Hi,
First I would like to say hi to all because of my first post. I have no idea how to calculate something. Hope you people can help me.
My question (and problem):
-holdem
-2 players go to flop

I found (online) some calculations for part of my problem:
One player will pair at least one of his unpaired hole cards: 32,4%.

My question is: What is percentage for at least one of 2 players to pair at least one of his unpaired hole cards?


It is not 32,4% *2=64,8%, right? Because this does not make any sense.

Hope my question was understandable. Thanks

Hint: calculate the chance neither of them do, and subtract that from 100%.

Still not sure what is answer to my question.

Using your 32.4% to hit, that means 67.6% that he won't hit. Then ignoring the very small dependency between the two players and assuming independence, we can estimate that 67.6% squared, or 45.7% is the chance netiher of them hit. Thus 1 - .457 = 54.3% is the chance that at least one of them hits.

This answer will be slightly off because sometimes the two players will have a card of the same rank (which reduces the chance for both of them), and even more rarely they will have two cards of the same rank. The exact calculation is more work than I want to put into it.

The same way that the 32% is calculated can be used to calculate it for two players. It's just 12 cards that can hit instead of 6, so instead of 1-P(6 cards miss) it's 1-P(12 cards miss)

= 1 - C(36,3)/C(48,3) = 58.72%

Or 1 - (36/48)(35/47)(34/46)

I don't do this stuff enough any more and should leave it to the experts.

I would say however, that it might be less than 12 cards when the players have ranks in common, which will be fairly often. We know it is 6 cards for one player. With two players, there are 3 cases. No ranks shared, 1 rank shared, 2 ranks shared.

And a 4th very rare case, where all 4 hole cards are the same rank (but the OP conditions do exclude this one).

Good point, OP didn't say there can't be shared ranks. My bad - you had already mentioned that in your other post but I missed that part.

So if there can be shared cards, we need the weighted average.

N(no shared) = 52*48*44*40/4! = A

N(one shared) = (2/3)13*C(12,2)*C(4,2)*4² = B

N(same unpaired hand) = (2/3)*C(13,2)*C(4,2)² = C

A + B + C = 239824

Next we need the conditional probabilities of hitting the flop given each scenario:

P(hit flop | no shared) = .5872

P(hit | one shared) = 1 - C(40,3)/C(48,3) = 927/2162 ≈ .4288

P(hit | same unpaired hand) = 1 - C(44,3)/C(48,3) = 1013/4324 ≈ .2343

Finally, P(hit | no pocket pairs) = (.5872*A + .4288*B + .2343*C) / 239824 = 54.82%

NewOldGuy's estimate turned out to be accurate; ignoring card removal largely compensated for ignoring the possibility of shared cards.

Nice work.

Thank you both.