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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?
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If this problem upsets you then you should probably read this Wikipedia article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradox
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.
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You're talking about -your- race, and that you got Zerg once. I don't see how anything else matters here.
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I don't see why the fact that you were Zerg in the first game changes the possibility for any of the races in the second game.
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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
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On June 10 2011 10:19 ComusLoM wrote:Show nested quote +On June 10 2011 10:16 theDreamStick wrote: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.
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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.
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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?
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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
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.
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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.
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Answered 1/3 and I'm an idiot, misread the question entirely lol.
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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
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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.
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1/2, based on my battlenet experience
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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.
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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
o0
I must be stupid..
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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
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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.
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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 
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EPT
