Probability Question

My thinking:

if $p_{ij}$ is the chance of horse i finishing in position j, you require

All horses finish in some position:
$\sum_j p_{ij} = 1 ~~~~ \forall i$

All positions are occupied by a horse
$\sum_i p_{ij} = 1 ~~~~ \forall j$

All probabilities have sensible values
$0 \leq p_{ij} \leq 1 ~~~~ \forall i,j$

Suppose that you have a solution $p*$ which satisifes the conditions. Is it unique?

it's easy (ish) to show that it is not unique for the case where any horse can finish in any position (Pij >0). You didn't specify this was the case but it seems pretty sensible given the problem you are describing.

Spoiler:

Write $p*$ as a matrix (i in rows, j in columns). The probabilities you gave for each horse winning make up the 1st column.

Note that the constraints mean that every row and column in the matrix must sum to 1. All we need to do is change some terms in the matrix while preserving the column/row totals and keeping all matrix values in the range [0,1]

http://quicklatex.com/cache3/ql_2447...61a5687_l3.png

consider the 2x2 square in the middle:
http://quicklatex.com/cache3/ql_dd2b...4933795_l3.png

None of the (unknown) values in P* can be 1, because this would make the row total be greater than 1. Now use the assumption that any horse can finish in any position ( $P_{ij}>0)$ . Combining these results, we can replace our 3rd constraint with strict inequalities:

$0 < p_{ij} < 1 ~~~~ \forall i,j~~~~~~~~ [A]$

Let $\epsilon$ be the half of the smallest nonzero number for which $0 \leq p_{ij} \pm \epsilon \leq 1$ . Hence $0 < p_{ij} \pm \epsilon < 1$ . Epsilon is always nonzero due to the strict inequality [A].

Now change the values of the middle square as follows:
http://quicklatex.com/cache3/ql_3026...18067e6_l3.png

The column and row totals are preserved within the middle square, and hence for the whole matrix. All values are still in the range (0,1) because of the definition of $\epsilon$ .

Since more than one set of probabilities satisfy your contraints, you cant deduce what the real probabilities are.

It'll be more tricky to prove if you cant assume that any horse can finish in any position. You can generalise slightly: This method would still work for the case where at least two horses can finish in at least 2 overlapping positions (without loss of generality, let those horses and positions be 3 and 4, then do almost the same steps again.).

Ineresting problem! Cant promise i got it right though.