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
 A: 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.
A: 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?
