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I am playing in a tennis tournament, and I am up against a player I have watched but never played before. Based on what I have seen, I consider three possible models for our relative strengths:

• Model A: We are evenly matched, so that each of us is equally likely to win each game.

• Model B: I am slightly better, so that I win each game independently with probability 0.6.

• Model C: My opponent is slightly better and wins each game independently with probability 0.6.

Before we play, I consider each of these possibilities to be equally likely. In our match, we play until one player wins three games. I win the second game, but my opponent wins the first, third and fourth games. After the match, with what probability should I believe in model C (i.e., that my opponent is slightly better than me)?

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    Since neither has any consideration for serve and return games I wouldn't believe in either.2011-11-12

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As cardinal says this can be answered using Bayesian techniques, if you have probabilities for the three hypotheses before the games, say $\Pr(A)=a$, $\Pr(B)=b$ and $\Pr(C)=c$ with $a+b+c=1$.

Then your answer is $\Pr(C|G)=\frac{\Pr(G|C)\Pr(C)}{\Pr(G|A)\Pr(A)+\Pr(G|B)\Pr(B)+\Pr(G|C)\Pr(C)} $ and in this case $\frac{0.6\times 0.4\times 0.6\times 0.6\times c}{0.4\times 0.6\times 0.4\times 0.4\times a+0.5\times 0.5\times 0.5\times 0.5\times b+0.6\times 0.4\times 0.6\times 0.6\times c}.$

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    @John.Mathew In this instance, since $a=b=c=\frac{1}{3}$, the maximum-likelihood decision rule that I suggested in my comment on your question gives the same _decision_ as the Bayesian decision rule explained in detail by Henry, though, of course, frequentists cannot calculate the probability of $C$ given the outcome.2011-11-12