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I've played two games. Then the possible combinations are:
ZZ, ZP, ZT, PZ, PP, PT, TZ, TP, TT.
However, I've said I played Zerg. Then that eliminates PP, PT, TP, TT.
Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices, and that is the only which corresponds to "The other game is Zerg."
Then the correct answer is 1/5.
The elimination only occurs on the first game.
You said you're talking about your race in the second game, so I don't get why you exclude PP, PT, TP, TT. You made no assumptions about the second game, therefore your chance of getting Zerg is 3/9, or 1/3
That's my guess
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On June 10 2011 22:57 Tektos wrote:
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)
If you are asking "what are the chances that I get zerg if I play random" then the answer is 1/3. I don't think that was ever in dispute.
But the original question is: "You play two games, I ask you to give me a zerg replay if you have one. What race is the other replay?"
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On June 10 2011 23:01 Llama wrote:Show nested quote +On June 10 2011 22:57 Tektos wrote:
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) If you are asking "what are the chances that I get zerg if I play random" then the answer is 1/3. I don't think that was ever in dispute. But the original question is: "You play two games, I ask you to give me a zerg replay if you have one. What race is the other replay?"
Yes, the probability of getting zerg if you play random is 1/3, and because there are no constraints and the second replay is independent of the first replay it is still 1/3.
Now add this to the equation: Between that first zerg replay I've handed you, and the replay I still have, at least one of them is of zerg.
I've already handed you one zerg replay, so the condition of "At least one replay being zerg" is fulfilled. Therefore the probability of the other replay that I still hold being zerg is still 1/3.
But the original question is: "You play two games, I ask you to give me a zerg replay if you have one. What race is the other replay?"
That is not the original question, as saying "IF YOU HAVE ONE" implies that TT, TP, PT, PP are all still options.
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On June 10 2011 23:06 Tektos wrote:Show nested quote +But the original question is: "You play two games, I ask you to give me a zerg replay if you have one. What race is the other replay?" That is not the original question, as saying "IF YOU HAVE ONE" implies that TT, TP, PT, PP are all still options.
They are options if you random twice--that's the nature of random. Obviously they get eliminated if we get handed a zerg replay, but not until then.
The point is that the asking for one zerg replay out of two (if there is at least one) is different from simply receiving a zerg replay. The first applies the condition of at least one being zerg before the question is asked, the second is a statement of fact after gaining the relevant information.
Please understand that as a question of probability the two are distinct.
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And you completely ignored the first part of my post which was 99% of the point.
On June 10 2011 23:14 Llama wrote:Show nested quote +On June 10 2011 23:06 Tektos wrote:But the original question is: "You play two games, I ask you to give me a zerg replay if you have one. What race is the other replay?" That is not the original question, as saying "IF YOU HAVE ONE" implies that TT, TP, PT, PP are all still options. They are options if you random twice--that's the nature of random. Obviously they get eliminated if we get handed a zerg replay, but not until then. The point is that the asking for one zerg replay out of two (if there is at least one) is different from simply receiving a zerg replay. The first applies the condition of at least one being zerg before the question is asked, the second is a statement of fact after gaining the relevant information. Please understand that as a question of probability the two are distinct.
Okay so everything is an option.
Now before handing you a replay I tell you that at least one of them is of a zerg, that eliminates the possibilities of the games down to just being:
Z Z
Z T
Z P
T Z
P Z
Now I hand you one zerg replay. There are 6 zerg replays in those 5 possible outcomes (this is where your error comes in).
I hand you one of those 6 zerg replays. Of those 6 possible replays that I could have handed you, can you tell me what the 6 possible other race outcomes that I played in that pair are?
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On June 10 2011 23:20 Tektos wrote:And you completely ignored the first part of my post which was 99% of the point. Show nested quote +On June 10 2011 23:14 Llama wrote:On June 10 2011 23:06 Tektos wrote:But the original question is: "You play two games, I ask you to give me a zerg replay if you have one. What race is the other replay?" That is not the original question, as saying "IF YOU HAVE ONE" implies that TT, TP, PT, PP are all still options. They are options if you random twice--that's the nature of random. Obviously they get eliminated if we get handed a zerg replay, but not until then. The point is that the asking for one zerg replay out of two (if there is at least one) is different from simply receiving a zerg replay. The first applies the condition of at least one being zerg before the question is asked, the second is a statement of fact after gaining the relevant information. Please understand that as a question of probability the two are distinct. Okay so everything is an option. Now before handing you a replay I tell you that at least one of them is of a zerg, that eliminates the possibilities of the games down to just being: Z Z Z T Z P T Z P Z Now I hand you one zerg replay. There are 6 zerg replays in those 5 possible outcomes (this is where your error comes in). I hand you one of those 6 zerg replays. Of those 6 possible replays that I could have handed you, can you tell me what the other race I played in that pair is?
