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I have been interested in calculating a specific horse's chance of finishing in nth place given every horse's chance of winning in a particular race.

i.e. Given the following:

Horse    Chance of winning
A        0.35
B        0.25
C        0.15
D        0.10
E        0.09
F        0.05
G        0.01

Calculate for any horse it's chance of finishing in nth place. For instance, calculate the chance of HorseC finishing in 2nd place, or calculate the chance of HorseB finishing in 3rd place.

I thought that this is something I could get help with online. All of the journal articles I found discuss chances of a pair of horses winning, i.e. the chance of horse A winning and horse B coming second, which is obviously different to this question.

This: http://forum.sbrforum.com/handicapper-think-tank/526381-win-v-place-odds-value-math-question.html#post5076725 is the closest thing I have found to what I am looking for, but I believe he assumes that given one horse wins, the others have an equal chance of placing second, which is clearly not the case.

UPDATE

I just had time to have a go at this and I am trying to come up with a formula for P(i, n) as @ThanosDarkadakis suggested.

I am unsure of whether the odds of HorseZ finishing 3rd is:

sum of HorseXWin * HorseY2nd * (HorseZWin/(1-HorseXWin-HorseY2nd)), for each X/Y

or

sum of HorseXWin * HorseY2nd * (HorseZWin/(sum of remaining win probabilities)), for each X/Y

or

sum of HorseXWin * HorseYWin * (HorseZWin/(1-HorseXWin-HorseYWin)), for each X/Y

Where HorseX is the winner of the race, HorseY comes 2nd and HorseZ comes 3rd (for each X/Y).

I'm sure that with a formula for P(i, 3) it would be trivial to write a formula for P(i, n). Any suggestions are greatly appreciated.

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    $\begingroup$ You need to know or assume more than you have here, and in particular (a) the distribution of each horse's time or speed and (b) whether the times or speeds of each horse are independent of those of the other horses $\endgroup$
    – Henry
    Commented Jan 3, 2014 at 7:11
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    $\begingroup$ about the update: it's the first one. still it's not that easy to get the result. because this formula is a SUM, and it contains HorseY2nd, which is also a SUM. $\endgroup$
    – Thanos Darkadakis
    Commented Jan 17, 2014 at 8:25

2 Answers 2

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First of all lets assume that if horseD has twice the chances of horseF to finish 1st in a race of 7 horses, then it also has twice the chances to finish first when there are 100 horses or 2 horses or n horses. (It is not defined whether this is true or not. I'll continue assuming this.)

Let's take your example: "calculate the chance of HorseC finishing in 2nd place"

If horseA finishes 1st (0.35), then the probability of horseC finishing 2nd (1st amongst the others) is $\frac{0.15}{0.25+0.15+0.10+0.09+0.05+0.01}=\frac{0.15}{1-0.35}$. So this probability is $0.35\frac{0.15}{1-0.35}$.

Likewise if horseB finishes 1st, then the probability of horseC finishing 2nd is $0.25\frac{0.15}{1-0.25}=P_B\frac{P_C}{1-P_B}$.

The total probability of horse i to finish 2nd is the sum of these:

$P(i,2)=\sum_{i\neq j}{P_j\frac{P_i}{1-P_j}}$.

Can you continue in order to find a general formula for $P(i,n)$?

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  • $\begingroup$ Thanks for your answer, I only just got around to looking at this. Please see the update. $\endgroup$
    – janderson
    Commented Jan 16, 2014 at 13:58
  • $\begingroup$ This is the Harville model. See my answer below for a discussion of shortcomings. $\endgroup$
    – Peter Cotton
    Commented Nov 4, 2021 at 13:23
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I recently published an article in the SIAM Journal on Quantitative Finance that addresses this issue, and also provides a discussion of why other answers in this thread, past, present, and perhaps future, are likely to be inaccurate. Those answers typically assume Luce's axiom of choice and thereby re-invent the formula named for Harville. Unfortunately there isn't a dreadfully compelling theoretical or empirical reason to believe that the conditional probability of a horse finishing second should be equal to its renormalized probability of winning (re-normalized by removal of the winner, of course).

In the paper I instead consider the general problem of calibrating a model for performance to winning probabilities. In a continuous horse race problem we might assumes a density $f^*$ with distribution $F^*$ and then seek parameters $(a_1,\dots,a_n)$ modifying $f^*$ in some way (typically by scale or location) to satisfy \begin{equation} \label{eqn:continuous} p_i = \int f^*(x;a_i) \Pi_{j\neq i}^n \left( 1- F^*(x;a_j)\right) dx \end{equation} for some specified winning probabilities $p_i$. I pay particular attention to the case where each horse's performance distribution is a translation of every other. I provide a fast numerical algorithm and open source code solving this inversion problem.

Once one has backed into a reasonable performance model, all other quantities can be computed including the odds of finishing 2nd and 3rd. You can find more details in the paper and I'm happy to copy more here if that's requested. However, I suspect you'll want the code and the README explains usage.

The result is a performance model like the following:

enter image description here

By the way I'm editing this after several years so I'll leave here the fact that this follows on from an earlier discussion of the general case at http://finmathblog.blogspot.com/2013/09/the-horse-race-problem-general-solution.html

Update 2021. I'm now running a live contest to see if there are better ways. See What is the Probability a Horse Finishes Last?

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