I can't sleep, so I thought I'd come math at the forum for a while. This post may be of interest to gremlinn and probably nobody else. Stephen and Dave will read this far, then discover it's more about math than poker and probably hang up. Anyhow.

Earlier tonight, I was playing poker. More earlier, I was discussing with Stephen, Dave, and gremlinn why a flush beats a straight, given that four to a flush is easier than four to a straight. This all got me to thinking about various probabilities of various hands in Hold 'Em, and how to figure them out.

The obvious first step is to figure out how many possible hands there are, which is a simple enough operation. Assuming that you're playing by yourself (in which case you're being killed to death by the rake), the total number of possible spreads is (52 * 51 ... 46), or 674274182400. (Note to gremlinn: Is there a more convenient notation for (52 * 51 ... 46)? Like maybe 52!46? If there isn't, I'm going to write to the Federal Bureau of Math Notation and demand one.)

Now we'll go with a pair. The total number of two-card combinations that will yield a pair is 78 (six per rank, thirteen ranks). I incorrectly assumed that the total number of possible pairs was therefore (78 * 50 ... 46), because I forgot that any hand that forms better than a pair no longer is a pair. So it gets sticky.

Let's say we have pocket deuces, and decide to stay in on that (hey, we're getting killed by the rake no matter what). There's our 78. Then comes the flop. If we flop 268, we have trips and that's not a pair. So the first card of the flop cannot be a 2, which means our first number is 48 rather than 50. The second card of the flop also can't be a 2, but it also can't be the same as the first flop card, or we have two pair. So this is where it starts to get tricky. Looking at it, there are five cards remaining in the deck that this card cannot be if we are to retain our pair. So instead of 49, our number here must be 44. The third card plays out almost the same: it can't be a 2, and it can't be the same as either other card, which means we have (48 - 2 - 3 - 3) = 40 possible card at this point.

So now the flop's down, and we have a total number of combinations of (78 * 48 * 44 * 40) = 6589440. Next comes the turn. Of the 47 cards left in the deck, the turn cannot be the same as any card already down on the table, so (47 - 2 - 3 - 3 - 3) = 36. At this point, straights and flushes become an issue, but I'm not going to worry about them yet (you'll see why in a minute). So 36 is the number we need on the turn, and, likewise, (46 - 2 - 3 - 3 - 3 - 3) = 32 on the river. That gives us a running total of (78 * 48 * 44 * 40 * 36 * 32) = 7591034880.

Now we worry about straights and flushes. Given any five cards, there are 10240 possible straights and 8580 possible flushes, for a total of 18820. From this we subtract 40 (there are 40 possible straight flushes, and they're obviously invluded in each set), which leaves 18780. Now, of our seven cards, how many different five-card combinations do we have? The answer is made slightly tricky by the fact that no valid combination can include both pocket cards - they're a pair, remember, so they can't be serial or suited. So there are eleven possible combinations of the available cards that could yield a straight or a flush. (11 * 18780) = 35268840 possible straights or flushes. Therefore, (7591034880 - 35268840) = 7555766040 possible pair hands exist, which gives a 7555766040 / 674274182400 chance of having a pair (roughly 1.12%).

Those odds seem very low to me, and I'm not so full of myself as to think that I can do that much math at this hour of the morning and not make any mistakes. So can anybody (read: gremlinn) find something wrong in my calculations or in my logic? I'm reasonably sure I miscalculated the possible number of flushes, to start. And am I correct in assuming that, since the numbers I'm using preclude two pair or trips, I don't have to consider boats or quads, since the former are prerequisite?