The reader is advised that the description provided in the article for the contents of the second table is poorly worded and outright misleading. The probabilities given in the column headed ``flop'' are the probabilities that a player holding a pocket pair of the given rank will flop a pair of larger rank given that she flops trips. The probabilities given in the column headed ``board'' are the probabilities of at least one pair of larger rank on board given that she has exactly one card of the same rank as her pocket pair on board, but neither trips nor quads are on board.

In the preceding article on overcards - given a player holding a pair in hold'em - I showed how to calculate the probability of an overcard coming on the flop and on the board. Now we are going to refine this to more complicated situations. First, let's see how we would determine the probability of more than one higher rank showing up thereby increasing the chance someone has made a higher pair.

We illustrate this with the player holding a pair of nines. There are C(50,3) = 19,600 flops. There are 20 cards larger than a nine, 28 cards smaller than a nine, and two nines comprising the remaining 50 cards. Thus, there are C(48,3) = 17,296 flops without a nine, 2C(48,2) = 2,256 flops with one nine, and 48 flops with two nines. Let's now look at the flops without a nine.

If all of the cards on the flop are smaller than a nine, we are choosing three cards from 28 so there are C(28,3)=3,276 flops with no overcards to the nines. The probability of this is 3,276/17,296 = 0.130. If there is one card larger than nine and two cards smaller than nine, this can happen in 20C(28,2) = 7,560 ways giving a probability of 7,560/17,296 = 0.437. If there are two cards larger than nine of different ranks, then there are two choices out of five for the ranks of the two cards, four choices for each of the two chosen ranks, and 28 choices for the smaller card. This yields flops of this type. The probability of such a flop is then 4,480/17,296 = 0.259. Finally, if there are three cards all of different ranks larger than nine, there are C(5,3) = 10 choices for the ranks, and four choices of each rank giving 640 flops of this type. The probability of such a flop is 640/17,296 = 0.037.

The preceding paragraph completely describes the technique used to calculate the probabilities of zero, one, two or three overcards, all of different ranks, coming on the flop given that a player has a pocket pair. The calculations do not distinguish between what happens with the smaller cards - there is no consideration being given to whether or not a smaller pair or, heaven forbid, a smaller three-of-a-kind comes on the flop. The next table gives the probabilities of the above information for each pair. The assumption is that the player does NOT catch another card of her pair and the column headed k larger cards means the flop contains k larger cards no two of which form a pair.

There is a wide variety of probabilities along these same lines about which one can ask. Let's do one more as an illustration of how these kinds of probabilities are computed. We now suppose the player with the pocket pair ends up with three-of-a-kind and we want to know about the likelihood of a larger pair on the board. This creates an interesting playing situation for the player with the hidden pocket pair - since she has a full house.

First we count the flops giving trips to a player with a pocket pair (let's ignore flopping four-of-a-kind). There are two choices for the card making three-of-a-kind. The other two cards can be any two cards chosen from 48 cards. This can be done in C(48,2)= 1,128 ways and multiplying by two yields 2,256 flops giving the player three-of-a-kind. Of these flops, the number with a larger pair is obtained by multiplying six times the number of ranks greater than the rank of the pocket pair since there are six ways of choosing a pair of cards of a fixed rank. We then multiply this by two to account for the number of choices of the card to form three-of-a-kind with the pocket pair.

To count the number of boards with a third card of the rank of the pocket pair and a larger pair, we note that we have two choices for the card forming trips reducing the problem to finding the number of ways of choosing four cards from the remaining 48 cards so that there is a pair of larger rank. The latter problem breaks into two parts: Four cards forming two pair at least one of which is a pair of larger rank, or four cards containing exactly one pair of larger rank. The total number of boards is 2C(48,4)=389,160.

In the following table we are giving the probabilities under the following assumptions. The left column shows the pocket pair held by the player. The middle column is the probability of the flop containing a card of the rank of the pocket pair and a pair of larger rank. The right column is the probability of the board ending up with one card of the rank of the pocket pair and at least one pair of larger rank.