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Say you have two people playing first to 11 points for a game of whatever. A has a 0.55 chance to get a point and B has a 0.45 chance. How would I go about working out the probability that A will win the match?

The way I think you'd work it out is by running it say, 10000 times and then do :

(number of games A wins) / (total number of games). Is this correct?

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possible duplicate of If a player is 50% as good as I am at a game, how many games will it be before she finally wins one game? (The question is phrased a bit differently, but answering it also requires working out the probability you want, and this is what the answers do.) –  joriki Dec 14 '12 at 9:31
    
Thanks. Yeah, setting the probability to win at 50% each came out with them winning 50% of the time. –  Arktri Dec 14 '12 at 9:36
    
That's a misunderstanding. That somewhat ambiguous title is intended to mean that one player wins $2$ out of $3$ games. Anyway, the precise probability isn't important; the answers can readily be adapted to an arbitrary probability. –  joriki Dec 14 '12 at 9:50

1 Answer 1

Your simulation would provide a reasonable approximation.

You could get a more direct result by looking at playing 21 points and seeing if A wins 11 points or more. Truncating the game when A or B have 11 will not affaect this probability. You can use the binomial distribution

So you want to know $$\sum_{i=11}^{21} {21 \choose i} p^i (1-p)^{21-i}$$ where $p=0.55$. This gives about $0.679$ while your simulation is likely to give an answer between $0.67$ and $0.69$.

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