Brian Alspach

Poker Digest Vol. 2, No. 5, February 26 - March 11, 1999

In the preceding article, I verified Bib Ladder's calculation of 0.371 as the probability of a flushing board occurring given no information about any of the players' hands. We now are going to look at the probability of a flushing board occurring from an individual player's viewpoint. Since there are many similar subcases to consider, we carry out the first few in detail and leave the rest to be verified by the reader. Full details of the computations may be found at my website http://www.math.sfu.ca/ alspach.

Since a given player has two cards, the total number of possible boards is

because we are choosing five cards from the 50 remaining cards. As before, the possible suit distribution types are (0,0,0,5), (0,0,1,4), (0,0,2,3), (0,1,1,3), (0,1,2,2), and (1,1,1,2), where, for example, the type (0,0,0,5) indicates the board is made up of 5 cards in the same suit. It is easy to see the first four types describe flushing boards and the last two types describe non-flushing boards.

Suppose our fixed player is holding two suited cards. We now enumerate the possible boards by type with subcases arising depending on the suits represented.

The type (0,0,0,5).
If the suit on board is the same as the suit held by the player, there are possible boards. This gives 462 boards with five cards in the same suit as the suit held by the player.

If the suit on board is different than that held by the player, there are three choices for suit and choices for the five cards of the chosen suit. This yields boards with five cards in the same suit where the suit is different from the suit held by the player.

The type (0,0,1,4).
First suppose the suit with four cards in the board is the same as the suit held by the player. There are choices for the four cards of the player's suit, three choices for the suit of the remaining card, and 13 choices for the card in the latter suit. This yields boards whose type is (0,0,1,4) such that the four suited cards are in the same suit held by the player.

Suppose next the suit of the singleton in the board is the same as the suit held by the player. There are 11 choices for the card from the players suit, three choices for the other suit in the board, and choices for the four cards of this suit. This gives us boards of type (0,0,1,4) for which the suit of the singleton card is the same as the suit of the player's cards.

Finally, suppose the player's suit does not occur amongst the cards in the board. Then there are three choices for the suit of the four cards, choices for the four cards of this suit, two choices for the remaining suit, and 13 choices for the card of that suit. We have boards of type (0,0,1,4) which have no suits in common with the player's suit.

The type (0,0,2,3).
There are 38,610 boards of type (0,0,2,3) for which the three suited cards are in the same suit as the player's suit. There are 47,190 boards of type (0,0,2,3) for which the two suited cards are in the same suit as the player's suit. Finally, there are 133,848 boards which have no suits in common with the player's suit.

The type (0,1,1,3).
There are 83,655 boards for which the three suited cards are in the same suit as the player's cards. There are 245,388 boards for which the suit of the player's cards is the same as one of the single suited cards in the board. Finally, there are 145,002 boards for which the player's suit does not appear in the board.

The type (0,1,2,2).
There are 334,620 boards for which the player's suit is the same as one of the two suited cards, 200,772 boards for which the player's suit is the same as the suit of the single suited card, and 237,276 boards for which the player's suit does not appear in the board.

The type (1,1,1,2).
There are 120,835 boards for which the player's suit is the same as the suit of the two suited cards. Finally, 435,006 boards for which the player's suit agrees with the suit of one of the single suited cards.

Adding the numbers of boards of the various subcases yields 2,118,760 boards as it should. There are several interesting probabilities we may compute at this point.

There are four subcases above in which the board produces a flush in the suit the player is holding. The number of boards doing this is

462 + 12,870 + 38,610 + 83,655 = 135,597

which implies the probability that a player holding two suited cards will end up with a flush is

Thus, the odds against finishing with a flush given that a player has two suited cards are about 14.6:1.

There are seven subcases in which the board produces a potential flush in a suit other than the suit being held by the player. The number of such boards is

3,861 + 23,595 + 55,770 + 47,190 + 245,388 + 145,002 = 654,654.

This means the probability the board creates the possibility of a flush in a suit other than the suit held by the player is

Altogether, given that a player has two cards in the same suit, the number of flushing boards is 135,597 + 654,654 = 790,251 making the probability of a flushing board

Now we consider the case of a given player having two unsuited cards. Again there are subcases to consider and more of them because the player has two suits in her hand. The counting principles remain the same so that we only summarize the results.

The type (0,0,0,5).
There are 1,584 boards with the board suit being one of the suits in the player's hand, and there are 2,574 boards with the board suit not being one of the player's suits.

The type (0,0,1,4).
There are 11,880 boards where both suits in the board agree with the player's two suits, there are 25,740 boards for the subcase in which the four suited cards agree with one of the player's suits and the other suit in the board does not appear in the player's hand, there are 34,320 boards for which the four suited cards are not in one of the player's suits, and there are 18,590 boards whose suits do not overlap either of the player's suits.

The type (0,0,2,3).
When the player's suits and the board's suits are the same, there are 29,040 boards. There are 68,640 boards when the player's suits and the board's suits agree in the suit that has three suited cards, there are 75,504 boards when the player's suits and the board's suits agree in the suit with two suited cards, and there are 44,616 boards when the board's suits and the player's suits do not overlap.

The type (0,1,1,3).
There are 137,280 boards when the player's suits agree with the three suited cards and one of the other cards in the board, there are 74,360 boards when the three suited cards agree with one of the player's suits and the other two cards in the board have no suit in common with the player, there are 82,368 boards when the player's two suits agree with the two suits in the board which are single suited, and there are 178,464 boards when the player's two suits agree with only one of the single suited cards in the board.

The type (0,1,2,2).
There are 113,256 boards when the two double suited cards agree with the player's suits, 247,104 boards when one of the double suited cards agrees with one of the player's suits and the single suited card agrees with the other player's suit, 267,696 boards for the subcase of the double suited cards agreeing with one of the player's suits and the other two suits in the board not agreeing with the player's other suit, and 146,016 boards when neither of the suits of the two double suited cards agree with either of the player's suits.

The type (1,1,1,2).
There are 267,696 boards if the double suited cards agree with one of the player's suits, and if the player's two suits agree with two of the single suited cards in the board, there are 292,032 boards.

If we sum all of the above numbers we obtain 2,118,760 as required. There are three subcases in which there is a flush in one of the suits the player has in her hand. The number of such boards is

1,584 + 11,880 + 25,740 = 39,204.

The probability of a player with two unsuited cards making a flush is given by

There are 11 subcases which result in a flushing board such that the player with two unsuited cards does not have a flush. This sum is 745,756. The probability of having a flushing board in a suit different from either suit held by a player with two unsuited cards is

Altogether there are 784,960 flushing boards for a player holding two unsuited cards. The probability of a flushing board occurring in this case is

The reason the two probabilities do not sum to 0.371 is due to round-off error but it is occurring only in the third decimal place.

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