[Math] Calculating probabilities in horse racing!

probabilitypuzzle

I've seen a few similar threads to this on different forums but they don't seem to conclude to a satisfactory answer. My question is this:

If you have 3 horses, A, B, and C and you know the winning probabilities of each horse racing against each other as a pair, how do you work out their winning probabilities if all 3 horses race together? Is there a formula for n number of horses?

So you know the probabilities of:
A beating B
A beating C
BA
BC
CA
CB

I initially thought that: p(A winning) = p(A beat B). p(A beat C)
But this is clearly wrong.

Can someone please help!!
Many thanks
Matt

Best Answer

Expanding on Tony K's comment, here's an (extreme) example of why it's impossible.

Let's say that we have three very strange horses:

  • Horse A runs the race in $1$ second with probability $1/2$, and in $9$ seconds with probability $1/2$.
  • Horse B runs the race in $3$ seconds with probability $1/2$, and in $7$ seconds with probability $1/2$.
  • Horse C always runs the race in $5$ seconds.

Then it's not hard to check that, in head-to-head races, $P(A \text{ beats } B) = P(A \text{ beats } C) = P(B \text{ beats } C) = 1/2$.

In a three-way race, A will win whenever it runs fast, so $P(A \text{ wins})=1/2$. If A doesn't run fast, B will win if it runs fast, and C will win if B runs slowly. So $P(B \text{ wins})=P(C \text{ wins})=1/4$.

On the other hand, let's say we have three more normal horses; in fact, horses D, E, and F are so ordinary that they're identical to each other in every way. Then it must be true that $P(D \text{ beats } E) = P(D \text{ beats } F) = P(E \text{ beats } F) = 1/2$ again. But in this case, each horse must have a $1/3$ probability of winning the entire race.

That is, we've found two sets of three horses such that:

  • The head-to-head probabilities are identical between them.
  • The probabilities of winning a 3-way race are different.

So there can't possibly be any way of calculating the 3-way probability, given only knowledge of the head-to-head probabilities.

EDIT:

The problem is that the three-way win probabilities depend not just on the average abilities of the horses, but also on how reliable they are. A horse that's very unreliable (sometimes runs really fast and sometimes runs really slow) will generally do better in a 3-way race than in a 2-way race, because if they run really fast against one of their opponents they'll run really fast against both of them! Notice that my three horses all finish the course in the same amount of time on average, but horse A still wins out by being unreliable.

In short, if you want to calculate 3-way win probabilities, then you'll need to either know or assume something about how a horse's performance varies, as well as about its average performance. There's no way to get this data from 2-way win probabilities.

Related Question