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?
 A: 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 $(1/2) + (1/2)(9/17) + (1/2)(8/17)(9/16) = 61/68$.
A: 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 $$(1/2)+9p^8(1-p)+36p^7(1-p)^2$$ 
A: 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.
