**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.

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