# Exotic 6-horse race betting probabilities

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.

• I would not use the equal sign in $2!4!/6!=6.7\%$ as they are not equal-it is rounded. I would either leave it as $\frac 1{15}$ or use \approx to get $\approx$ Feb 11 '14 at 17:15
• You're absolutely right. I haven't used LaTeX in years (or whatever TeX they're using on this site). It's fixed. Feb 11 '14 at 18:19
• Yes, we use $\LaTeX$ enclosed in dollar signs. A tutorial is here Feb 11 '14 at 18:23

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.