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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?
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.
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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.
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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.
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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?
Uhhh, 60%?
Do you mean YOU playing Z or a Z in the game? If it's YOU it's 60%, if it's in the game it's 100%
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On June 10 2011 22:27 Tektos wrote:Show nested quote +On June 10 2011 22:25 Dimagus wrote: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:
[quote]
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...
Who cares...? Not what's being asked.
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On June 10 2011 22:35 Llama wrote:Show nested quote +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.
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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.
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On June 10 2011 22:37 Tektos wrote:Show nested quote +On June 10 2011 22:35 Llama wrote: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?
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On June 10 2011 22:37 Tektos wrote:Show nested quote +On June 10 2011 22:35 Llama wrote: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.
It's 60% chance you are playing Z. It's 100% chance there is a Z in the game.
Can we stop talking about stupid questions?
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On June 10 2011 22:27 Llama wrote:Show nested quote +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.
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.
[edit for clarity]
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On June 10 2011 22:39 Llama wrote:Show nested quote +On June 10 2011 22:37 Tektos wrote:On June 10 2011 22:35 Llama wrote: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.
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Conditional Probability at its finest. Ill go back to my mathematical structures course now! Seems to be way much arguing for a very simple problem.
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On June 10 2011 22:41 Tektos wrote:Show nested quote +On June 10 2011 22:39 Llama wrote:On June 10 2011 22:37 Tektos wrote:On June 10 2011 22:35 Llama wrote: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.
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On June 10 2011 22:33 Tektos wrote:Show nested quote +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 
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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...
![[image loading]](http://i4.photobucket.com/albums/y107/tblinkies/Icons/Movies/megashark.gif)
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On June 10 2011 22:45 Llama wrote:Show nested quote +On June 10 2011 22:41 Tektos wrote:On June 10 2011 22:39 Llama wrote:On June 10 2011 22:37 Tektos wrote:On June 10 2011 22:35 Llama wrote: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.
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Misclicks galore, gah
What is the probability of hitting quote instead of edit?
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On June 10 2011 22:45 DarkPlasmaBall wrote:Show nested quote +On June 10 2011 22:33 Tektos wrote: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.
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On June 10 2011 22:49 Tektos wrote:Show nested quote +On June 10 2011 22:45 Llama wrote:On June 10 2011 22:41 Tektos wrote:On June 10 2011 22:39 Llama wrote:On June 10 2011 22:37 Tektos wrote:On June 10 2011 22:35 Llama wrote: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.
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On June 10 2011 22:53 Llama wrote:Show nested quote +On June 10 2011 22:49 Tektos wrote:On June 10 2011 22:45 Llama wrote:On June 10 2011 22:41 Tektos wrote:On June 10 2011 22:39 Llama wrote:On June 10 2011 22:37 Tektos wrote:On June 10 2011 22:35 Llama wrote: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)
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EPT
