I'm gearing up for horse racing season, and I'm trying to teach some fellow engineering friends how to bet "exotic" bets by using colored dice to simulate horses. So, the odds for each horse winning are the same. I'm keeping it simple with just 6 horses (colored dice).
That being said, I can't figure out how to do the math to calculate the probabilities of a 3-horse Exacta box or a 4-horse Trifecta box. I believe that a typical Exacta box (betting on two horses, in either combination, to come in 1st and 2nd) is 2!4!/6! = $\frac{1}{15}$. However, I can't figure out how to mathematically account for a 3-horse Exacta box (betting on 3 horses, in any combination, to come in 1st and 2nd). Likewise for adding an additional horse into a box for a Trifecta (which is betting on 1st, 2nd & 3rd).
My initial thought is that a 3-horse Exacta box and a 4-horse Trifecta box have the exact same probability, 3!4!/6! = 4!3!/6! = 20%. However, that doesn't seem intuitive.
Any help? Feel free to correct me if I'm way off. Thanks.
Best Answer
Your 3-horse Exacta would be ${3 \choose 2}\frac{2!4!}{6!}$. You choose two of the three horses to be the two that win, then the same calculation you did for the 2-horse Exacta. You are correct that this is $\frac 15$
If your 4-horse Trifecta is choose four and three of them have to be the top three, that is ${4 \choose 3}\frac{3!3!}{6!}=\frac 15$ I don't see an intuitive reason that should match the above.