Progress

The next things I'm going to try to compute are the number of pair hands with flush draws, pair hands with open-ended straight draws, no-pair hands with flush draws and no-pair hands with straight draws. Maybe I'll try to compute the number of hands in different no-pair categories and numbers for gutshot draws too, but I think there are very few situations in 5-card draw where it makes sense to draw to inside straights.

categories of two pairs

I calculated earlier that there are 1584 threes-up hands:

n(threes up hands) = 1584
n(fours up hands) = 1584 * 2 = 3168
n(fives up hands) = 1584 * 3 = 4752
n(sixes up hands) = 6336
n(sevens up hands) = 7920
n(eights up hands) = 9504
n(nines up hands) = 11088
n(tens up hands) = 12672
n(jacks up hands) = 14256
n(queens up hands) = 15840
n(kings up hands) = 17424
n(aces up hands) = 19008

flush categories

There are no ordinary 5-high or 6-high flushes because they would be straight flushes. I'm going to calculate the number of ordinary flushes in each category. One of the rank combos for each category (two for aces) will be a straight combo, so it wouldn't count as an ordinary flush.

n(seven-high flushes) = suit combos * (rank combos - straight combos) = 4 suits * ((5*4*3*2/4*3*2*1)-1) seven-high flushes/suit = 16 seven-high flushes

I'll explain the computation for rank combos in a little more detail. I pretend like I've looked at the 7 and I somehow know that I have a 7-high flush. There are five ranks below a 7. Once I've taken the 7 out of my hand, the first of the remaining cards can be any of the 5 cards that would give me a chance to make a 7-high flush. The next card can be any of the four remaining cards and so on. How these cards are arranged doesn't matter, so I divide the arranged rank combos by the number of possible arrangements (4*3*2*1).

n(eight-high flushes) = 4 suits * ((6*5*4*3/4*3*2*1)-1) eight-high flushes/suit = 56 eight-high flushes

n(nine-high flushes) = 136
n(ten-high flushes) = 276
n(jack-high flushes) = 500
n(queen-high flushes) = 836
n(king-high flushes) = 1316

n(ace-high flushes) = 4 suits * ((12*11*10*9/4*3*2*1)-2) ace-high flushes/suit = 1972

There are two straight combos for aces because the ace is interpreted as the highest ranking card of a flush, but it would be smallest ranking card of a five-high straight. We already counted the 5-high straight flush as a straight-flush, so we can't count a2345 suited as an ace-high flush.

Summary

Here are the number of hands in each category:

Straight flush: 40
Quads: 624
Full house hands: 3 744
Flushes: 5 108
Straights: 10 200
Trip hands: 54 912
Two pair hands: 123 552
Pair hands: 1 098 240
No pair hands: 1 302 540

You can find the odds of being dealt any of these hands by dividing the number by 2 598 960

Odds of no pairs

A "no pair hand" can be defined as a hand that doesn't fit into any other category. The number of no pair hands can be found by subtracting the number of hands in the other categories from the total number of hands. If you do this you get the following result:

n(no pair hands) = n(total hands) - n(hands that beat ace-high) = (2 598 960 - 1 296 420) no pair hands = 1 302 540 no pair hands


NB: You could also calculate the number of no pair hands by calculating kicker combos where all five cards have to have different ranks, and subtract the number of straights and flushes from this. 4 out of the 1024 possible suit combinations for hands with five cards of different ranks will be flushes.

Odds of two pairs

Two pair hands are defined by the rank combos of the pairs, the suit combos of the biggest and the smallest pair, and the kicker combos. The first pair can be bigger or smaller than the second pair. We're not interested in the arrangement of the pairs, so we divide the arranged rank combos by two.

n(two pair hands) = rank combos * suit combos(big pair) * suit combos(small pair) * kicker combos = (13*12/2) pair combos * 6 suit combos/pair combo * 6 suit combos/suit combo * 44 two pair hands/suit combo = 123 552 two pair hands

Aces up is two pair where the biggest pair is a pair of aces, kings up is two pair where the biggest pair is a pair of kings and so on. These categories might not be enough to decide if one hand beats another, since two players might have two pair in the same category.

n(threes up hands) = small pair combos * suit combos * kicker combos = 1 small pair * 36 suit combos/small pair * 44 threes up/suit combo = 1584 threes up

I just lumped together the suit combos here, because there are always 6 suit combos for each pair and these combos are independent from each other. 1584 is actually the number of possible hands for any combination of two specific pairs (1584 "fours and deuces hands", 1584 "aces and kings hands" and so on). The only pair that is smaller than threes, are deuces. Fours up can be either "fours and threes" or "fours and deuces", so there are twice as many fours-up hands as trees-up hands.

Odds of a full house

A full house hand consists of trips of a certain rank, and a pair of a different rank. I have already stated the parameters that define trips and pairs. When you have a full house, there are no kickers so you ignore kicker combos. Trip aces and a pair is called aces full, trip kings with a pair is called kings full and so on.

n(aces full hands) = suit combos(trips) * pair combos * suit combos(pair) = 4 suits * 12 pairs/suit * 6 aces full hands/pair = 288 aces full hands

There are also 288 kings full hands, 288 queen full hands and so on.

n(full house hands) = trip combos * suit combos(trips) * pair combos * suit combos(pair) = 13 trips * 4 suits/trip * 12 pairs/suit * 6 full house hands/pair = 3744 full house hands

NB: Once you have trips of a certain rank, you can't have an additional pair of the same rank. There are only 12 possible pairs for a certain category (jacks full for instance) of full house hands.