**Brian Alspach**

**Poker Digest Vol. 5, No. 2, January 11-24,
2002**

You are playing in a 10-handed hold'em game and find that you have been dealt pocket 8s. When thinking about possible hands for other players, it is natural to wonder about the possibility of other players having been dealt pocket pairs.

In my previous article, I determined the probability of another player also holding pocket 8s. What we learned is that the probability of this occurring is about 1/136. Thus, it is an unlikely event.

Given that you have been dealt a pocket pair, contemplating the possibility
of another player holding a pair of the same rank is mostly a matter of
curiosity. What is far more interesting, and more important, are the
chances that one or more opponents have been dealt pocket pairs of *larger* rank. This is the topic of this article.

The probabilities I am about to present should be of interest to all players. As far as I know, these calculations have not been carried out before. I cannot be certain no one has done them before, but given the general level of probabilistic results commonly available, it is no surprise I haven't seen this information. The methods employed in the popular poker books are incapable of handling this problem.

If someone has worked on this problem, I would not be surprised to learn simulations were used. I want to emphasize these probabilities are the outcome of exact computations. Thus, the results are exact to within the decimal approximations of rational numbers.

pair | one rank | two ranks | three ranks |

K-K | .0439 | - | - |

Q-Q | .08412 | .001863 | - |

J-J | .12161 | .005435 | .0000768 |

10-10 | .15519 | .01044 | .0002844 |

9-9 | .1857 | .01662 | .0007109 |

8-8 | .2132 | .02397 | .001367 |

7-7 | .2380 | .03218 | .0023 |

6-6 | .2603 | .04114 | .003538 |

5-5 | .2801 | .05071 | .0051 |

4-4 | .2977 | .06076 | .007 |

3-3 | .3133 | .07118 | .009251 |

2-2 | .3269 | .08186 | .01186 |

The table above contains the information. I'll now describe the entries of the table. The column headed ``Pair'' tells us the rank of the pair held by the player. The column headed ``One Rank'' tells us the probability there is exactly one larger rank for which either one or two pairs of that rank have been dealt to other players. The column headed ``Two Ranks'' tells us the probability there are exactly two larger ranks for which either one or two pairs of each rank have been dealt to other players. There is an analogous meaning to the column headed ``Three Ranks''.

Now consider some examples. Look at the row corresponding to a player holding pocket kings. We find a probability of .0439 under the column headed ``One Rank.'' That is the probability there is either one or two players holding pocket aces. This translates into odds of about 21.8-to-1 against someone having pocket aces. Note that there are no entries under the other columns for pocket kings because there is only one rank larger than king.

Now look at the row corresponding to a player holding 8-8. We see there is a probability of .2132 that there is exactly one larger rank with one or two players holding pocket pairs of the larger rank. This is slightly bigger than one-fifth, so that approximately one in every five times you find 8-8 in your hand, you are going to be going against a pocket pair (or two) of one larger rank.

It's now appropriate to mention one fact about the probabilities given in the table. I have not bothered to filter out situations when you are facing two people holding pocket pairs of the same larger rank. That is, if you hold 8-8, then a deal with one person holding 10-10 against you, and a deal with two people holding 10-10 against you have not been separated. The reason for this is that two people holding pocket pairs of the same rank is rare as we saw in part I. Thus, most of the contribution to the probabilities given in the table comes from each larger rank having only one pair of that rank.

Of course, as a player holding 8-8, for example, having one player with 10-10 against you is worse than having two people holding 10-10 against you. Two people with 10-10 prevent the larger pair obtaining trips which is to your advantage. On the other hand, a 10 cannot come on board, so that your chance for a straight has diminished.

Now look at the row corresponding to 2-2. We see the probability of a pair of larger rank having been dealt is .3269. In other words, about one-third of the time, a pair of larger rank will be dealt. Not only does pocket deuces have to worry about overcards possibly pairing another player, about one-third of the time he already is up against a larger pocket pair.

Looking in the third column, we find a probability of .08196 that there are pairs of two larger ranks against pocket deuces. Thus, the odds against two larger pocket pairs is only about 11.2-to-1. This is significant and indicates just how weak pocket deuces are.

Let me conclude with a few words on the derivation of these probabilities. A fairly detailed version is available at my web site (http://www.math.sfu.ca/~alspach) under a file entitled ``Multiple Pocket Pairs'' in the Poker Computations directory. I always am happy to have someone check the details since elimination of errors is desirable.

The first step is to calculate the total number of ways of completing a given fixed player hand. We are choosing 18 cards from 50, and this can be done in ways. Once 18 cards are chosen, they can be broken into nine hands in ways. The product of those two numbers is the number of completions.

The next step is to choose the rank of the pair held by the player. Call this rank . Now count the number of ranks larger than .

We then calculate how many completions have 1, 2,..., larger ranks with pocket pairs of those ranks. After doing this, we use inclusion-exclusion to get the exact number of completions with 1, 2 and 3 larger ranks. Dividing by the total number of completions gives the entries in the table.

What makes the computations more intricate than normal is the fact there can be either one or two pocket pairs of the same ranks.

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