I agree with @whuber's comment on the original post. However, we can ask what additional assumptions, with the given information, could lead to an estimate.
Additional Assumptions:
- The stated win probabilities are good estimates of the outcome of a game for each team.
- The teams are playing two independent games
- The desired win probabilities are conditional on team A exclusively-or team B winning.
Essentially this is turning the fact that they are playing each other (where we don't have enough information) into a different situation with a result that can be estimated.
- $P(A) = 0.6$
- $P(B) = 0.9$
- $P(A \cap \bar{B}|A\ \mathbf{xor}\ B) = \frac{P(A)P(\bar{B})}{P(A)P(\bar{B}) + P(\bar{A})P(B)} = 0.143$
- $P(B \cap \bar{A}|A\ \mathbf{xor}\ B) = \frac{P(B)P(\bar{A})}{P(A)P(\bar{B}) + P(\bar{A})P(B)} = 0.857$
Why Condition on XOR?
If you think about the truth table for two independent events,
| A | B | P if A,B independent | Allowed in a head-to-head game? | A xor B |
|---|---|---|---|---|
| W | W | 0.54 | N | F |
| W | L | 0.06 | Y | T |
| L | W | 0.36 | Y | T |
| L | L | 0.04 | N | F |
How is the conditional probability constructed?
$$P(A \cap \bar{B}|A\ \mathbf{xor}\ B) = \frac{P((A \cap \bar{B}) \cap (A\ \mathbf{xor}\ B))}{P(A\ \mathbf{xor}\ B)} = \frac{P(A \cap \bar{B})}{P(A\ \mathbf{xor}\ B)} = \frac{P(A)P(\bar{B})}{P(A)P(\bar{B}) + P(\bar{A})P(B)}$$