**Brian Alspach**

**10 January 2000**

We determine the probabilities of a board occurring which allows the possibility
of a flush depending on a player's hand.

In hold'em, where 5 cards are displayed in the middle
for everyone to use in her or his hand, we refer to the 5 cards in the middle
as the *board*. In order for some player to be able to make a flush, there
must be at least 3 cards in the same suit in the board. We shall call such a
board a *flushing* board. We are interested in determining the probability
of having a flushing board depending on what a fixed player holds in her hand.

If a fixed player has 2 cards, then
the total number of possible boards is

because we are choosing 5 cards from the 50 remaining cards. Let's express the

Suppose our fixed player is holding two cards in the same suit. 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 5 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 3 choices for suit and choices for the 5 cards of the chosen suit. This yields boards with 5 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 4 cards in the board
is the same as the suit held by the player. There are
choices for the 4 cards of the player's suit, 3 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 4 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, 3 choices for the other suit in the board, and choices for the 4 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 3 choices for the suit of the 4 cards, choices for the 4 cards of this suit, 2 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 again 3 subcases for boards of this type.
First suppose the player's suit is the same as the 3 suited cards in the board.
This means there are
choices of the player's suit, 3 choices
for the remaining suit in the board, and
choices for the
2 cards of that suit. This produces
boards of
type (0,0,2,3) for which the 3 suited cards are in the same suit as the player's
suit.
Next suppose the player's suit is the same as the 2 suited cards in the board. Then there are choices for the 2 cards of the player's suit, 3 choices for the remaining suit in the board, and choices for the cards of the latter suit. This gives us boards of type (0,0,2,3) for which the 2 suited cards are in the same suit as the player's suit.

Finally, suppose the player's suit does not occur amongst the suits in the board. There are 3 choices for one of the suits, 2 choices for the other suit, choices for the 3 suited cards, and choices for the 2 suited cards. Altogether we have boards of type (0,0,2,3) which have no suits in common with the player's suit.

**The type (0,1,1,3).**- First suppose the 3 suited cards are in the same
suit as the player's cards. There are
choices for the
3 suited cards, 3 choices for the other 2 suits (we are choosing 2 suits from
3 available), and 13 choices for each card of the other 2 suits. This gives
us
boards of type (0,1,1,3) for which the
3 suited cards are in the same suit as the player's cards.
Next suppose the player's suit is the same as one of the single suited cards. This means there are 11 choices for the card of the player's suit, the are 3 choices for the suit of the 3 suited cards, 2 choices for the other suit, choices for the 3 suited cards, and 13 choices for the other card. This yields boards of type (0,1,1,3) for which the suit of the player's cards is the same as one of the single suited cards in the board.

Finally, suppose the player's suit is not represented in the board. Then there are 3 choices for the suit of the 3 suited cards, choices for the 3 cards of that suit, and 13 choices for each of the cards of the other two suits. This gives us boards of type (0,1,1,3) for which the player's suit does not appear in the board.

**The type (0,1,2,2).**- There are 3 subcases as usual. Consider the subcase
in which the player's suit is the same as one of the suits of the 2 suited
cards. There are
choices for the 2 cards in the player's
suit, there are 3 choices for the suit of the other 2 suited cards, 2 choices
for the remaining suit,
choices for the other 2 suited
cards, and 13 choices for the remaining card. This produces
boards of type (0,1,2,2) for which the player's
suit is the same as one of the 2 suited cards.
The next subcase is when the player's suit is the same as the suit of the single suited card. There are 11 choices for the card of this suit, 3 choices for the 2 suits both of which are 2 suited, and choices for the cards of each of the latter 2 suits. This yields boards of type (0,1,2,2) for which the player's suit is the same as the suit of the single suited card.

The final subcase is when the player's suit does not appear in the board. There are 3 choices for the suit of the single suited card, 13 choices for the card of this suit, and choices for the 2 cards of each of the other 2 suits. This gives boards of type (0,1,2,2) for which the player's suit does not appear in the board.

**The type (1,1,1,2).**- There are 2 subcases to consider. First, suppose
the player's suit is the same as the suit of the 2 suited cards. The there
are
choices for the 2 cards of that suit, and there are
13 choices for each of the cards of the remaining suits. This produces
boards of type (1,1,1,2) for which the player's suit
is the same as the suit of the 2 suited cards.
The final subcase is when the player's suit is the same as the suit of one of the single suited cards. In this subcase there are 11 choices for the card of the player's suit, 3 choices for the suit of the 2 suited cards, 13 choices for each of the cards of the other 2 single suited cards, and choices for the 2 cards which are double suited. This produces boards of type (1,1,1,2) for which the player's suit agrees with the suit of one of the single suited cards.

There are 4 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 2 suited cards will end up with a flush is

Thus, the odds against finishing with a flush given that a player has 2 suited cards is about 14.6:1.

There are 7 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 2 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 2 unsuited cards. We again go through all the subcases of the 6 types.

**The type (0,0,0,5).**- If the suit in the board is the same as one of
the player's suits, there are 2 choices for the suit and
choices for 5 cards in the suit. This gives us
boards of type (0,0,0,5) with the board suit being one of the suits in the
player's hand.
If the suit in the board is not one of the player's suits, there are 2 choices for the suit and choices for the 5 cards. Thus, there are boards of type (0,0,0,5) with the board suit not being one of the player's suits.

