A combinatoric/probability puzzle A horse race has four horses A, B, C and D. The probabilities for each horse to finish first or second are P(A)=80%, P(B)=60%, P(C)=40% and P(D)=20%. The probabilities add up to 200% because one horse will finish first with 100% probability and one horse will finish second with 100% probability. What are the probabilities of all possible six combinations of results for first and second place, i.e. * Horse A and horse B in 1st and 2nd place? * Horse A and horse C in 1st and 2nd place? * Horse A and horse D in 1st and 2nd place? * Horse B and horse C in 1st and 2nd place? * Horse B and horse D in 1st and 2nd place? * Horse C and horse D in 1st and 2nd place? I've got a little monte carlo simulation that tells me the answer, but surely there's a simple analytic formula?

There is no unique solution.

For example:

* $P(\\{A,B\\})=60\%$, $P(\\{A,C\\})=20\%$, $P(\\{C,D\\})=20\%$
* $P(\\{A,B\\})=40\%$, $P(\\{A,C\\})=40\%$, $P(\\{B,D\\})=20\%$



each meet the requirements of the question