Brian Alspach

Poker Digest Vol. 2, No. 8, April 9 - 22, 1999

We now finish our discussion about probabilities associated with flushes by looking at a player who has a flush made and is worried about other players perhaps having a flush as well. There are many settings for this so let us consider a player who has two suited cards in her hand and is facing a board with exactly three cards in her suit. The reasoning behind these calculations is similar to what we condsidered in the last issue so we shall outline part of the computation and leave the rest to the reader.

There are eight flushing cards (cards of the same suit as the three suited cards in the board) unaccounted for amongst the 45 unseen cards. Thus, there could be as many as four other players with a flush. We shall see just how likely this is.

The remaining nine players receive 18 cards so we break the sets of cardinality 18 into parts according to how many cards of the flushing suit are in the set. Among the 45 unseen cards there are eight cards of the flushing suit and 37 not in the flushing suit. This means there are C(37,18)= 17,672,631,900 sets of 18 cards not having any card of the flushing suit. There are 8C(37,17) = 127,242,949,680 sets of 18 cards having precisely one card of the flushing suit. There are C(37,16) C(8,2) = 360,521,690,760 sets of 18 cards having precisely two cards of the flushing suit. Continuing in this way, there are 395,186,560 sets of 18 cards having precisely three cards of the flushing suit, 496,076,000 sets of 18 cards with precisely four flushing cards, 199,498,168,800 sets with precisely five flushing cards, 51,869,523,888 sets with precisely six flushing cards, 6,839,937,216 sets with exactly seven flushing cards, and 348,330,136 sets with exactly eight flushing cards. Summing all these numbers yields C(45,18) as it should.

We outline how to handle the case of exactly one other player having a flush. The first subcase is when there are two flushing cards among the 18 cards comprising the semi-deal. They must go to one hand and the remaining 16 cards are partitioned in 15!! = 2,027,025 hands (see previous articles to see where the 15!! is arising). Multiplying by the number of sets of 18 cards with two flushing cards, we obtain 730,786,480,212,789,000 semi-deals with one other player having a flush in this subcase.

Suppose there are three flushing cards among the 18 cards. Choose two to go into one hand and partition the remaining 16 cards in 15!! ways. This gives us 6,081,075 semi-deals with a flushing hand. Now multiply by the number of sets of 18 cards with three flushing cards to obtain 3,188,886,459,110,352,000 semi-deals for this subcase.

Now we move to four flushing cards among the 18 cards. We choose two cards from the four to go into the same hand, 14 and 13 cards to join the other two flushing cards, respectively, and partition the remaining 12 cards into hands. This gives us . Multiply this by the number of sets of 18 cards with four flushing cards to get 4,852,653,307,341,840,000 semi-deals for this case of exactly one other player having a flush.

When there are five flushing cards, we choose two of them to go in the same hand, choose three cards in succession from the non-flushing cards to join the other flushing cards, and partition the remaining 10 cards into hands. This can be done in ways. We multiply by the number of sets of 18 cards with five flushing cards and get 3,235,102,204,894,560,000 semi-deals for this particular subcase.

If there are six flushing cards, there are 15 ways to choose two of them to go into one hand. We then have ways to choose cards to join with the leftover flushing cards so that no more hands have two flushing cards. Finally, there are 7!! = 105 ways to partition the remaining eight cards into four hands. This gives us 18,711,000 semi-deals from the given set of 18 cards which produces exactly one person with a flush. Now we multiply by the number of sets of 18 cards having six flushing cards and obtain 970,530,661,468,368,000 semi-deals for this subcase.

Moving to seven flushing cards, we have 21 ways to choose two cards from the seven, and the product for the number of choices of cards to join the flushing cards. There are 5!! = 15 ways to partition the remaining six cards. Altogether this gives 17,463,600 semi-deals from the given set of 18 cards. Multiplying by the number of sets of 18 cards with seven flushing cards yields 119,449,927,565,337,600 semi-deals for this case.

The last subcase involves eight flushing cards. There are 28 ways to choose two cards from the eight to go in the same hand. We then choose cards to join the leftover flushing cards. There are three ways to partition the remaining four cards. We obtain 12,700,800 semi-deals for the given set of 18 cards. Multiplying by the number of sets of 18 cards with eight flushing cards, we obtain 4,424,071,391,308,800 semi-deals for this final subcase. Summing all of these numbers produces 13,101,833,111,984,555,400 semi-deals for which precisely one other player has a flush. Dividing by the total number of semi-deals yields a probability of 0.222 for exactly one other player having a flush.

Arguing in a smilar fashion, we find the probability of exactly two other players having a flush to be 0.0161, the probability of exactly three other players having a flush to be 0.000286, the probability of four other players to be 0.00000058, and the probability of no one else having a flush to be 0.762.

Bib Ladder then said, Well, professor, I find the numbers you just showed me interesting, but is knowing them going to do anything for my game?''

Bib, even though I am a mathematician, I do not inflate the importance of mathematics in poker. There are some mathematical aspects of the game which are important, but some of the topics we have discussed are mathematical curiosities and little more than that. I believe people skills are far more important in terms of being successful. For example, there are some people who frequent this club who are in love with any two suited cards. If they check and call to the river and then come out firing when the river card produces a third card in some suit, you can be almost certain they have made a flush. You should play accordingly and throw away top two-pair or trips or a straight. There is no need to emulate others who will say, I know you have a flush, but I'm going to call'' and then get shown the flush. The kind of player I am talking about is not clever enough to camouflage a weaker hand with the preceding betting pattern in order to try to induce another play to fold after the river.

On the other hand, given the right mix of players left after the flop and with two suited cards in the flop, I sometimes have check-raised trying to convince any callers I am on a flush draw. If the third suited card shows up on the river, a bet sometimes will get the remaining players to fold. You can not do this often and only when the circumstances are appropriate. This kind of play has little to do with probability and much more to do with knowing the players.''

So can I use these numbers in any way,'' asked Bib?

We saw last time that the probability of someone making a flush given a board with three suited cards and neither of your two cards are in that suit is 0.3605. What does this number mean? Remember how we calculate it. We take all possible semi-deals in which one or more hands has two cards of the flushing suit and divide by the total number of semi-deals. This is equivalent to assuming all players stay until the river card, but that is not what happens in a real game. The texture of the game you are in will have a dramatic effect on the proportion of flushes.

If the game is fairly tight, many of the semi-deals which would have led to a flush because of someone having two weak suited cards will not produce a flush because such hands will be discarded pre-flop by many of the players. Hence, the actual proportion of flushes will be substantially smaller than 0.3605. On the other hand, if the game is loose, the proportion will be near 0.3605.

So, Bib, if you are in a tight game with trips made, a flushing board with three suited cards, and an aggressive opponent, who is capable of putting a move on you, coming after you, the fact that the odds against a flush, had everyone stayed in, are slightly less than 2:1 is not going to be much help. You will have to respond to your opponent according to your knowledge of her and the likely hands you put her on.''

I know knowledge of my opponents is extremely important. Still, I enjoy now knowing something about the likelihood of flushes.''

And I enjoyed working them out, Bib. See you later.''

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