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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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    dup: http://math.stackexchange.com/questions/44579/p-chance-of-winning-tennis-point-what-fp-chance-of-winning-game2012-06-22
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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.2012-06-22
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    @leonbloy It isn't a duplicate because the probabilities don't add up to 1. 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?2012-06-22
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    @Mark I supposed there was an implicit assumption that $A$ and $B$ had accumulated these records by playing against the same pool of opponents.2012-06-22
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    @MarkBennet: I would put that in as an answer. I think it is the best we will get.2012-06-22
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    @MarkDominus I gave an extreme example to show that the answer would depend on the assumptions. How often have they played each other?2012-06-22
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    @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.2012-06-22
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    @Mark (both) I agree that I was wrong, it's quite a different (and not well posed) question.2012-06-22
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    @MarkDominus: that would be true if they play some of the same people, but we aren't even given that.2012-06-22
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    @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.2012-06-22
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    Sorry to have caused so much confusion, I'm pretty new to this and still learning how to pose questions correctly. Part of the problem with not knowing how to figure something out is that I don't have the insight/experience to realise that the source of the records is important. I was indeed assuming that they had accumulated these records by playing against the same pool of opponets... should I edit the question to include this assumption? Many thanks for the Bradley-Terry model too, I had never heard of it. Sorry once again, and thanks to everybody that took time to respond to my question.2012-06-23
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    @MarkBennett Regarding your comment "How often have they played each other?" I now see this could be very important, are the resulting probabilities significantly different? Perhaps I should pose this as a separate question...2012-06-23

2 Answers 2

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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.2012-06-22
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    Sabermetricians (baseball statistics people) know the Bradley-Terry model as the "log5 model": http://www.chancesis.com/2010/10/03/the-origins-of-log5/2012-06-22
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    Let $\Lambda$ denote the skill of a player uniformly chosen at random. One would expect $9/10$ to be $E(\lambda_A/(\lambda_A+\Lambda))$ but you say this is $\lambda_A/(\lambda_A+E(\Lambda))$. These do not coincide. What makes it possible to use the latter?2012-06-22
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    @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.2012-06-22
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    It seems the data provided by the OP (namely, the overall proportions of wins by A and B) is not enough to compute the proportions of wins of A over B, even assuming the Bradley-Terry model holds. One should appreciate the ingeniosity displayed by the introduction of a parameter $\lambda_I$ to reach, nevertheless, a result, but, a definition of $\lambda_I$ being still elusive, one may have some doubts about the validity of this result. Assume for example exactly three players A, B and C, with respective skills $a$, $b$ and $c$. Then the system of .../...2012-06-23
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    .../... equations $\frac12\frac{a}{a+b}+\frac12\frac{a}{a+c}=\frac9{10}$, $\frac12\frac{b}{b+a}+\frac12\frac{b}{b+c}=\frac6{10}$ models **exactly** a large number of games between A, B and C, in equal proportions. The solution $a=4$, $b=1$, $c=0$ leads **entirely rigorously** to the probability $\frac{a}{a+b}=\frac45$ that A wins a game against B. Considering 4 players yields still other **exact** solutions, and probably whole families of these (including solutions where every skill is nonzero, in case this aspect of the 3-players solution is a trouble), .../...2012-06-23
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    .../... each leading **entirely rigorously** to still other values $\frac{a}{a+b}$. To escape this conundrum, one could wish to consider the mean-field limit, that is, infinitely many games between infinitely many players, whose skills would be distributed like some random variable $\Lambda$. But then, each distribution of $\Lambda$ would lead to a different solution. Furthermore, the value of $\frac{a}{a+b}$ might become difficult to compute, except numerically, hence this route is not very engaging either.2012-06-23
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    With 4 players A, B, C, D of respective skills 17.3893, 4.1119, 1, and 1, A wins 90% of her games, B wins 60% of her games and A wins 80.26% of her games against B. But with players A, B and infinitely many identical other players of respective skills 54 (for A), 9 (for B) and 6 (for every other player), A wins 90% of her games, B wins 60% of her games and A wins 6/7=85.7% of her games against B (this is the approximation in the post above).2012-06-23
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    P(A beats I)=9/10=λA/λA+λI seems to imply λA=9 and P(B beats I)=6/10=λB/λB+λI seems to imply that λB=6 so I am confused how P(A beats B)=λA/λA+λB=6/7 and not 9/152012-06-23
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    @pglove I am not sure why you think $\lambda_A/(\lambda_A + \lambda_A) = 9/10$ implies $\lambda_A = 9$. The first equation has two variables and the second has only one; how did you get rid of $\lambda_I$?2012-06-23
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    @MarkDominus The first equation states P(A beats I) = 9/10 = λA/(λA+λI) this seems to say that the top of the equation, λA is 9 and the bottom, λA+λI is 10 ... and that λI would therefore be equal to 1. Similarly, for P(B beats I) = 6/10 = λB/(λB+λI) it looks like λB equals 6 and λI equals 4. I suspect I am being incredibly stupid.2012-06-23
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    @did My analysis only makes sense if $A$ and $B$ have played have played against similarly skillful opponents. We've already discussed the fact that one can't compare $A$ and $B$ at all in the absence of this assumption. But in your example, $B$'s opponents have *not* been similarly skillful; they have been much more skillful than $A$'s opponents overall. Since $B$ had better opponents than $A$, your analysis underestimates $B$'s skill compared to $A$ and therefore $B$'s chances of beating $A$.2012-06-23
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    @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.2012-06-23
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    In other words, you do not want to take into account the games between A and B when evaluating the proportions of games won by A and by B. This seems illogical but OK, let us do that. Thus, consider again 4 players A, B, C, D, this time with respective skills 18, 3, 2, 2. Then, A wins 90% of her games, B wins 60% of her games, and A wins 6/7 of her games against B, just as in .../...2012-06-23
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    .../... 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).2012-06-23
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    @MarkDominus Thanks Mark, now I understand that the first does not imply λA=9 and the second does not imply λB=6. Thanks for the clear explanation. Am I right in my new working; λA/λB=6 tells me that λA=6λB and so λA/(λA+λB) is the same as writing 6λB/(6λB+λB) = 6/7 ?2012-06-24
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    @pglove: Yes, that looks good.2012-06-24
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    If somebody wants a condensed, final formula of this answer's model, it is: Probability of A winning, `Pa = (A - AB)/(A + B - 2AB)`, where A is win % of first player, B is win % of second player.2014-10-17
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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.2012-06-22