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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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    @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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    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