There are six replays but that doesn't mean that each one has the same probability. If each of the pairs has equal possibility then each "Z" in the first pair is halved since you can only pick one of them. Each "Z" does not represent a unit of probability. If you reframe the question in terms of, say, Z and NOTZ then you get:
Z Z
Z NOTZ
NOTZ Z
but that doesn't imply anything about the respective probabilities of each Z or even each pairing.
edit: to clarify further, there aren't six replays at all, there are one or two. Each letter does not represent an actual replay but the possibility of a replay.
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lol if he plays random, there are three races that he could possibly be.. however he can only land as one of those three. if the question is simply asking what the probability that his next game (worded as if the first game has any relevance what so ever) then the answer is one of three or 1/3
edit : busted i didnt read it right.. what are the odds that he is zerg in BOTH games.. well he had a 1/3 chance to be zerg in the first game and a 1/3 chance in the second game.. 1/9?
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to those who say 1/5: flip two coins, then tell me one of the results in any way that does not exclude the possibility of a pair. you would give me 1/3 odds on a pair, while the odds clearly havent changed and really are 1/2. now we do this 100 times and I make alot of money.
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There's 9 possibilities total for TLO:
TT, TZ, TP, ZT, ZZ, ZP, PT, PZ, PP
Out out those, we already know that something with zerg has happened, which is TZ ZT ZZ ZP PZ. Out of those, only ZZ means two mirror matches, so it's 1/5 to get two ZvZ when we already know there's at least one ZvZ.
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On June 10 2011 23:41 Dlok wrote:
to those who say 1/5: flip two coins, then tell me one of the results in any way that does not exclude the possibility of a pair. you would give me 1/3 odds on a pair, while the odds clearly havent changed and really are 1/2. now we do this 100 times and I make alot of money.
If we do the test 100s of times the result will in fact be 1/3rd and I would gladly take you up on such odds at something less than 1/2 but greater than 1/3rd as it'd be free money. You could even write a very simple program to test it.
Not understanding probability (which is very counter-intuitive) doesn't change reality.
I'll also just leave this here:
http://www.ted.com/talks/peter_donnelly_shows_how_stats_fool_juries.html
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On June 10 2011 23:41 Dlok wrote:
to those who say 1/5: flip two coins, then tell me one of the results in any way that does not exclude the possibility of a pair. you would give me 1/3 odds on a pair, while the odds clearly havent changed and really are 1/2. now we do this 100 times and I make alot of money.
Uhm, if you get told one outcome, it's 1/2 for a pair? Possible outcomes are HH, HT and TT. Being told that one coin came up heads would exclude TT and thus it's 1/2 for a pair.
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On June 10 2011 23:54 Hittegods wrote:Show nested quote +On June 10 2011 23:41 Dlok wrote:
to those who say 1/5: flip two coins, then tell me one of the results in any way that does not exclude the possibility of a pair. you would give me 1/3 odds on a pair, while the odds clearly havent changed and really are 1/2. now we do this 100 times and I make alot of money. Uhm, if you get told one outcome, it's 1/2 for a pair? Possible outcomes are HH, HT and TT. Being told that one coin came up heads would exclude TT and thus it's 1/2 for a pair.
No, HT and TH aren't the same.
Or they are the same, but have 2x the chance of appearing.
So it's:
HT TH HH TT
and we remove TT.
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Your wrong. The orders are considered different match ups because you always put the person your referring to race first. Jinro doesn't play ZvT, but he does play TvZ.
Since NesTea is always Z that means it is XvZ so you eliminate either ZvP and ZvT or PvZ and TvZ (either set not a mix and match)
What you are calling a paradox is actually your failure to understand starcraft terminology.
Just read the original problem and you still fail. You don't make any mention of a match up only the race you play, the games are indeed independent events of one another. If order mattered then the odds would be different, but you never specify that. You just suck at trying to sate a question correctly.
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On June 11 2011 00:01 wswordsmen wrote:
Your wrong. The orders are considered different match ups because you always put the person your referring to race first. Jinro doesn't play ZvT, but he does play TvZ.
Since NesTea is always Z that means it is XvZ so you eliminate either ZvP and ZvT or PvZ and TvZ (either set not a mix and match)
What you are calling a paradox is actually your failure to understand starcraft terminology.
They aren't talking about matchups. They're talking about the 2 games.
ZZ is TLO getting zerg twice (so 2 ZvZ's)
ZP is TLO getting zerg then protoss (ZvZ ZvP)
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I'm not into Math, but I think I figured it out by drawing:
![[image loading]](http://www.abload.de/img/brainteaserotie.jpg)
So there are 5 possible answers (playing 2 games rnd of which 1 or more as Z) but only one that leads to ZZ
so it's 1/5
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We're given that he played two games and that he played Z at least once.
The question is: what is the probability that in the other game he also played Z? If he played Z in the other game as well, then hes played Z twice overall. So the question is equivalent to: what is the probability he played Z twice.
Combining the question and our givens into the same sentence, we get:
What is the probability he played Z twice, given that he played Z at least once out of two games.
First lets define some events:
Event A = He played Z twice (out of two games)
Event B = He played Z at least once (out of two games)
P(A|B) = Probability of event A occurring given event B has occurred.
