Join Date: Jan 2009
Posts: 5,038
Assuming you mean odds of at least one two, or three or four on the flop, the easiest way is to figure out the chance none of these ranks fall. Then one minus this probability is the chance at least one does.
There are 12 cards of interest. Therefore there are 52-12 = 40 that are not a 2, 3 or 4.
The number of ways of selecting 3 of those 40 cards is C(40,3). The total number of possible flops is C(52,3).
Therefore the probability of no 2, 3, or 4 is C(40,3)/C(52,3) = 9880/22100 =0.447. Then the probability at least one of 2,3 or 4 falls on the flop is 0.553.
No, it doesn't matter if it is Hold'em or Omaha. Also, note this calculation was done assuming you know nothing about opponent's or your hand. In fact, we have seen data from millions of hands where it is actually more likely that a flop will have a 2, 3 or 4 than straightforward combinatorial theory says. That is because if a flop was seen, it is more likely the players still in the hand have higher cards, thus slightly increasing the chances the flop will have lower cards.
