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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.

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  • $\begingroup$ 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 $\approx$ $\endgroup$ Feb 11 '14 at 17:15
  • $\begingroup$ You're absolutely right. I haven't used LaTeX in years (or whatever TeX they're using on this site). It's fixed. $\endgroup$
    – roberree
    Feb 11 '14 at 18:19
  • $\begingroup$ Yes, we use $\LaTeX$ enclosed in dollar signs. A tutorial is here $\endgroup$ Feb 11 '14 at 18:23
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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.

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  • $\begingroup$ Thank you! That's just not what I expected. Continuing that math, a 5-horse Superfecta box is (5,4)4!2!/6! = 1/3. In that case, I'd very much like to go to the horse track and find a 6 horse race... $\endgroup$
    – roberree
    Feb 11 '14 at 17:24
  • $\begingroup$ You can confirm that calculation easily-you picked one horse to leave out of your list, and it has to come in 5th or 6th for you to win, so it is 1/3 $\endgroup$ Feb 11 '14 at 17:27
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I see the four-horse exacta box as being an application of hypergeometric math, i.e.,the same math used to calculate lotto odds, with N = field; K = horses "drawn" (4); B = two "winning" horses, 1 & 2. If my view is incorrect, then how does the four-horse box differ from the lotto model?

Even if the hypergeometric model is correct, how can odds be accounted for in the standard model? If your four horses have the worst odds, the probability is going to be lower that if they were the highest odds. Also, the size of field must be considered. In your math shown, there is no consideration for how many horses are in the race.For example in the Derby of 20 horses, I figure the probability of picking two correct of 4 would be: .1486.

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