Originally Posted by WarEagle1266
Well, to hold a pair before the flop is 16 to 1; so, in a nine handed game, 1 player would have a pocket pair (16/9) to 1 which is 1.78 to 1; therefore, 3 would have a pocket pair (1.78*3) to 1, which is 5.33 to 1. The probability of making a set with a pocket pair on the flop is 7.51 to 1, so three people making a set would be (7.51^3) to 1, which is 423.7 to 1, which is then multiplied by the 5.33 to 1 odds we got earlier, giving us 2257.6 to 1.
So by my calculations, and I hope someone corrects me if I'm wrong: In a 9 handed game 3 players with pocket pairs will make their set on the flop 1 time out of 2257.6 hands.
*Odds for holding a pair preflop and making a set on the flop were taken from Doyle Brunson's Super System.
It's a good try, but not quite correct. You forget to take into account the fact that one player flopping a set affects the odds of the next player flopping his (because there are only 2 cards that can make his set). You also made a similar error preflop.
It is going to become pretty complex, so I'm going to do this in two parts.
Right now we'll figure out the odds of 3 players hitting sets when there are 3 players going to the flop, all with pocket pairs.
Obviously, we must assume they have different pocket pairs, or this would be impossible. Basically, we need to have one of each card flop. For example, say the players have 44 55 and 66. We need the flop to be 4 5 6 or some rearrangement of that.
First the chances that the first card is a 4 5 or 6 is 6/46. Then the second card must be one of the other two numbers. There will be 4 outs out of 45 unknown cards so the chances is 4/45. Similarly the chances that the third flop card will be the remaining pair is 2/44.
Multiplying these gives 6/46 * 4/45 * 2/44 = .000527, or 1 in 1897.5.
So one of every 1897.5 times three players all have pocket pairs and see the flop
, all three will hit their set.
More to come.