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First or not, you base your answer on knowing which game has been played as zerg, and this is simply not what was asked.
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On June 10 2011 22:08 Dimagus wrote:Show nested quote +On June 10 2011 22:04 Tektos wrote:On June 10 2011 22:03 Dimagus wrote:On June 10 2011 22:02 Tektos wrote: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.
NOT 1/5
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On June 10 2011 22:09 Tektos wrote:Show nested quote +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.
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Abandon all hope...
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.
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On June 10 2011 22:13 Tektos wrote:Show nested quote +On June 10 2011 22:08 Dimagus wrote:On June 10 2011 22:04 Tektos wrote:On June 10 2011 22:03 Dimagus wrote:On June 10 2011 22:02 Tektos wrote: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....
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On June 10 2011 12:58 lyAsakura wrote:Show nested quote +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
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On June 10 2011 22:17 Dimagus wrote:Show nested quote +On June 10 2011 22:13 Tektos wrote:On June 10 2011 22:08 Dimagus wrote:On June 10 2011 22:04 Tektos wrote:On June 10 2011 22:03 Dimagus wrote:On June 10 2011 22:02 Tektos wrote: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?
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This:
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!
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On June 10 2011 22:18 Tektos wrote:Show nested quote +On June 10 2011 22:17 Dimagus wrote:On June 10 2011 22:13 Tektos wrote:On June 10 2011 22:08 Dimagus wrote:On June 10 2011 22:04 Tektos wrote:On June 10 2011 22:03 Dimagus wrote:On June 10 2011 22:02 Tektos wrote: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...?
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Wow, TL is not really good at math, or reading. 
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. :/
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On June 10 2011 22:20 Dimagus wrote:Show nested quote +On June 10 2011 22:18 Tektos wrote:On June 10 2011 22:17 Dimagus wrote:On June 10 2011 22:13 Tektos wrote:On June 10 2011 22:08 Dimagus wrote:On June 10 2011 22:04 Tektos wrote:On June 10 2011 22:03 Dimagus wrote:On June 10 2011 22:02 Tektos wrote: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.
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United States10328 Posts
On June 10 2011 22:18 Tektos wrote:Show nested quote +On June 10 2011 22:17 Dimagus wrote:On June 10 2011 22:13 Tektos wrote:On June 10 2011 22:08 Dimagus wrote:On June 10 2011 22:04 Tektos wrote:On June 10 2011 22:03 Dimagus wrote:On June 10 2011 22:02 Tektos wrote: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.
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On June 10 2011 22:21 Tektos wrote:Show nested quote +On June 10 2011 22:20 Dimagus wrote:On June 10 2011 22:18 Tektos wrote:On June 10 2011 22:17 Dimagus wrote:On June 10 2011 22:13 Tektos wrote:On June 10 2011 22:08 Dimagus wrote:On June 10 2011 22:04 Tektos wrote:On June 10 2011 22:03 Dimagus wrote:On June 10 2011 22:02 Tektos wrote: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?
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On June 10 2011 22:21 Tektos wrote:Show nested quote +On June 10 2011 22:20 Dimagus wrote:On June 10 2011 22:18 Tektos wrote:On June 10 2011 22:17 Dimagus wrote:On June 10 2011 22:13 Tektos wrote:On June 10 2011 22:08 Dimagus wrote:On June 10 2011 22:04 Tektos wrote:On June 10 2011 22:03 Dimagus wrote:On June 10 2011 22:02 Tektos wrote: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.
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On June 10 2011 22:21 ]343[ wrote:Show nested quote +On June 10 2011 22:18 Tektos wrote:On June 10 2011 22:17 Dimagus wrote:On June 10 2011 22:13 Tektos wrote:On June 10 2011 22:08 Dimagus wrote:On June 10 2011 22:04 Tektos wrote:On June 10 2011 22:03 Dimagus wrote:On June 10 2011 22:02 Tektos wrote: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.
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On June 10 2011 22:24 Tektos wrote:Show nested quote +On June 10 2011 22:21 ]343[ wrote:On June 10 2011 22:18 Tektos wrote:On June 10 2011 22:17 Dimagus wrote:On June 10 2011 22:13 Tektos wrote:On June 10 2011 22:08 Dimagus wrote:On June 10 2011 22:04 Tektos wrote:On June 10 2011 22:03 Dimagus wrote:On June 10 2011 22:02 Tektos wrote: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.
And.... there's the flaw.
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On June 10 2011 22:25 Dimagus wrote:Show nested quote +On June 10 2011 22:24 Tektos wrote:On June 10 2011 22:21 ]343[ wrote:On June 10 2011 22:18 Tektos wrote:On June 10 2011 22:17 Dimagus wrote:On June 10 2011 22:13 Tektos wrote:On June 10 2011 22:08 Dimagus wrote:On June 10 2011 22:04 Tektos wrote:On June 10 2011 22:03 Dimagus wrote:On June 10 2011 22:02 Tektos wrote:
[quote]
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.
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On June 10 2011 22:19 ControlMonkey wrote:
This:
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.
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On June 10 2011 22:25 Dimagus wrote:Show nested quote +On June 10 2011 22:24 Tektos wrote:On June 10 2011 22:21 ]343[ wrote:On June 10 2011 22:18 Tektos wrote:On June 10 2011 22:17 Dimagus wrote:On June 10 2011 22:13 Tektos wrote:On June 10 2011 22:08 Dimagus wrote:On June 10 2011 22:04 Tektos wrote:On June 10 2011 22:03 Dimagus wrote:On June 10 2011 22:02 Tektos wrote:
[quote]
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.
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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 :/
Chance = 1 / (races^(matches-1) + (races-1)*(races^(matches-1) - (races-1)^(matches-1)))
or if we call
chance = C
races = R
matches = M
C = 1 / (R^(M-1) + (R-1)*(R^(M-1) - (R-1)^(M-1)))
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EPT
