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On June 11 2011 05:04 Kikimiki wrote:Show nested quote +On June 11 2011 04:51 Logo wrote:On June 11 2011 04:29 Kikimiki wrote:On June 11 2011 03:47 L3g3nd_ wrote:On June 11 2011 03:43 Kikimiki wrote:
Probability of getting zerg two consecutive times is 1/9..Given that probability to get zerg is 1/3 the reason this is wrong is on this page, read a little OK ill explain it to you,why you are wrong. This is right if the question was "whats the probability of getting zerg 2 consecutive times ?" This situation could be described using joint probability rules.... If "A" was used to denote getting zerg at game 1, P(A)= 1/3 If "B" was used to denote getting zerg at game 2, P(B)= 1/3 The probability of occurrence of both of A and B is denoted ![[image loading]](http://upload.wikimedia.org/math/1/c/5/1c5070056d7351f8090e9436778a2cf5.png) (for independant variables or probabilities, which means the result of event A (getting zerg in game 1) in the 1st try doesn't alter event B) Which is also equal to P(A) x P(B) = 1/9If the question was "Whats the probability of occurrence of event be given that A had occured ?" That would be denoted by: P(B|A) which is also equal to P(B)= 1/3, for independent event.. Thats because getting zerg in the 1st try doesn't change your chances of getting zerg the 2nd try.. But B does not equal getting zerg on your first try. It's getting Zerg on one of the two tries. The events are not independent. The original problem is :
"So I was playing random today, and I played 2 games of Starcraft 2!
I played as Zerg at least once. What is the probability that my other game was as Zerg as well?"2 Games played. A = getting zerg in 1st game B = getting zerg in 2nd game My translation of these words would be P(B|A), the probability of getting zerg in the 2nd try is independent of the 1st try.....In other words getting zerg the 1st game doesnt change the probability of getting zerg 2nd game, and thus events are independent.
but that's not reading English correctly because he never said the first game was played as Zerg.
If I said, "I played the lottery a million times and won at least once" you wouldn't assume I won the lottery on my very first try then played it 999,999 more times (in fact you're likely to assume the exact opposite). You're reading more into his statements than he's said because you consider playing Zerg a common event and he's only done 2 trials.
Your probability is correct if you assume he meant he got Zerg the first game, but that's not what he said if you ready precisely what he said and not what you think he's implying.
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The order of the games doesn't matter though. One of them being Zerg is given and the other is not so you are only calculating the probability of the one that isn't given.
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On June 11 2011 05:04 Kikimiki wrote:Show nested quote +On June 11 2011 04:51 Logo wrote:On June 11 2011 04:29 Kikimiki wrote:On June 11 2011 03:47 L3g3nd_ wrote:On June 11 2011 03:43 Kikimiki wrote:
Probability of getting zerg two consecutive times is 1/9..Given that probability to get zerg is 1/3 the reason this is wrong is on this page, read a little OK ill explain it to you,why you are wrong. This is right if the question was "whats the probability of getting zerg 2 consecutive times ?" This situation could be described using joint probability rules.... If "A" was used to denote getting zerg at game 1, P(A)= 1/3 If "B" was used to denote getting zerg at game 2, P(B)= 1/3 The probability of occurrence of both of A and B is denoted (for independant variables or probabilities, which means the result of event A (getting zerg in game 1) in the 1st try doesn't alter event B) Which is also equal to P(A) x P(B) = 1/9If the question was "Whats the probability of occurrence of event be given that A had occured ?" That would be denoted by: P(B|A) which is also equal to P(B)= 1/3, for independent event.. Thats because getting zerg in the 1st try doesn't change your chances of getting zerg the 2nd try.. But B does not equal getting zerg on your first try. It's getting Zerg on one of the two tries. The events are not independent. The original problem is :
"So I was playing random today, and I played 2 games of Starcraft 2!
