Ok that looks a lot more realistic I think that with multiple players in a hand you are still looking at 5% or less of set over set. It is pretty rare to have more than 3 people with PP so we can exclude that from the equation. The numbers with 3 people in a hand is going to be approximatly 3/2 * the odds with two people in a hand. It will be off just a little bit since you do know tow more cards so the calculation is out of 46 instead of 48 but you could approximate it to be just slighly larger than 3/2 maybe 1.75 * the original value. So even with 4 or 5 peolple with PP your odds are still going to be less than 5%.

how about the odds of set over set over set on the flop?

I played a live limit game on fri night where there was a set over set over set on the flop.

the flop was 6 4 2. betting was capped on the flop...then none of three bet again on the turn or river!! what are the odds of that?

[the top set held up]

Assuming none that all three players are drawing at a both cards, the chances are
1:1728 of all three flopping a set. So was it a 5 or 3 that hit on the turn that killed all the action on future streets, other than than I cant possibly see why this wouldnt get bet again.

Wow, i take AP stats and K3nt hit it right on the knob... were going through this right now. All I do is talk about poker in that class, half the reason i took the class. My teacher loves foxwoods so we talk about odds a lot.


Haha, kind of off topic. But I talked to the other stats teacher. He heard how much money I make online and he made up a rumor that I cheat, because he is jelous no doubt. I talked to him and exaplained how I played. He was baffled, "you wont play KJs to a raise?" drirect qupte, he told me how him and him friends talk on the phone while playing and tell each other their hands. Once i figure out his name and the site he plays on, I'm reporting him... simply beacuse I dont like him,

Copied and pasted from: http://www.math.sfu.ca/~alspach/mag86/

Brian Alspach

Poker Digest Vol. 4, No. 26, December 14 - 27, 2001

The definition of the term ``set'' in hold'em is three-of-a-kind for which two of the cards are the player's hole cards and the third card of the same rank is on board. Sometimes you will hear players say someone has a set when there is a pair on board and someone has a third card of that rank in her hand. This is an incorrect use of the term. When I use the term ``set'', I am referring to the correct definition above.

Some of the most dramatic hands in hold'em arise when two players have sets. The main reason for the drama is that a set is about as well hidden as a strong hold'em hand can be. If you see a board with ranks A-2-6-8-Q and no flush possible, then two players holding, say, a set of 6s and a set of 8s are going to be confident of being ahead. If both players flopped sets, there likely is going to be some action. I've seen players eliminated early in no-limit tournaments without rebuys because of two players having flopped sets.

There was a message posted on rec.gambling.poker in November in which the poster asked about the chance of two players flopping sets, given that each player has been dealt a pocket pair. I'm going to answer this question, along with some related questions, in this column.

Let's suppose one player has been dealt a pair of rank x and another player, a pair of rank y, different from x. We make the assumption we know nothing about anyone else's hand. Thus, we are assuming the flop is coming from 48 unseen cards. The total number of possible flops is C(48,3) = 17,296. The number of flops of the form x-y-z, where z is different from both x and y, is 176 because there are two choices for each of the cards of ranks x and y, and there are 44 choices for the remaining card. Dividing by the total number of flops yields a probability of .0102, which is approximately 1/98.

Below is a table of some other flops of interest for two players holding pocket pairs of ranks x and y, respectively.

flop probability flop probability
x-x-y .00012 y-y-x .00012
x-x-z .0025 y-y-z .0025
x-z-z .0076 y-z-z .0076
x-y-z .0102 x-z-w .1018
y-z-w .1018 z-z-z .0025
z-z-w .1526 z-w-u .6105

Next, let's consider a few board probabilities. The total number of boards is C(48,5) = 1,712,304. Exactly 44 of those boards result in both players making quads. This gives a probability of .000026, or about 1 in 38,916.

There are 10,560 boards of the form x-y-z-z-w. Both players have full houses in this case. The probability of this outcome is .00617 or about 1 in 162.

There are 42,240 boards of the form x-y-z-w-u. Both players have trips (we are not going to worry about straights and flushes on board) in this case. The probability of this outcome is .02467, or about 1 in 41.

All of the above probabilities were derived under the assumption two players have been dealt pocket pairs. Most people probably are more interested in how often deals occur in which two players flop set over set. In order to calculate this, we must take into account how often exactly two players are dealt pocket pairs.

In my article entitled ``I'm In ... No, I'm Out: Part 5'' (Poker Digest, Vol. 2, No. 18), I gave a table of probabilities for various numbers of pocket pairs. In a hold'em game with nine players, the probability for precisely two players being dealt pocket pairs is .082. For 10 players it is .096, and for 11 players it is .11.

Let's concentrate on the 10-handed game. There is a probability of .096 that precisely two players are dealt pocket pairs, but some of the time the two players have pairs of the same rank. Clearly, in the latter situation, no one can make a set. What we really need is the probability two players are dealt pocket pairs of distinct ranks. We can use the .096 figure and proportionality to get a very good approximation.

I like proportionality arguments because of their simplicity and elegance. If you go back and look at my articles deriving the .096 figure, you will discover the method is complicated. In order to get the exact probability of two players being dealt pocket pairs in a 10-handed game, we would have to go through a modified version of the complicated process.

There are 13 ranks and three ways of choosing two pairs of a given rank. Thus, there are 39 ways of dealing two pocket pairs of the same rank. There are C(13,2) = 78 ways of choosing two distinct ranks, and there are six pairs of any given rank. Thus, there are 2,808 ways of dealing two pocket pairs of distinct ranks. This means there are 2,847 ways of dealing two pocket pairs of which 2,808 have distinct ranks. So the proportion of two pocket pairs of distinct ranks to two pocket pairs is 72/73. Multiplying .096 by 72/73 gives the approximate probability precisely two players have been dealt pocket pairs of distinct ranks. The reason this proportion is not exact is fairly subtle and I leave it up to readers to figure out the reason.

The product of 72/73, .096 and .0102 gives us the probability of a deal with a set over set flop in a 10-handed game, where precisely two players have been dealt pocket pairs. The product is .0009658 which is approximately 1/1,035.

The preceding number may strike you as being too small, but remember, there are other ways for a set over set flop to occur. Going back to the same table referred to above, we find a probability of .016 that precisely three players have been dealt pocket pairs. It is obvious that the chances of a set over set flop occurring when three players hold pocket pairs of distinct ranks is going to be greater. In fact, following the same type of arguments used above, we find the probability of a set over set flop deal with precisely three players holding pocket pairs to be .00492, or about 1/203. So we see, there is a much bigger contribution to the probability of a set over set flop coming from three players being dealt pocket pairs than from two players being dealt pocket pairs.

In the same way, we determine the probability of a set over set flop deal, where exactly four players hold pocket pairs, to be .00011886.

Adding the three probabilities gives us .00600466 or about 1/167. There will be a small contribution for deals with five or more players being dealt pocket pairs, but it will not change the number by much. So if you are playing in a game where players holding pocket pairs stay to see the flop, you should be seeing flops with two or more players making sets on the flop about once every 167 hands.