In a horse race there are 10 horses. Bob wants to make a "trifecta Box bet". A trifecta box bet is when you choose the first three horses that finish the race in ANY order. What is the probability to win a single trifecta box bet assuming every horse has equal chances to win.
My solution: $\ P = {10!/(7!3!) \over 10!}$
Given solution: $\ P = {7!3! \over 10!}$
What am i doing wrong?
Thanks!
$\endgroup$ 11 Answer
$\begingroup$JMoravitz showed that you can use your probability space as all possible trifectas (ignoring specific orders of the horses). You attempted the problem by setting the probability space as all possible orders of the horses. For the numerator, you chose all possible ways of choosing three horses to finish in the first three spots. However, you do not win just because three horses finish in the first three spots. You only win if your chosen horses finish in the first three spots in some order.
So, a winning outcome is your chosen three horses come in the first three spots in some order in $3!$ ways. The remaining seven horses come in the last seven places in $7!$ ways, and the total probability is:
$$\dfrac{3!7!}{10!}$$
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