**Poker Digest Vol. 3, No. 18, August 25 - September 7, 2000**

In the previous issue of *Poker Digest*, I initiated a series to
compare probabilities of at least one player having a flush or straight
in seven-card stud and hold'em. The groundwork was established, and we
now are ready to examine the flush probability for hold'em. Presently we
are making the important assumption that we have a ten-handed game and
all players stay until the end no matter what they hold.

We are interested in the probability of at least one player making
a flush and not which player gets a flush. We are going to work
from the suit structure of the possible boards. Altogether there
are
*C*(52,5) = 2,598,960 possible boards. Of these, 5,148 consist
of five cards in the same suit (including straight flushes); 111,540
have four cards in the same suit; 267,696 have three cards in one
suit and two cards in a different suit; and 580,008 have three cards
in one suit and one card in each of two other suits.

Details on the
derivation of these numbers can be found at my web site (http://www.math.
sfu.ca/~alspach) under Poker Computations. They also are discussed in
Vol. 2, No. 4 of *Poker Digest*.

If there is a flush on board, then everyone has a flush. Of course, one or more players may also have cards in the flushing suit so that there may be only one winner. The probability of a flush on board is obtained by dividing 5,148 by 2,598,960. This yields .001981.

As stated above, there are 111,540 boards with four cards in one suit and one card in a different suit. If any player has a card in the suit represented by the four suited cards in the board, then at least one player has a flush. To determine the probability of at least one player having a card of the magic suit, it is easier to determine the probability that no one has a card of that suit and then subtract the latter probability from one.

The simplest way of calculating the probability that no one receives
a card of the magic suit may seem slightly devious, but if you think
about it, you will see that it works. We are going to deal 20
cards from 47 to the 10 players. This means there are
*C*(47,20) possible different sets of 20 cards to be distributed.
If any card of the magic suit is among the 20 cards, someone
has to get it leading to that someone having a flush. Thus, the only
way we can avoid having a flush is if all 20 cards are chosen from
the 38 remaining cards not in the magic suit. There are
*C*(38,20) such choices. So the probability of no one having a flush
is obtained by dividing *C*(38,20) by *C*(47,20). Doing the messy
arithmetic yields a probability of .003439 that no one has a flush.

Thus, there is a probability of .996561 that someone has a flush given four suited cards on the board. The probability of four suited cards in the board is obtained by dividing 2,598,960 into 111,540. Altogether we get a contribution of .042770 towards the probability of a flush occurring.

There are 847,704 boards containing precisely three cards from the same suit; that is, almost one-third of the time there will be three cards in the same suit on board. There will be a flush only if some player has both cards in the magic suit. There are several ways to determine the probability of at least one player having two cards in the magic suit, but I shall not go into the details here. I shall post them at my web site. It turns out the probability of a flush occurring given that there are precisely three cards in the same suit on board is .365617. The probability of a board with precisely three suited cards is obtained by dividing 2,598,960 into 847,704 and then multiplying by the number in the preceding sentence. This gives us .119253 as the final contribution for the probability of a flush occurring.

Adding the three probabilities .001981, .042770 and .119253, we obtain .164 as the probability of at least one player having a flush under the conditions of randomly dealing 10 two-card hands, followed by a random selection of five board cards. Let me point out that this calculation is an exact calculation and not a result of random trials. Also, straight flushes have been included in the counting.

We shall continue with a similar calculation for seven-card stud in the next issue.

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