Are You Facing Trips?

You are playing in an eight-handed seven-card stud game and find that you have been dealt a pair among your first three cards. You wonder what the chances are that one of the other seven players has been dealt trips (three-of-a-kind). The preceding question, in fact, was posted by Linda Sherman this past October on the newsgroup rec.gambling.poker.

I am about to show you a straightforward way to answer the question followed by variations on the basic question.

Assume you have been dealt a pair so that the rank multiset of your hand has the form x-x-y. Let P(i) denote the probability that a specific collection of i of the other seven players have been dealt trips for their first three cards. It is important to note that we make no assumptions about what has been dealt to the other 7 - iplayers. The probability that no one has been dealt trips is given by the formula

The coefficients in the expression arise from the number of ways of choosing i objects from seven. To see that the above expression does what I claim, consider a deal in which exactly two players are dealt trips. This deal contributes once to the first term, twice to the second term, and once to the third term. Because of the alternating signs of the expression, the deal contributes zero to the final answer. Similarly, any deal in which at least one player receives trips contributes zero to the expression. What is left are the deals for which no one receives trips and they contribute exactly once. So the claim follows.

All that remains to do is calculate P(i) for each i. These are easy to do and I shall do the first two to illustrate.

When i = 1, we are asking for the probability a given individual player, say A, is dealt trips. The number of possible hands for A is C(49,3) = 18,424. He can be dealt trips of rank y in one way, and for each of the remaining 11 ranks (rank x is not possible), he can be dealt trips in four ways. Altogether he can be dealt trips in 45 ways. This gives P(1) = 45/18,424. Then 7P(1) = 45/2,632

When i = 2, we are asking for the probability two given players, say A and B, both are dealt trips. The total number of hands they can be dealt is C(49,3)C(46,3) = 279,676,320. Determining the number of ways both players can be dealt trips breaks into two subcases: One of the players has trips of rank y, or neither of the players has trips of rank y. For the first subcase, there are two choices of the player having trips of rank y and there are 44 choices of trips for the other player. This yields 88 ways both can be dealt trips with one of the players holding trips of rank y.

For the other subcase, there are 44 choices of trips for player A and 40 choices of trips for player B. Multiplying gives 1,760 ways both players can be dealt trips of ranks different from y. Altogether, there are 1,848 ways both can be dealt trips. Dividing by 279,676,320 gives P(2) = 1/151,340. Then 21P(2) = 3/21,620.

We perform similar calculations for the remaining values of i and obtain a probability of .98304 that no one has been dealt trips. Thus, the probability one or more of the other seven players has been dealt trips is .01696, or about odds of 58-to-1 against someone having been dealt trips.

The preceding probability tells us it is unlikely someone has been dealt trips. However, the player holding the pair has additional information available to her. For one thing, she is aware of how the other people have bet, but betting after the first three cards have been dealt does not give much information about the possible existence of a player holding trips.

The really important information available to her is the seven upcards she has seen. Some patterns of upcards make it impossible that anyone has been dealt trips. For example, if she sees three cards of rank y (recall one of her cards has rank y) and four kings, she knows no one can have trips.

Other patterns of upcards make it highly unlikely someone has trips. For example, if she sees one card of rank x plus two threes, two fours, and two sixes, she knows it is going to be rare for someone to have trips.

Let's take a look at the particular pattern which clearly is the most likely to involve someone having trips. This is the pattern for which she sees seven cards of distinct ranks, where none of the ranks is x or y. Someone has trips only if he possesses a pair of downcards of the same rank as his upcard.

The calculation is performed in a fashion similar to that above. Upon doing so, we find the probability that someone has trips, given that the seven players have seven cards of distinct ranks different from x and y, is .02411, or slightly more than 40-to-1 against someone having trips.

Let's change the scenario slightly and suppose she sees no upcards of rank x or y, but she does see two cards of the same rank among the seven opponents' upcards. How much does this reduce the probability of someone having rolled up trips?

Doing the same computation yields a probability of .01957 that one or more opponents has trips. That is, the odds are about 50-to-1 against someone having trips.

Finally, let's look at the somewhat extreme condition mentioned above. Suppose she sees one card of rank y, and three pairs of distinct ranks for her opponents' upcards. Our intuition leads us to believe it is highly unlikely someone is holding trips. Upon doing the computation, we find there is a probability of .0081 that someone has trips. In other words, the odds against someone having trips is about 122-to-1.

The latter two computations just completed above require more than a routine application of inclusion-exclusion. The first of the two requires some adjustment in the weight given to trips involving the paired rank she sees. The second requires adjustments to the coefficients used in the inclusion-exclusion expression. Thus, these examples are a good task for someone wishing to understand inclusion-exclusion beyond an elementary textbook explanation.