a little help with this probability problem

[jba]

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!

[rufus]

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.

[BruceZ]

[ 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).

[jay_shark]

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.

[jba]

[ 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..