The binomial coefficient calculator calculates C(n,k), or pronounced n choose k. This basically means, how many ways can k be chosen from n?

For example, how many possible hold'em hole card combinations are there? That would be choosing two cards from the possible fifty-two cards in the deck, or C(52,2) = 1326. Note: The order of the cards does not matter... AhKd is the same as KdAh. If the order of cards did matter, you would be wanting a permutation P(n,k) instead of a combination. In poker, the order of cards doesn't matter so C(n,k) is exactly what we want.

Another example, how many possible hold'em flops are there? Since you are holding two cards, that leaves fifty cards for the flop. There are three cards on the flop, so C(50,3) = 19600. Note: We can't omit any other players hole cards since we don't know what they are, so we have to assume they are all available.

Now for something a little more complex. Holding a pocket pair, what are the chances of flopping a set? We are holding two cards, so that leaves fifty in the deck. Of those fifty, there are two cards that will make our set (our outs). There are three cards we are going to draw for the flop. So to make up our flop, one of them must be one of our two outs C(2,1), and the other two cards can be any of the remaining 48 cards C(48,2). We know that there is a total of fifty cards, and we are drawing three C(50,3). Given this information, we can compute C(2,1)*C(48,2)/C(50,3) = 0.1151, or about 7.7:1. This is the kind of computation the poker calculator does for us. You supply the outs (number of outs in the deck), the required outs (number of outs we need), drawn (number of cards we are drawing), total (total cards drawn from).

To do more complex calculations, we do the computations individually and combine them. So for instance, holding two unpaired cards, what are the chances of flopping two pair or trips? Figure those separately, then add them together to get the total probability.