# If a horse has 50% chance to win a 10 runner race, what is the chance to finish in Top 3?

There are 10 horses running! If a horse has 50% chance to win, what is the chance of that same horse to finish in Top 3?

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Not possible to determine given the information provided. – Thomas Andrews Jul 9 '12 at 18:48
It is possible to determine to a degree: it is between 50% and 100%. :) – tomasz Jul 9 '12 at 19:04
If you specify his odds of individually beating each of the other 9 horses a paired comparison model (e.g. Bradley-Terry) could be used to estimate the probability of ranking in the top 3. – Michael Chernick Jul 9 '12 at 20:04
This question is underspecified but I think it's valuable to try to think about how one could come up with an answer, as Gerry Myerson has done. – Michael Lugo Jul 10 '12 at 2:15
In horse racing, odds are set by parimutuel betting which is a market-based voting system. - There is no "event algebra" – alancalvitti Jul 10 '12 at 13:58

Here's a somewhat plausible model. Suppose we have independent exponential random variables $X_j$ corresponding to the horses, each with its rate $r_j$. The order of finishing is the order of the $X_j$. The "lack of memory" property of the exponential distribution makes calculations easy. When the horses with rates $r_{j_1}, \ldots, r_{j_k}$ have not yet finished, the probability that a given one (say $j_1$) is the next to finish is $r_{j_1}/\sum_{k} r_{j_k}$. For simplicity, let's say our horse has rate $r_1$ and the other $9$ horses all have rate $r_2$. The probability that our horse wins is $r_1/(r_1 + 9 r_2)$, so if this is $1/2$ we must have $r_1 = 9 r_2$. Given that our horse doesn't come first, the probability that it comes second is then $r_1/(r_1+8 r_2) = 9/17$. Given that it doesn't come first or second, the probability that it comes third is $r_1/(r_1 + 7 r_2) = 9/16$. So the total probability that our horse is in the top 3 is $(1/2) + (1/2)(9/17) + (1/2)(8/17)(9/16) = 61/68$.

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EDIT: Here's a different model. The horses are independent random variables; nine of them are uniform on $[0,1]$, while our horse is uniform on $[0,9/5]$. It is not hard to compute that our horse wins with probability $1/2$, and finishes in position $j$ with probability $1/18$ for $2\le j\le10$. So the probability our horse finishes in the top 3 is $11/18$.

[Previous model, note objections in the comments]
As noted in the comments, more information is needed - so let's make some up. Let's assume that for each other horse, the chances that our horse will beat that horse is $p$. Let's further assume that beating any one set of horses is independent of beating any other (disjoint) set of horses. Now we can get somewhere.

First of all, the probability of winning is $p^9$, so we know $p^9=1/2$, which you can solve to find $p$.

Then, the probability of finishing second is $9p^8(1-p)$, since there are 9 ways to choose the winning horse, probability $1-p$ of being beaten by that horse, and probability $p^8$ of beating the others.

Similarly, the probability of finishing third is $36p^7(1-p)^2$. So, the probability of finishing in the top three --- given the assumptions we've made --- is $$(1/2)+9p^8(1-p)+36p^7(1-p)^2$$

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These assumptions are very doubtful. The events of one horse beating another are not independent: given that $A$ beats $B$ and $B$ beats $C$, $A$ must also beat $C$. Imagine that $n$ horses are all equally matched, so any given horse has probability $1/2$ of beating any other. Your reasoning would say that each horse has probability $1/2^{n-1}$ of finishing first. The expected number of horses that finish first would then be $n/2^{n-1}$. But exactly one horse has to be the winner! – Robert Israel Jul 10 '12 at 1:35
@Robert, maybe all I need is that the event of our horse beating any other given horse is independent of the event of our horse beating any other other given horse? – Gerry Myerson Jul 10 '12 at 1:42
Still very unlikely to be true. In order to beat a given horse, you have to run fast. If you run fast, you're more likely to beat another horse too. – Robert Israel Jul 10 '12 at 2:45
To corroborate Robert Israel's remark: beating $8$ of the $9$ other horses will almost certainly improve the chances of beating the remaining horse. Indeed, unless that remaining horse is actually the fastest of the nine (which in absence of further information would have only a $1$ in $9$ probability), beating that final horse is even implied by beating the $8$. – Marc van Leeuwen Jul 10 '12 at 13:21

P(event) = P(W10) + [P(L10)][P(W9/L10)] + [P(L10)][P(L9/L10)][P(W8/L9)]

P(event) = (1/2) + (1/2)(1/2) + (1/2)(1/2)(1/2)

P(event) = 7/8

i.e. P(3) = 7/(2^3)

In general, P(n) = (2^n - 1)/(2^n)

where n is a positive integer less than 10

P(10) = 1

P(Wi) stands for the probability of a win among i horses. P(Li) stands for the probability of a loss among i horses. ASSUMPTION : Probability P(Wn) of a win in a race among n horses remains the same for every horse.

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Does this mean the probability of finishing in the top 10 in a 10-horse race is less than 1? – Gerry Myerson Jul 10 '12 at 13:01
@GerryMyerson: you might not finish the race. :-D – tomasz Jul 10 '12 at 13:12
The present edition takes care for n = 10 – Rajesh K Singh Jul 10 '12 at 13:43
P(10) = P(W10) + [P(L10)][P(W9/L10)] + ....... + [P(L10)][P(L9/L10)][P(L8/L9)][P(L7/L8)][P(L6/L7)][P(L5/L6)][P(L4/L5)][P(L3/L4)][‌​P(L2/L3)][P(W1/L2)] = 1/2 + 1/4 + 1/8 + 1/2^4 + 1/2^5 + 1/2^6 + 1/2^7 + 1/2^8 + 1/2^9 + (1/2^9)(1/2^0) – Rajesh K Singh Jul 10 '12 at 15:00