Not sure what you're asking. His 1755 number counts those three as the same flop, A72 rainbow. And actually the first two are the exact same cards, just listed in a different order. If you want to count each specific three card combination separately, that's 52c3 = 22,100. And if for some unknown reason you care about the order the cards come down, that's 52*51*50 = 132,600 ordered three card sequences.

Then what is the median flop (if you have no hole cards)? Looking to do a little over/under gambling and kind of burnt out on red/black. Thx.

I know this is very very old but I just wanted to say that if you were observing flops in holdem games, the stats may be biased by the hands that most players see a flop with. Players will usually see a flop with pocket pairs and will always fold many unpaired hands. So the flops you see will have a slightly reduced probability to show up with a pair on the board. This may account for a slight difference from theoretical and observed flop textures.

This same issue came up in a fairly recent thread. Pairs are tricky in poker. If I have a pocket pair, that increases the chances that another player has a pocket pair, not decreases. If you think about it for a little bit, the reason becomes obvious.

Same logic applies to flops as well. So if I have a pocket pair, the chance that a pair appears on the flop actually is increased compared to if I do not have a pocket pair.

It is not hard to work through the actual tallies via combinatorics:

Number of flops having at least a pair given that I hold a pocket pair = 12*C(4,3)+12*C(4,2)C(46,1)+1*C(2,2)*C(48,1)
= 3,408

Number of flops having at least a pair given that I do not hold a pocket pair = 2*C(3,3)+11*C(4,3)+2*C(3,2)C(47,1)+11*C(4,2)C(46,1 )
= 3,364

Of course, to derive the respective probs the common denominator is C(50,3) = 19,600