Calculating Probability of a horse winning based on expected position

Question

My question is this:

If you have 4 horses (A , B , C and D ) in a race:

• Horse A is expected to get position 2
• Horse B is expected to get position 2
• Horse C is expected to get position 3
• Horse D is expected to get position 3

Or any distribution of the positions which sums 10.

Is there any formula or procedure to obtain the probability of each horse winning?

May someone please help! Many thanks Sergio.

EDIT: Edited positions after joriki clarification.

Answer 1

This is impossible. By linearity of expectation, the expected value of the sum of the positions would have to be $2+2.5+3+3.5=11$, but the sum of the positions is the constant $10$, independent of the outcome of the race.

Answer to the edited question:

Those expected positions are possible, but they don’t fix the winning probabilities. For instance, those expected positions would result if the results ABCD and ABDC each have probability $\frac3{10}$ and the results CBAD, CBDA, DBCAand DBAC each have probability $\frac1{10}$, or if the results ACBD and BDAC each have probability $\frac12$; but in the first case C and D have non-zero probability to win the race and in the second case they don't. The distribution of the results has $24-1=23$ degrees of freedom and the set of expected positions only has $4-1=3$, so you can't fix the former by fixing the latter.