# probability of a horse winning a race.

Lets suppose ten horses are participating in a race and each horse has equal chance of winning the race. I am required to find the following:

(a) the probability that horse A wins the race followed by horse B.
(b) the probability that horse C becomes either first or second in the race.

I know there are $10 \cdot 9 \cdot 8$ ways of having first, second or third.

Since each horse has an equal chance of winning, each has probability of 1/10. Would I be right in saying that the probability that A wins followed by B is $\frac{1}{10} \cdot \frac{1}{10}$?

Is it okay if I do this for (b)? $\frac{1}{10} +\frac{1}{10}$?

• Imagine there are only 2 horses. Would the probability that A wins, followed by B, be $(1/2)(1/2)=1/4$? Jun 5 '12 at 6:36

Probability horse $A$ wins the race is $\frac {1} {10}$. Then, there are $9$ horses left, so the probability that $B$ will come second is $\frac {1} {9}$, not $\frac {1} {10}$. So the probability for $(a)$ is $\frac {1} {90}$.

For $(b)$, horse $C$ has an equal chance of being first, second, third, etc. or tenth. Thus, your $\frac {1} {10} + \frac {1} {10}$ answer is sufficient.

• seeing your answer for (a), I'm beginning to doubt my answer for (b). If C is second, then some other horse already came first. so C would have prob 1/9 of being second. So why is the final answer not $\frac{1}{10} + \frac{1}{9}$? Jun 5 '12 at 6:53
• There's a huge difference. In $(a)$, you are selecting one horse, then selecting another in a way such that order matters. In $(b)$, you are not selecting a particular horse to be in first place. You are selecting 1 out of 10, we haven't eliminated anything. Why should $C$ be more likely to come in second than first? Does that mean that $C$ is guaranteed to come in tenth, since if that were so, then the other 9 horses had to take places 1-9?
– A.S
Jun 5 '12 at 7:22

You can double-check yourself by counting outcomes. There are $10!$ possible finish orders, and it’s implied that they’re all equally likely. If $A$ wins and $B$ finishes second, there are $8!$ equally likely possible orders for the remaining $8$ horses. Thus, there are $8!$ outcomes in which $A$ wins and $B$ comes second out of a total of $10!$ possible outcomes, for a probability of $$\frac{8!}{10!}=\frac{8!}{10\cdot9\cdot8!}=\frac1{10\cdot9}=\frac1{90}\;,$$ just as Andrew Salmon argued directly from the probabilities.

For the second question, if $C$ finishes first, there are $9!$ possible finish orders for the other $9$ horses. The same is true if $C$ finishes second. Since $C$ can’t finish both first and second, these $9!+9!$ outcomes are all distinct, and the probability that $C$ finishes first or second is therefore $$\frac{9!+9!}{10!}=\frac{2\cdot9!}{10\cdot9!}=\frac2{10}=\frac15\;,$$ again just as Andrew argued directly from the probabilities.

I don't think we have enough information to answer this question. The fact that each horse has equal change of winning doesn't mean that each horse have equal change of being second, or third and so on.

In fact in real life they wouldn't have the same probabilites most of the time. For example some horses/jockeys might use win-or-bust approach. Ie. they either win the whole race or will be last etc.

But if this is homework then the given answers are most likely correct.

The answer to A is 1/90 because of non replacement method. Horse A has 1/10 chance to be 1st and horse b would then be 1 of 9 with a chance to be 2nd. Multiply 1/10 times 1/9 gets 1/90th

The answer to b is different as its an addition problemas in he has a chance to be first and/or 2nd so 1/10 plus 1/9...common denominator of 90 so 9/90 plus 10/90 or 19/90 chance. This is just a statistical question not a real question because percentages change based on post position, horse ability, jockey, trainer etc.