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On June 10 2011 16:01 Tektos wrote:Show nested quote +On June 10 2011 15:58 OneOther wrote:On June 10 2011 15:51 Tektos wrote: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.
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On June 10 2011 16:06 Tektos wrote:Show nested quote +On June 10 2011 16:03 qrs wrote:On June 10 2011 15:59 Tektos wrote:On June 10 2011 15:57 qrs wrote:On June 10 2011 15:51 Tektos wrote: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.
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On June 10 2011 16:10 Ivs wrote:Show nested quote +On June 10 2011 16:06 Tektos wrote:On June 10 2011 16:03 qrs wrote:On June 10 2011 15:59 Tektos wrote:On June 10 2011 15:57 qrs wrote:On June 10 2011 15:51 Tektos wrote: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). Thanks; that's a better analogy than mine.
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Stop ignoring me, goddamnit.
I roll a 1d100 twice.
I tell you, "one of the numbers was a 19."
I ask, "what's the chance that the other number is a 19?"
Poll: Chance that the other number is 19?1/199 (1) 100% 1/201 (0) 0% 1 total votes Your vote: Chance that the other number is 19? (Vote): 1/199
(Vote): 1/201
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On June 10 2011 15:58 OneOther wrote:Show nested quote +On June 10 2011 15:51 ZombiesOMG wrote:On June 10 2011 15:47 OneOther wrote:
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? ^^
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On June 10 2011 16:16 EmeraldSparks wrote:Stop ignoring me, goddamnit. I roll a 1d100 twice. I tell you, "one of the numbers was a 19." I ask, "what's the chance that the other number is a 19?" Poll: Chance that the other die is 19?1/199 (1) 50% 1/100 (1) 50% 2 total votes Your vote: Chance that the other die is 19? (Vote): 1/199
(Vote): 1/100
Who cares this has nothing to do with the OP
make your own thread if you want to talk about a different situation
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On June 10 2011 16:17 Clog wrote:
this has nothing to do with the OP
But that's wrong.
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United States10774 Posts
On June 10 2011 16:14 qrs wrote:Show nested quote +On June 10 2011 16:06 Tektos wrote:On June 10 2011 16:03 qrs wrote:On June 10 2011 15:59 Tektos wrote:On June 10 2011 15:57 qrs wrote:On June 10 2011 15:51 Tektos wrote: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.
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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.
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On June 10 2011 16:18 EmeraldSparks wrote: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.
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This does not deserve a thread.
this is middle school math.
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On June 10 2011 16:20 Clog wrote:Show nested quote +On June 10 2011 16:18 EmeraldSparks wrote: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.
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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.
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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 ^^
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On June 10 2011 16:22 EmeraldSparks wrote:Show nested quote +On June 10 2011 16:20 Clog wrote:On June 10 2011 16:18 EmeraldSparks wrote: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.
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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: Show nested quote +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.
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On June 10 2011 16:32 Eknoid4 wrote:Show nested quote +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.
What did you put?
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On June 10 2011 16:20 Clog wrote:Show nested quote +On June 10 2011 16:18 EmeraldSparks wrote: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.
The actual question asked was "I played as Zerg at least once. What is the probability that my other game was as Zerg as well?"
Brain teaser messed up the OP so badly that he posted it in a retarded manner.
Wrong Clog.
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On June 10 2011 16:29 Clog wrote:Show nested quote +On June 10 2011 16:22 EmeraldSparks wrote:On June 10 2011 16:20 Clog wrote:On June 10 2011 16:18 EmeraldSparks wrote: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.
No, I'm arguing against you, and you are wrong.
On June 10 2011 16:32 Eknoid4 wrote:Show nested quote +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.
I'm decent at math. You're wrong.
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On June 10 2011 10:19 oxidized wrote:
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.
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EPT
