O great minds -- flush Q

[gergery]

How often will a flush be possible on a five card board?

When playing Omaha, how often will 1 of your opponent have 2 flush cards that match the board, if you are against 3 opponents (assume 3-flush on board)?

I’m more interested in methodology than the answer, but I’d love both.

--thx,

Greg

[Paul2432]

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How often will a flush be possible on a five card board?

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This question can be answered using combinations. I am assuming for this question no knowledge of hole cards. That is prior to the deal, will the upcoming board contain three cards of the same suit.

The number of boards possible is 52C5 (in MS Excel this would be =combin(52,5) )

52C5 = 2,598,960

A three flush on board occurs whenever any combination of three flush cards are combined with any combination of two side cards. In other words, the number of three flush boards of a particular suit is 13C3 x 39C2. This number then needs to multiplied by four because there are four suits.

No of three flush boards = 4 x 13C3 x 39C2 = 847,704

Hence the frequency of a three flush board is 847704/2598960 = 32.6%.

If you want to include four flush boards and five flush boards, the approach is similar.

Note that this approach will not work for two flush boards. This is because you will double count all boards that contain two two flushes.

Also note if you want to assume particular hole cards for two players, simply adjust the number of flush cards from 13 to the number remaining in the stub, similarly for side cards.

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When playing Omaha, how often will 1 of your opponent have 2 flush cards that match the board, if you are against 3 opponents (assume 3-flush on board)?

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This question cannot readily be answered because it depends on how people play. An answer could be derived assuming noone ever folds, but that would not be very useful.

Paul

[gaming_mouse]

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[ QUOTE ]
When playing Omaha, how often will 1 of your opponent have 2 flush cards that match the board, if you are against 3 opponents (assume 3-flush on board)?

[/ QUOTE ]

This question cannot readily be answered because it depends on how people play. An answer could be derived assuming noone ever folds, but that would not be very useful.


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A good point, Paul. However, we can obtain an approximation by, eg, assuming that our opponents will play any hand in any of the groups of the HPFAP starting charts. We'd then go through those groups and count what percent of those starting hands are suited. Call this percentage x. Then the chance that an opponent has suited cards that match the board is x/4. Since the opponent's hands are only loosely dependent, we can now assume independence to get out approximate answer:

1 - (1 - x/4)^3

More refined approximations can be obtained by writing a program that executes various preflop systems -- eg, you might write one that plays the HPFAP system, another that plays a looser system, etc. Then decide the compositoin of your field -- eg, 30% loose, 20% tight, 50% average -- and run a simulation.