Thread: Match odds versus overall odds of winning in horse racing

1. Match odds versus overall odds of winning in horse racing

The problem I have is as follows. Take a horse race with multiple entrants.

Assume that I can accurately assess the probability of horse X beating horse Y for all combinations within the race. I’m doing this using a chess-based Elo rating.

What I need to do is calculate the probability in a race with N horses that horse A will win.

Let’s write the probability of A beating B overall as pAB. As stated above, assume I know every single combination pXY for all combination of horses XY.
My initial solution was that to calculate the probability of horse A winning, all I needed to do was multiply pAB x pAC x pAD ... pAN, as A must win each contest. However, a quick sense check says this can’t be so.

The sense check involved looking at a race with three horses, each of which has a 0.5 probability of beating the others. In that case, my maths outlined above says that the probability of A winning would be 0.5 x 0.5 = 0.25. Do that for all three horses and it would suggest the total probability of one of the horses A,B or C winning would be 0.75 – clearly nonsensical, as it must be 1.00.

I’m utterly stuck. I think I might be able manually to calculate the true probability in this case by simply dividing each result by the total, i.e.1/4 divided by 3/4 = 1/3, but I doubt this would work when each horse doesn't have an equal chace of winning.

I’m not necessarily looking for the complete solution, just a direction I can head to get towards the answer (though I will, of course, accept the answer if anyone has it).

I hope I’ve explained this clearly; if not, please do feel free to ask for clarification.

Thanks,

Onionman

2. Further info

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?

Onionman

3. well, regarding your second "further info", there are six combinations with three horses in the race:
ABC, ACB
BAC, BCA
CAB, CBA

to calculate the probability of A winning the race, you need to add, not multiply:
pABC=1/6,pACB=1/6 so pA winning = 1/6+1/6 = 1/3

but this is a long way to achieve a simple result.
start the problem again, there are three horses, so only three possibilities
either A can win
or B can win
or C can win
so, to calculate pA, it is 1/3 (simple!)

similarly, to calculate pA with N horses, i think the answer will be 1/N

4. Bourne,

First of all, thanks for the response. I can see exactly what you're saying. At the same time, I need to stress that the special case where each horse has a 1/2 probability of beating the others was only being used to prove that the initial solution I came up with was flawed. It's not the problem I'm trying to solve.

What I'm trying to find is a general solution for any probabilities of pairs beating one another, so, for example, in an N horse race where the probabilities between any pair of horses can be any number.

Onionman

5. The reason you can't do it the way you describe is because the pairs of outcomes are not independent in a race with more than 2 horces. Let's stick with the simplified case for a bit--3 horses, 1/2 chance for each in a two horse race. But now suppose A beats B and B beats C. The chance that A beats C isn't 1/2 anymore, it's 1. So p(A>C) can't be independent of p(A>B) and p(B>C). If the outcomes aren't independent, you can't just multiply the probabilities.

Horse racing is different from chess--in a round robin chess tournament everone plays each other and then you have final standings, but it's certainly possible that A beats B, B beats C, and C beats A. This can't happen in horse racing.

I suspect that just knowing the probability of victory over each individual horse is insufficient. Here's why:

What's happening in a horse race is that each horse's time is a realization of a random variable, and each horse has a different distribution. For a pair of horses with any two distributions, you can compute the probability that one will have a faster time. But to compute overall winning probabilities for more than two at a time, you need to know the underlying distributions.
Here's a case that might make this clear:

Suppose each of three horses' times is distributed uniformly between 80 seconds and 85 seconds. They're evenly matched, so the probability that one will beat another is 1/2 and the chance that one will win is 1/3.

Now suppose two of the horses have the same distribution but the third runs half his races in 70 seconds and the other half in 90. The chance that any horse will beat any other is still 1/2, but now the third horse wins the whole race 1/2 the time, not 1/3 of the time. (He wins whenever he runs in 70s.) Think through some scenarios here to make sure you get this.

So we have two cases that look identical in terms of the one-on-one probabilities, but are very different in the overall probabilities, and the only way to tell the difference is by knowing the underlying distribution.

6. Ahhh, that's much clearer.

Several words come to mind, though none of them are really suitable for a nice, polite forum like this. Looks like I'm unlikely to come up with a formula to resolve this. that was a beautiful explanation of why not.

Thanks for all your help, chaps.

O

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