# Tennis Probability

Player A wins 90% of matches

Player B wins 60% of matches

If they play each other what is the probability that;

Player A will win

Player B will win

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@leonbloy I don't think that is at all the same question; that is much more straightforward. Games are made up of points. This question asks how to compare two players based on their previous records playing other people. – MJD Jun 22 '12 at 17:36
@Mark I supposed there was an implicit assumption that $A$ and $B$ had accumulated these records by playing against the same pool of opponents. – MJD Jun 22 '12 at 17:39
@Ross I think we can do better. This sort of problem is well-studied in sports statistics. For example, football teams $A$ and $B$ have the known won-lost records at the end of the season; whom do we expect to win when they are matched up in a postseason game? Someone who knows something about it could at the very least cite the existing literature. – MJD Jun 22 '12 at 17:42
@Mark (both) I agree that I was wrong, it's quite a different (and not well posed) question. – leonbloy Jun 22 '12 at 17:49
@ross As I said, I supposed there was an implicit assumption that $A$ and $B$ had accumulated these records by playing against the same pool of opponents. Otherwise Mark Bennett's trivial objection holds. But why interpret the question in the least interesting way possible when there is a reasonable question lurking inside? Presumably OP is not interested in comparing Roger Federer against Joe Schlobb based only on their win-lose records. – MJD Jun 22 '12 at 17:59

The widely-studied Bradley-Terry model assumes that there is some underlying parameter $\lambda_i$ for each player that measures their skill or intrinsic quality, and that the $\lambda_i$ parameters can be used to rank the players and to predict the likelihood that one player beats another according to the following formula:

$$P(i\ \mathrm{beats}\ j) = {\lambda_i\over \lambda_i + \lambda_j}$$

Here we suppose that $A$ and $B$ have both played a large number of games against the same pool of other players. (Without this assumption, as others have observed, the question does not really make sense.) Let $\lambda_I$ be the skill of a hypothetical "typical" player $I$ who is beaten by $A$ 0.9 of the time and by $B$ 0.6 of the time. Then we have

$$P(A\ \mathrm{beats}\ I) = \frac9{10} = {\lambda_A\over \lambda_A + \lambda_I} \\ P(B\ \mathrm{beats}\ I) = \frac6{10} = {\lambda_B\over \lambda_B + \lambda_I}$$

We want to calculate $P(A\ \mathrm{beats}\ B)$. From the two equations above, we get:

$${\frac{\lambda_I}{\lambda_B} = \frac46}\qquad {\frac{\lambda_I}{\lambda_A} = \frac19}$$

Then, dividing one by the other, $${\lambda_A\over\lambda_B} = 6,$$ and finally:

$$P(A\ \mathrm{beats}\ B) = {\lambda_A\over \lambda_A + \lambda_B} = \frac67$$

For a probability of about 86%; the probability that player $B$ will win is then about 100%-86%=14%.

This is a reasonable result: since $B$ is only a little bit above average, $B$ is beaten by $A$ just a little bit less than the average player is.

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An interesting addendum, not directly related to the original question: It has been empirically known for some time that you can predict a baseball team's win-lose record quite accurately by calculating $s^k / (s^k + a^k)$, where $s$ and $a$ are their runs scored and runs allowed, respectively, and $k$ is a constant that is approximately 2, now known to be closer to 1.83; in basketball the $k$ is 14 when $s$ and $a$ are points scored and points allowed. – MJD Jun 22 '12 at 18:11
Sabermetricians (baseball statistics people) know the Bradley-Terry model as the "log5 model": chancesis.com/2010/10/03/the-origins-of-log5 – Michael Lugo Jun 22 '12 at 18:35
@did: $\lambda_I$ is not intended to be $E(\Lambda)$. It's not the mean skill of the opponents, but rather the skill of an abstract "typical" opponent whom $A$ can beat 0.9 of the time and $B$ can beat 0.6 of the time. There might need to be an argument that such $\lambda_I$ exists, but the problem has enough degrees of freedom that that should be straightforward. I think this makes sense, but if it still sounds wrong to you, I would be grateful for any help you could render. – MJD Jun 22 '12 at 20:46
@pglove: $\def\la{\lambda_A}\def\li{\lambda_I} \la/(\la + \li) = 9/10$ doesn't necessarily mean that $\la=9$ and $\li = 1$. There are many possibilities; for example, $\la=18$ and $\li=2$ also works just as well. In fact this equation is not enough to pin down either $\la$ or $\li$. But it is enough to pin down $\la / \li$, which is 9. Similarly the other equation, $\lambda_B/(\lambda_B + \li) = 6/10$ does not determine either $\lambda_B$ or $\li$ individually. But the two equations together allow us to compare $\la$ and $\lambda_B$ even though we don't know either one separately. – MJD Jun 23 '12 at 17:21
.../... your answer. But for skills 90, 9, 1 and 21, A wins 90% of her games, B wins 60% of her games, and A wins 10/11 of her games against B! And in fact, with 4 players and for some suitable skills, A may win against B any proportion of games between 6/7=85.7% and 16/17=94.1% (and if one allows for more players than 4, the range of possible proportions widens). – Did Jun 23 '12 at 19:27

From comment.

The answer is indeterminate. If player A is the best in the world and player B is a decent 15-year-old playing for school teams, for example ... . Or player A could be a school champion, and player B a decent club player. Who knows?

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What we do know is that they don't spend all their time playing each other. – Mark Bennet Jun 22 '12 at 17:42