I played as Zerg at least once. What is the probability that my other game was as Zerg as well?"2 Games played. A = getting zerg in 1st game B = getting zerg in 2nd game My translation of these words would be P(B|A), the probability of getting zerg in the 2nd try is independent of the 1st try.....In other words getting zerg the 1st game doesnt change the probability of getting zerg 2nd game, and thus events are independent.
If your first game is Protoss or Terran, then you know your second game will be Zerg no matter what, since the axiom is that at least one of your games must be Zerg. Your second game is indeed dependent on the first in those cases. If your first game is Zerg, then your second game is truly random (can be Zerg, Terran, or Protoss). Therefore, there are 5 total outcomes which give you your result of at least one Zerg game. One of those cases has double Zerg. 1/5 is the answer.
Being a Random player only means you're able to roll each race unless restricted by axioms; you're actually omniscient in these mathematical situations (you'll know what you're going to get, based on certain restrictions).
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+ Show Spoiler +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.
I was going to flame and say "why didn't you also eliminate ZP and ZT, your opponent is only playing zerg, then I realised you made it all confusing with your wording.
Same quote, made clearer:
I've played two games. My opponents race doesn't matter. The possible combinations of races I got are:
(Z,Z), (Z,P), (Z,T), (P,Z), (P,P), (P,T), (T,Z), (T,P) and (T,T)
However, I've said I played as Zerg at least once. 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 "I got zerg both games"
Then the correct answer is 1/5.
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+ Show Spoiler +Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever.
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On June 11 2011 05:18 Lomak wrote:+ Show Spoiler +Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever.
He listed 5 pairs of games (not 6), and they aren't matchups.
ZP =/= ZvP
ZT =/= TZ, ZP =/= PZ because of the fact that they're pairs of games (getting Zerg then Terran, as opposed to Terran then Zerg).
You clearly didn't even attempt to read the OP. You literally could not be more wrong with the brain teaser's information.
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I've played thousands of games of random in SCBW, and I've gotten streaks of like 18 in a row as terran, maybe like 11 in a row as protoss, etc. As well as streaks of just not getting one race at all for like 30 games.
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On June 11 2011 05:23 DarkPlasmaBall wrote:Show nested quote +On June 11 2011 05:18 Lomak wrote:+ Show Spoiler +Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever. He listed 5 pairs of games (not 6), and they aren't matchups. ZP =/= ZvP ZT =/= TZ, ZP =/= PZ because of the fact that they're pairs of games (getting Zerg then Terran, as opposed to Terran then Zerg). You clearly didn't even attempt to read the OP. You literally could not be more wrong with the brain teaser's information.
Player 1 is Nestea (always playes zerg).
So player 1 is constant as Z
so the Zv? is the only question there is. IF one player is playing zerg and one is playing random, the chances of it being ZvZ are 1/3 not 1/5. Its not my fault this thread is full of condescending people bickering over a poorly written "brain teaser"
If the OP wanted people to understand WTF he was asking he would say.
2 players played 2 starcraft games as Random Vs. Random. The first game was ZVZ, AND THEN ask his question, it would make some sense. But he doesnt. He lead's you to believe that ONE player is ZERG and the OTHER is RANDOM. It's not a brain teaser if he can't even ask the question properly.
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On June 11 2011 05:27 Lomak wrote:Show nested quote +On June 11 2011 05:23 DarkPlasmaBall wrote:On June 11 2011 05:18 Lomak wrote:+ Show Spoiler +Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever. He listed 5 pairs of games (not 6), and they aren't matchups. ZP =/= ZvP ZT =/= TZ, ZP =/= PZ because of the fact that they're pairs of games (getting Zerg then Terran, as opposed to Terran then Zerg). You clearly didn't even attempt to read the OP. You literally could not be more wrong with the brain teaser's information. Player 1 is Nestea (always playes zerg). So player 1 is constant as Z so the Zv? is the only question there is. IF one player is playing zerg and one is playing random, the chances of it being ZvZ are 1/3 not 1/5. Its not my fault this thread is full of condescending people bickering over a poorly written "brain teaser"
That's not the question though. The question is over a 2 game series what's the chance of there being 2 ZvZs played given that at least one game was a ZvZ.
