What is the probability of three players each obtaining a set on a flop of holdem in a 9-handed game with no one folding before the flop? I'm going to guess next-to-impossible.

What is the probability of five players each obtaining a set in a hand of holdem in a 9-handed game with no one folding before the river? I'm going to guess even more nearly impossible than next-to-impossible.

P(flop set > set > set) = [C(13,3) * 4^3 / C(52,3)] * C(9,3)*3^3 / C(49,6) / 5!! ≈ 1 in 111,666

P(set > set > set > set > set) = [C(13,5) * 4^5 / C(52,5)] * C(9,4)*3^5 / C(47,10) / 9!! ≈ 1 in 315,169,169

This maybe should be stickied.

Thanks, heehaw...

And when you think about the large percent of pocket pairs that are usually raised preflop (AA, KK, QQ, JJ at least) in a 9-handed game, which often chases out an equally large percent of pocket pairs (22, 33, 44, 55 at least), especially when stacks are not deep...and it takes only one such big pair to do this...and when two of the pairs are in that big four cohort...and that's just preflop...it approaches true impossibility.

I'm sure if enough hyperturbos are played it could happen, but then those games are not 9-handed usually, and even if they are, they aren't for long!

Impossible is not possible.

Yes, and your point?

Did you understand how heehaw did it? I'm guessing not.

Here is a method that anyone who understands simple parlays can follow.

First calculate the chances that the flop contains no pair. The second card has a 48/51 chance that it won't match the first. Then there is a 44/50 chance that the third card matches neither. Multiply and get about 82.8%

If that happens the next step is to imagine a three handed game rather than a nine handed one. MR A's chances of having a set of one of the board cards is 9/49 for his first card multiplied by the 2/48 that his second card matches his first.

If so, the chances that Mr B has a set is 6/47 times 2/46. And if they both have sets, Mr C's chances of having a set of the remaining board card is 3/45 times 2/44.

Multiply the chances the board isn't paired by all those fractions and you get the chances that all three players in a three handed game flop sets. But because it is a nine handed game there are many different triplets. 9C3 which is 84. So its 84 times easier. (You can't use this technique if it was possible for more than three players to flop a set but that's not the case here.)

The same technique works for the second question as well. Multiply (48/51 x 44/50) x (15/49 x 2/48) x (12/47 x 2/46) x (9/45 x 2/44) x (6/43 x 2/42) x (3/41 x 2/40) x 9C5.

I have an affection for probability, but am no good at math. I kind of understood heehaw on a basic level, but your explanation helped. Thanks again.

This reminds me of the concept that it is far easier to hit a bad beat jackpot when 9 or 10 handed than 5 or 6 handed, so you should not be paying nearly as much into the jackpot when you go short, but casinos don't care and they take the same buck every hand. Something to think about.

How about if you answer this first - What was the point of your original post?

Have a great day.

I'm betting some sort of math/stats class.

Doubt it. Because he thanked heehaw for an answer that I have doubts he understood.

I would lay heavy odds that it wasn't a homework question. I think he or a friend witnessed it and wanted to know the chance of it.

And I didn't explain my math because I assumed he only cared about the answer.

OK. But my math technique need little or no explanation if you know how to parlay. It is much closer to self explanatory. Do you disagree?