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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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    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.2012-12-14

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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 $0.69$.