# 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?

• Are A,B,C and D independent?
– GBQT
Oct 9 '15 at 9:48
• @GBQT: since exactly two horses must come "in 1st and 2nd place", independence seems unlikely. Oct 9 '15 at 9:57
• Since I posted this, I've realised that my monte carlo simulation is flawed! Oct 9 '15 at 10:12

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

• And indeed, my simulation turned out to be flawed :-(. Oh well! Oct 9 '15 at 10:13

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