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?

Although there is no unique solution, I think useful answers can be derived by searching for the maxiumum entropy propability distribution :-)
https://en.wikipedia.org/wiki/Maximum_entropy_probability_distribution
