Write $\displaystyle p*$ 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 ($\displaystyle P_{ij}>0)$. Combining these results, we can replace our 3rd constraint with strict inequalities:

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

Let $\displaystyle \epsilon$ be the

**half of** the smallest nonzero number for which $\displaystyle 0 \leq p_{ij} \pm \epsilon \leq 1$. Hence $\displaystyle 0 < p_{ij} \pm \epsilon < 1$. 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 $\displaystyle \epsilon$.

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.).