**The type (0,0,1,4).**- First assume the 2 player's suits are the same
as the 2 suits in the board. There are 2 choices for which of the player's
suits is 4 suited in the board,
choices for the 4
cards of that suit, and 12 choices for the remaining card. This gives us
boards of type (0,0,1,4) where both suits in
the board agree with the player's 2 suits.
Next assume the 4 suited cards in the board 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 2 choices for the suit of the 4 suited cards, choices of the 4 cards, 2 choices for the suit not in the player's hand, and 13 choices for the remaining card. This yields boards for this subcase.

Now switch the roles of the suits in the board with respect to the player's suits. We have 2 choices for the suit of the 4 suited cards, choices for the 4 cards of that suit, 2 choices for the suit of the remaining card, and 12 choices for a card of this suit. We obtain boards for which the 4 suited cards are not one of the player's suits and the other card is one of the player's suits.

Finally, we assume none of the suits in the board agrees with any of the player's suits. This gives us 2 choices for the suit of the 4 suited cards, choices for the 4 cards of that suit, and 13 choices for the remaining card. This gives us boards of type (0,0,1,4) whose suits do not overlap either of the player's suits.

**The type (0,0,2,3).**- There are 4 subcases here just as for the
previous type. So first we assume the player's suits and the board's
suits are the same. There are 2 choices for which of the player's suits
get 3 suited cards in the board,
choices for the
3 cards of that suit, and
choices for the 2 cards
of the other suit. This produces
boards
for this subcase.
Next assume the player's suits and the board's suits agree in the suit that has 3 suited cards. There are 2 choices for the suit of the 3 suited cards, 2 choices for the suit not amongst the player's suits, choices for the 3 cards in the one suit, and choices for the 2 cards of the other suit. This gives us boards for this subcase.

The next subcase is the player's suits and the board's suits agreeing in the suit with 2 suited cards. There are 2 choices for the common suit, choices for the 2 cards in the suit, 2 choices for the other suit, and choices for the 3 cards in the other suit. This produces boards for this subcase.

Finally, there is the subcase in which the board's suits and the player's suits do not overlap. There are 2 choices for the suit in which there are 3 cards, choices for the 3 cards, and choices for the remaining 2 cards. This yields boards for this subcase.

**The type (0,1,1,3).**- First we consider the case the player's suits
agree with the 3 suited cards and one of the other cards in the board.
There are 2 choices for the suit of the 3 suited cards,
choices for the 3 cards, 12 choices for the card in the player's
other suit, 2 choices for the suit of the remaining card in the board, and
13 choices for that card. This gives us
boards for this subcase.
Next we consider the subcase in which the 3 suited cards agree with one of the player's suits and the other 2 cards in the board have no suit in common with the player. We have 2 choices for the suit of the 3 suited cards, choices for the 3 cards, and 13 choices for each of the other 2 cards. This produces boards for this subcase.

Another subcase is when the player's 2 suits agree with the 2 suits in the board which are single suited. There are 2 choices for the suit of the 3 suited cards, choices for the 3 cards, and 12 choices for each of the other 2 cards. We have boards for this subcase.

The final subcase is when the player's 2 suits agree with only one of the single suited cards in the board. There are 2 choices for the suit of the 3 suited cards, choices for the 3 cards, 13 choices for the card of the other suit not in the player's hand, 2 choices for the remaining suit, and 12 choices for that card. This produces boards for this subcase.

**The type (0,1,2,2).**- Suppose the two double suited cards agree
with the player's suits. There are
choices for
each of the 2 cards, 2 choices for the suit of the remaining card, and
13 choices for that card. This gives
boards for this subcase.
Next suppose 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. This yields 2 choices for the suit of the double suited card which agrees with a player's suit, choices for the 2 cards, 12 choices for the card agreeing with the player's other suit, 2 choices for the suit of the other double suited cards, and choices for the 2 cards. Altogether there are boards for this subcase.

The next subcase is one 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. There are 2 choices for the suit in agreement with one of the player's suits, choices for the 2 cards, 2 choices for suit of the other double suited cards, choices for the 2 cards, and 13 choices for the remaining card. This produces boards for this subcase.

The final subcase is when neither of the suits of the 2 double suited cards agrees with either of the player's suits. This gives us choices for each of the double suited pairs, 2 choices for the suit of the single suited card, and 12 choices for the latter card. We have boards for this subcase.

**The type (1,1,1,2).**- If the double suited cards agree with one of
the player's suits, There are 2 choices for the suit,
choices for the 2 cards, there are 13 choices for each of the cards
not in the player's other suit, and 12 choices for the remaining card.
This yields
boards for this
subcase.
Finally, if the player's 2 suits agree with 2 of the single suited cards in the board, there are 2 choices for which of the player's non-suits is double suited in the board, there are choices for the 2 cards, 13 choices for 1 of the other cards, and 12 choices for each of the other 2 cards. This gives us boards for this final subcase.

If we sum all of the above numbers we obtain 2,118,760 as required.
There are 3 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 2 unsuited cards making a flush is given by

There are 11 subcases which result in a flushing board such that the
player with 2 unsuited cards does not have a flush. This sum is

2,574+34,320+18,590+29,040+68,640+75,504+44,616+137,280+74,360+82,368+
178,464 = 745,756.

The probability of having a flushing board in a suit different from either suit held by a player with 2 unsuited cards is

Altogether there are 784,960 flushing boards for a player holding 2
unsuited cards. The probability of a flushing board occurring in this
case is

The reason the two probabilities do not sum to 0.370 is due to round-off error but it is occurring in the third decimal place and is not significant.

Home | Publications | Preprints | MITACS | Poker Digest | Graph Theory Workshop | Poker Computations | Feedback

website by the Centre for Systems Science

last updated 10 January 2000