To do it out in long form notation for you here are the possible realities....
NesTea (Z) v TLO (R)
Game 1 Game 2
ZvZ ZvZ
ZvZ ZvT
ZvZ ZvP
ZvT ZvZ
ZvT ZvT
ZvT ZvP
ZvP ZvZ
ZvP ZvT
ZvP ZvP
Now we know that at least one game was ZvZ so we can remove all realities that don't have a ZvZ.
NesTea (Z) v TLO (R)
Game 1 Game 2
ZvZ ZvZ
ZvZ ZvT
ZvZ ZvP
ZvT ZvZ
ZvP ZvZ
1/5 of those is 2 ZvZ games.
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On June 11 2011 05:31 Logo wrote:Show nested quote +On June 11 2011 05:27 Lomak wrote:On June 11 2011 05:23 DarkPlasmaBall wrote:On June 11 2011 05:18 Lomak wrote:+ Show Spoiler +Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever. He listed 5 pairs of games (not 6), and they aren't matchups. ZP =/= ZvP ZT =/= TZ, ZP =/= PZ because of the fact that they're pairs of games (getting Zerg then Terran, as opposed to Terran then Zerg). You clearly didn't even attempt to read the OP. You literally could not be more wrong with the brain teaser's information. Player 1 is Nestea (always playes zerg). So player 1 is constant as Z so the Zv? is the only question there is. IF one player is playing zerg and one is playing random, the chances of it being ZvZ are 1/3 not 1/5. Its not my fault this thread is full of condescending people bickering over a poorly written "brain teaser" That's not the question though. The question is over a 2 game series what's the chance of there being 2 ZvZs played given that at least one game was a ZvZ.
Nope. The question is over a 2 game series what is the probability of getting zerg twice when playing as random, if at least one time you get zerg, and the order you play them in matters.
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the question IS worded properly, and the ops examples just add to the confusion. stating in your post that it isnt badly worded doesnt make it so. but if you understand what the op meant, yes its 1/5. but its easy to see why people are saying 1/3
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On June 11 2011 05:32 Sasashi wrote:Show nested quote +On June 11 2011 05:31 Logo wrote:On June 11 2011 05:27 Lomak wrote:On June 11 2011 05:23 DarkPlasmaBall wrote:On June 11 2011 05:18 Lomak wrote:+ Show Spoiler +Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever. He listed 5 pairs of games (not 6), and they aren't matchups. ZP =/= ZvP ZT =/= TZ, ZP =/= PZ because of the fact that they're pairs of games (getting Zerg then Terran, as opposed to Terran then Zerg). You clearly didn't even attempt to read the OP. You literally could not be more wrong with the brain teaser's information. Player 1 is Nestea (always playes zerg). So player 1 is constant as Z so the Zv? is the only question there is. IF one player is playing zerg and one is playing random, the chances of it being ZvZ are 1/3 not 1/5. Its not my fault this thread is full of condescending people bickering over a poorly written "brain teaser" That's not the question though. The question is over a 2 game series what's the chance of there being 2 ZvZs played given that at least one game was a ZvZ. Nope. The question is over a 2 game series what is the probability of getting zerg twice when playing as random, if at least one time you get zerg, and the order you play them in matters.
Uh that's the exact same thing as what I said except I worded it towards the example where he has NesTea playing TLO for 2 games.
The last part of your statement is not necessarily as well. It doesn't matter if you consider order important or not.
