But where's the fun in just being told?
Here's the math, for anyone interested in figuring it out from scratch:
You have a pair; there are 50 more cards. Two of them match your pair.
So, there's a 2/50 (or 1/25, or 24:1) probability of getting one of those two on the first flop card, 2/49 (or 23.5:1) on the second, and 2/48 (or 1/24, or 23:1) on the third.
We've double-counted a bit - if the first two cards on the flop are your card, for instance, we've counted it as both the 2/50 and 2/49. So we have to subtract (2/50*2/49)+(2/50*2/48)+(2/49*2/48) - a measly 2/1225 + 1/600 + 1/588 = .005 or .5%, but still statistically significant.
I don't know of any way of adding odds; I can only add probabilities, as fractions. (1*49*24)+(2*25*24)+(1*25*49)/25*49*24 = 3601/29400 = 12.24%. Subtract the .5% doublecounted, and get 11.74%. That's about 7.51 to 1.
If you still don't have it after the flop, it's a simple 2/47 on the turn (22.5:1) and 2/46 on the river (23:1).
In other news, I've been accepted to teacher's college in the fall! It's only a 9 month program, so sometime in the 2007/2008 school year you can expect to hear some kind of scandal about a Toronto teacher playing poker with her students, claiming to be teaching them probability.
MathBabe
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