Odds of 3 out of 4 getting flush

[aloiz]

In my home game before starting we always deal an open face hold'em hand to determine dealer. On one occasion there was four of us playing, and after dealing the hand three of us were suited in clubs. The flop comes down with two clubs, and the river brought the third giving three of us the flush. We wondered what the odds were so I took a stab.

Assuming the following. Four hands dealt and everyone in to the river. Only three of the four can make the flush (not all four). The board may only contain three of the flush cards.

I took a shot at it a came up with something, but I'm sure there is a better way to do it.

Case 1: Fourth player's hole cards contains one of the flush cards)
C(4,3)*4*C(13,6)/C(52,6) * C(7,1)*C(39,1)/C(46,2) [hole cards] * C(6,3)*C(38,2)/C(44,5) [board]

Case 2: Fourth player has none of the flush cards
C(4,3)*4*C(13,6)/C(52,6) * C(39,2)/C(46,2) [hole cards] * C(7,3)*C(37,2)/C(44,5) [board]

Adding both cases I get 2.533E-5

Is this right? Can anyone show me a better way to do it?

Thanks,
aloiz

[BruceZ]

[ QUOTE ]
Assuming the following. Four hands dealt and everyone in to the river. Only three of the four can make the flush (not all four). The board may only contain three of the flush cards.

I took a shot at it a came up with something, but I'm sure there is a better way to do it.

Case 1: Fourth player's hole cards contains one of the flush cards)
C(4,3)*4*C(13,6)/C(52,6) * C(7,1)*C(39,1)/C(46,2) [hole cards] * C(6,3)*C(38,2)/C(44,5) [board]

Case 2: Fourth player has none of the flush cards
C(4,3)*4*C(13,6)/C(52,6) * C(39,2)/C(46,2) [hole cards] * C(7,3)*C(37,2)/C(44,5) [board]

Adding both cases I get 2.533E-5

Is this right? Can anyone show me a better way to do it?


[/ QUOTE ]

It is right, and I can show you a better way. You can avoid having two separate cases if you consider the board first.

4*C(13,3)*C(39,2) / C(52,5) * [ C(4,3)*C(10,6)/C(47,6) - C(4,3)*C(10,8)/C(47,8) ]

= 2.533E-5 = 1 in 39,480.

The first term is for the board. The term in [] is for the hole cards. The first player term multiplies the probability of 3 players being dealt 6 suited hole cards by the number of ways to pick the 3 players C(4,3). This counts the cases where all 4 players have a flush C(4,3) times, so this is subtracted in the last term. This is the inclusion-exclusion principle again, except in this case we start with the probability of 3 players having flush cards. In this way it is not necessary to explicitly consider how many flush cards the 4th player has.

[aloiz]

Thanks, that makes sense. I wish I would of thought of that as it makes everything ten times easier, and much less prone to a stupid mistake. Also thanks for the link, great read.

aloiz