#1
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a little help with this probability problem
ok it's actually two problems.
two players are playing a game. They will win with frequency X and Y. What is the probability after N games that player A will win P more games than player B? for example: player A will win 60% of the time player B will win 55% of the time what is the probability that B will win 5 more games than A after they each player 30? The first question is how to calculate this if they are independent games (ie, player A and B can both win or lose any given game). The second question is how to calculate if they are dependent - ie, player A wins 60% and player B wins 40%, and if player A wins player B loses by definition and vice versa. thanks! |
#2
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Re: a little help with this probability problem
The second one is less work than the first, since it's just asking how often A will win 1/2 or more of the games, but AFAIK there's no closed form for it. You have to take the sum of the first k/2 terms of the binomial expansion of (X+Y)^k.
For the first question, you can generate the probabilities for A and B to win exact numbers of games, and then do summations. |
#3
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Re: a little help with this probability problem
[ QUOTE ]
two players are playing a game. They will win with frequency X and Y. What is the probability after N games that player A will win P more games than player B? The first question is how to calculate this if they are independent games (ie, player A and B can both win or lose any given game). [/ QUOTE ] I'll assume that you mean exactly P more games, not at least P more games. I'll also assume that each player plays N games. Then the possibilities are (A wins P, B wins 0), (A wins P+1, B wins 1) ... (A wins N, B wins N-P). We must sum the probabilities of each of these as follows: sum[k = P to N] { C(N,k) * X^k * (1-X)^(N-k) * C(N,k-P) * Y^(k-P) * (1-Y)^(N-k+P) } [ QUOTE ] The second question is how to calculate if they are dependent. [/ QUOTE ] Let A = number of games won by player A, and let B = number of games won by player B. Then we have: A + B = N A - B = P So A = (N+P)/2, and B = (N-P)/2. Note that A and B must both be odd or both be even. So your example of winning exactly 5 more games out of 30 is impossible. The probability that A wins exactly this many games is just C(N,A) * X^A * (1-X)^(1-A). |
#4
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Re: a little help with this probability problem
1)B may win 5 games and A wins 0
B may win 6 games and A wins 1 . . . . B wins k games and A wins k-5 . 30ck.55^k*.45^(30-k)*30C(k-5).6^(k-5)*0.4^(35-k){sum from k=5,k=30} 2)If A wins x% of the time and B wins (1-x)% of the time , then then the probability B wins 5 more games than A after 30 games is 0 [img]/images/graemlins/smile.gif[/img] . This one is easy since x+5+x=30 2x=25 x=12.5 which is a contradiction since x must be an integer . So we only have to consider when A or B wins by an even amount of games when n is even and vice versa . The probability A(60%) wins 4 more games than B(40%) is : 30c17*0.6^17*0.4^13. |
#5
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Re: a little help with this probability problem
[ QUOTE ]
I'll assume that you mean exactly P more games, not at least P more games. [/ QUOTE ] sorry, I am looking for *at least* P more games. I was hoping there was something more clever than a summation for the first one but I was afraid not.. |
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