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On June 11 2011 05:34 Logo wrote:Show nested quote +On June 11 2011 05:32 Sasashi wrote:On June 11 2011 05:31 Logo wrote:On June 11 2011 05:27 Lomak wrote:On June 11 2011 05:23 DarkPlasmaBall wrote:On June 11 2011 05:18 Lomak wrote:+ Show Spoiler +Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever. He listed 5 pairs of games (not 6), and they aren't matchups. ZP =/= ZvP ZT =/= TZ, ZP =/= PZ because of the fact that they're pairs of games (getting Zerg then Terran, as opposed to Terran then Zerg). You clearly didn't even attempt to read the OP. You literally could not be more wrong with the brain teaser's information. Player 1 is Nestea (always playes zerg). So player 1 is constant as Z so the Zv? is the only question there is. IF one player is playing zerg and one is playing random, the chances of it being ZvZ are 1/3 not 1/5. Its not my fault this thread is full of condescending people bickering over a poorly written "brain teaser" That's not the question though. The question is over a 2 game series what's the chance of there being 2 ZvZs played given that at least one game was a ZvZ. Nope. The question is over a 2 game series what is the probability of getting zerg twice when playing as random, if at least one time you get zerg, and the order you play them in matters. Uh that's the exact same thing as what I said.
Not really, you got the same result, but the question wasn't about playing a specific match like zerg vs zerg. It was about playing zerg as a race twice while playing random. It's not the same thing, despite the same answer.
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Nevermind. Not worth the effort. Have fun.
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On June 11 2011 05:36 Sasashi wrote:Show nested quote +On June 11 2011 05:34 Logo wrote:On June 11 2011 05:32 Sasashi wrote:On June 11 2011 05:31 Logo wrote:On June 11 2011 05:27 Lomak wrote:On June 11 2011 05:23 DarkPlasmaBall wrote:On June 11 2011 05:18 Lomak wrote:+ Show Spoiler +Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever. He listed 5 pairs of games (not 6), and they aren't matchups. ZP =/= ZvP ZT =/= TZ, ZP =/= PZ because of the fact that they're pairs of games (getting Zerg then Terran, as opposed to Terran then Zerg). You clearly didn't even attempt to read the OP. You literally could not be more wrong with the brain teaser's information. Player 1 is Nestea (always playes zerg). So player 1 is constant as Z so the Zv? is the only question there is. IF one player is playing zerg and one is playing random, the chances of it being ZvZ are 1/3 not 1/5. Its not my fault this thread is full of condescending people bickering over a poorly written "brain teaser" That's not the question though. The question is over a 2 game series what's the chance of there being 2 ZvZs played given that at least one game was a ZvZ. Nope. The question is over a 2 game series what is the probability of getting zerg twice when playing as random, if at least one time you get zerg, and the order you play them in matters. Uh that's the exact same thing as what I said. Not really, you got the same result, but the question wasn't about playing a specific match like zerg vs zerg. It was about playing zerg as a race twice while playing random. It's not the same thing, despite the same answer.
It breaks down to the same probability question of P(A|B) where A = at least one game as a zerg and B = both games as zerg since one side of the match-up is constant in the OP's example.
On June 11 2011 05:38 Lomak wrote:Show nested quote +Let's say IMNestea (always Z) is playing against, say, TLO, and TLO is playing random. They're going to play exactly two games (for simplicity). Your friend spoils the fun by saying "OMG Nestea's ZvZ is absolutely sick!" (implying he saw a ZvZ game). How is anyone supposed to read that and assume you are asking about a RANDOM vs RANDOM situation when you clearly state that one player is "always z". Ridiculous how condescending people are being. "oh YOU just don't understand this question I'm posing even though I"M sabotaging it myself by making it obnoxiously confusing with examples that I gave.
None one is assuming that. NesTea is always Z AS IT SAYS. It's a 2 game series with 1 player (TLO) playing random both games. Honestly you aren't even trying to read the problem.
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The most disturbing thing about this is that this thread is 21 pages long. This is not ambiguous. It is not a paradox. It is possible counterintuitive, but the answer is unquestionably right - get over it and get on with your lives.
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well since 1 variable is fixed to zerg (Nestea) then it would come out to be 1/5 right?