Translating this into a sentence with the events we've defined, we get:
P(A|B) = Probability he played Z twice, given that he played Z at least once out of two games.
Notice the two bolded parts are the same. So now we just need to find P(A|B).
I'm not going to prove Bayes' Theorem, but it is taught in intro stats courses and the proof is pretty straightforward. It states:
P(A|B)*P(B) = P(B|A)*P(A)
Therefore:
P(A|B) = P(B|A)*P(A)/P(B)
P(B|A) = The probability he played Z at least once, given that he played twice..well if we know he played Z twice, hes obviously played it at least once, making this probability = 1
P(A) = The probability he played Z twice out of two games = (1/3)^2 = 1/9
P(B) = The probability he played Z at least once out of two games = (1/3)*1 + (2/3)*(1/3) = 5/9
Explanation:
+ Show Spoiler +We can break this down into two independent events:
Event C: He gets Z the first game (probability=1/3) and gets anything the second game (probability=1). So P(C) = 1/3*1 = 1/3
This covers ZZ, ZT, and ZP.
Event D: He does not get Z in the first game(probability = 2/3) and gets Z the second game (probability=1/3). So P(D) = 2/3*1/3 = 2/9.
This covers TZ and PZ.
Because events C and D encompass all the ways he can get Z at least once (B is the union of C and D) and have no overlaps (the intersection is zero), we can find P(B) by adding P(C) and P(D).
P(B) = P(C) + P(D) - (C intersect D)
= 1/3 + 2/9 - 0
= 5/9
Plugging these numbers back into our equation we get:
P(A|B) = (1 * (1/9)) / (5/9)
= (1/9) / (5/9)
= 1/5
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On June 11 2011 00:08 TastyMuffins wrote:We're given that he played two games and that he played Z at least once. The question is: what is the probability that in the other game he also played Z? If he played Z in the other game as well, then hes played Z twice overall. So the question is equivalent to: what is the probability he played Z twice. Combining the question and our givens into the same sentence, we get: What is the probability he played Z twice, given that he played Z at least once out of two games.First lets define some events: Event A = He played Z twice (out of two games) Event B = He played Z at least once (out of two games) P(A|B) = Probability of event A occurring given event B has occurred. Translating this into a sentence with the events we've defined, we get: P(A|B) = Probability he played Z twice, given that he played Z at least once out of two games.Notice the two bolded parts are the same. So now we just need to find P(A|B). I'm not going to prove Bayes' Theorem, but it is taught in intro stats courses and the proof is pretty straightforward. It states: P(A|B)*P(B) = P(B|A)*P(A) Therefore: P(A|B) = P(B|A)*P(A)/P(B) P(B|A) = The probability he played Z at least once, given that he played twice..well if we know he played Z twice, hes obviously played it at least once, making this probability = 1 P(A) = The probability he played Z twice out of two games = (1/3)^2 = 1/9 P(B) = The probability he played Z at least once out of two games = (1/3)*1 + (2/3)*(1/3) = 5/9 Explanation: + Show Spoiler +We can break this down into two independent events:
Event C: He gets Z the first game (probability=1/3) and gets anything the second game (probability=1). So P(C) = 1/3*1 = 1/3
This covers ZZ, ZT, and ZP.
Event D: He does not get Z in the first game(probability = 2/3) and gets Z the second game (probability=1/3). So P(D) = 2/3*1/3 = 2/9.
This covers TZ and PZ.
Because events C and D encompass all the ways he can get Z at least once (B is the union of C and D) and have no overlaps (the intersection is zero), we can find P(B) by adding P(C) and P(D).
P(B) = P(C) + P(D) - (C intersect D)
= 1/3 + 2/9 - 0
= 5/9 Plugging these numbers back into our equation we get: P(A|B) = (1 * (1/9)) / (5/9) = (1/9) / (5/9) = 1/5
^ and I would like a tasty muffin.
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On June 11 2011 00:10 wswordsmen wrote:Show nested quote +On June 11 2011 00:04 Logo wrote:On June 11 2011 00:01 wswordsmen wrote:
Your wrong. The orders are considered different match ups because you always put the person your referring to race first. Jinro doesn't play ZvT, but he does play TvZ.
Since NesTea is always Z that means it is XvZ so you eliminate either ZvP and ZvT or PvZ and TvZ (either set not a mix and match)
What you are calling a paradox is actually your failure to understand starcraft terminology. They aren't talking about matchups. They're talking about the 2 games. ZZ is TLO getting zerg twice (so 2 ZvZ's) ZP is TLO getting zerg then protoss (ZvZ ZvP) I was wrong about that but it doesn't change the fact he never says order matters. TZ and ZT are still the same so he is double counting them.
They're not the same...
or more accurately if you ignore order you have the possibilities of:
TZ
ZZ
ZP
but they're not equal probability. If you random twice you are twice as likely to get TZ or ZP as you are ZZ.
So it's
ZZ - 1/5
ZT - 2/5
ZP - 2/5
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Isn't ZVP and PVZ the same thing? So aren't there really only 3 choices....?
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EPT
