What are the odd of winning if there are "X" horses in a race & there's a pay-out for holding top 3?
We are holding a Kentucky Derby party. We are paying out extra for anyone who holds bets on the top 3 horses IN ANY ORDER. How do you figure the odds of this happening (assuming the same number of bets are placed on every horse)?
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June 26th, 2010 at 2:14 am
The number of different combinations of k choices out of N can be computed using the formula:
N!/[(N-k)! * k!]
where ! denotes a "factorial":
i! = i * (i-1) * (i-2) * … * 2 * 1
In your case, k is 3 and you aren’t sure about N until Derby day. But the answer will simplify to
N * (N-1) * (N-2) / 6
because k! = 3! = 6 and N! / (N-k)! simplifies down to the numerator shown.
The find the odds on getting the top 3 horses in order (a permutation instead of a combination), omit the division by 6.
Obviously, these aren’t the real odds because the horses aren’t equal competitors.
June 26th, 2010 at 2:14 am
The probability of picking the top three horses in a race of X horses in any order is 1/ XC3
where XC3 = X! / [(X-3)!*3!]
where X! = X*(X-1)*(X-2)*…*2*1
For example, if there are 6 horses, then the probability of picking the top three in any order is:
1 / 6C3 = 1/20 = .05 = 5% (the odds would be 1:19)
Or if there are 10 horses, then the probability is:
1 / 10C3 = 1/120 = 0.008333333 = .833% (the odds would be 1:119)
June 26th, 2010 at 2:14 am
There are xC3 ways of choosing 3 horses out of x horses. There is 1 way of coosing the top 3 horses in any order.
1/xC3 is your odd.
nCr = n! / r! (n-r)!