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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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If you are a Bayesian, this question makes sense. If you are of a more frequentist bent, then it doesn't. In the latter case, either (a) one of the models is true or (b) none of them are. No probability is involved. Are you looking for a Bayesian solution? –  cardinal Nov 12 '11 at 1:47
    
Calculate the likelihood of the outcome LWLL under the three models, and choose the model which gives the largest likelihood of the observed result. But keep in mind that in reality there are several other matters to keep in mind, e.g. you are not as fit as your opponent and so your losses in the third and fourth game are attributable in part to the fact you are tired, and thus the independent trials model does not quite hold. –  Dilip Sarwate Nov 12 '11 at 1:54
    
@cardinal Yes, that's exactly what I am thinking about –  geraldgreen Nov 12 '11 at 6:04
1  
Since neither has any consideration for serve and return games I wouldn't believe in either. –  scineram Nov 12 '11 at 7:31

1 Answer 1

up vote 3 down vote accepted

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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shouldn't the second variable in the denominator be b instead of c? –  geraldgreen Nov 12 '11 at 6:06
    
yes it should - thank you –  Henry Nov 12 '11 at 13:18
    
@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. –  Dilip Sarwate Nov 12 '11 at 17:49

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