I've taken a further look and I can account for the discrepancy in this special case (where the probability between horses is 1/2). This results in the three probabilities adding up to 1 rather than 0.75.
One can look at the fact that A has beaten B (a probability of 1/2). The three cases where this is true are finishing orders ABC, ACB and CAB. Only in two thirds of these is A the winner, so the probability of A winning is 1/2 x 2/3 = 1/3 which we can see is the correct result using common sense. the problem I have is when the probability of A beating C or B beating C is not 1/2.
Any guidance as to how to account for this would be helpful. Or am I heading down a dead end?