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On June 10 2011 10:39 Keitzer wrote:
"D is for Diamond, E is for Everything Else"
even grand master? =O
Off topic, but I made that my sig before Master came out (I think it was during beta).
On June 10 2011 10:32 theDreamStick wrote:Show nested quote +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.
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On June 10 2011 10:31 garlicface wrote:Show nested quote +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)
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theres mathematical talent here, but page 1 is a bunch of retards
User was warned for this post
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EPT