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On June 11 2011 05:38 Logo wrote:Show nested quote +On June 11 2011 05:36 Sasashi wrote:On June 11 2011 05:34 Logo wrote:On June 11 2011 05:32 Sasashi wrote:On June 11 2011 05:31 Logo wrote:On June 11 2011 05:27 Lomak wrote:On June 11 2011 05:23 DarkPlasmaBall wrote:On June 11 2011 05:18 Lomak wrote:+ Show Spoiler +Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever. He listed 5 pairs of games (not 6), and they aren't matchups. ZP =/= ZvP ZT =/= TZ, ZP =/= PZ because of the fact that they're pairs of games (getting Zerg then Terran, as opposed to Terran then Zerg). You clearly didn't even attempt to read the OP. You literally could not be more wrong with the brain teaser's information. Player 1 is Nestea (always playes zerg). So player 1 is constant as Z so the Zv? is the only question there is. IF one player is playing zerg and one is playing random, the chances of it being ZvZ are 1/3 not 1/5. Its not my fault this thread is full of condescending people bickering over a poorly written "brain teaser" That's not the question though. The question is over a 2 game series what's the chance of there being 2 ZvZs played given that at least one game was a ZvZ. Nope. The question is over a 2 game series what is the probability of getting zerg twice when playing as random, if at least one time you get zerg, and the order you play them in matters. Uh that's the exact same thing as what I said. Not really, you got the same result, but the question wasn't about playing a specific match like zerg vs zerg. It was about playing zerg as a race twice while playing random. It's not the same thing, despite the same answer. It breaks down to the same probability question of P(A|B) where A = at least one game as a zerg and B = both games as zerg since one side of the match-up is constant in the OP's example.
Yeah I realize that, and stated that. Was just correcting you as the original question wasn't about matchups.
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On June 11 2011 05:41 Sasashi wrote:Show nested quote +On June 11 2011 05:38 Logo wrote:On June 11 2011 05:36 Sasashi wrote:On June 11 2011 05:34 Logo wrote:On June 11 2011 05:32 Sasashi wrote:On June 11 2011 05:31 Logo wrote:On June 11 2011 05:27 Lomak wrote:On June 11 2011 05:23 DarkPlasmaBall wrote:On June 11 2011 05:18 Lomak wrote:+ Show Spoiler +Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever. He listed 5 pairs of games (not 6), and they aren't matchups. ZP =/= ZvP ZT =/= TZ, ZP =/= PZ because of the fact that they're pairs of games (getting Zerg then Terran, as opposed to Terran then Zerg). You clearly didn't even attempt to read the OP. You literally could not be more wrong with the brain teaser's information. Player 1 is Nestea (always playes zerg). So player 1 is constant as Z so the Zv? is the only question there is. IF one player is playing zerg and one is playing random, the chances of it being ZvZ are 1/3 not 1/5. Its not my fault this thread is full of condescending people bickering over a poorly written "brain teaser" That's not the question though. The question is over a 2 game series what's the chance of there being 2 ZvZs played given that at least one game was a ZvZ. Nope. The question is over a 2 game series what is the probability of getting zerg twice when playing as random, if at least one time you get zerg, and the order you play them in matters. Uh that's the exact same thing as what I said. Not really, you got the same result, but the question wasn't about playing a specific match like zerg vs zerg. It was about playing zerg as a race twice while playing random. It's not the same thing, despite the same answer. It breaks down to the same probability question of P(A|B) where A = at least one game as a zerg and B = both games as zerg since one side of the match-up is constant in the OP's example. Yeah I realize that, and stated that. Was just correcting you as the original question wasn't about matchups.
Well yeah... sort of. The OP has posted 2 versions of the same question (after his edits) which is what's causing a lot of confusion.
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On June 11 2011 05:18 Lomak wrote:+ Show Spoiler +Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever.
Check that punctuation, one of those guys is a period!
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EPT
