Brain dead need help a sec

[shutupndeal]

/images/graemlins/confused.gif Can someone show me the forumla for how many times you will flop your suit for a four flush on the flop?
I know its easy but Im going blank here!

[DavidC]

Pretty sure it's something like this:

52 card deck, 2 dealt to you, 50 cards remaining.

Possible deals ?? (50*49*48)/(3*2*1) = PD

Cards in a suit: 13
Cards in your hand: 2
Cards left in your suit: 11

You want to be dealt ?? (11*10*9)/(3*2*1) = WD

And I think that leaves you with WD/PD as your chance.

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This is my best guess, and it's based on the fact that you can get a flop any of 6 different ways with the same three cards (the cards could be "shuffled" differently). If there's a mistake, someone please help me out.

[ohnonotthat]

There are 19,600 flops.

You will flop a made flush 165 times . . . (11x10x9) / 3!.

You will flop a flush draw 2,145 times . . . (11x10x39) / 2!.

You'll flop a flush OR flush draw 11.47% of the time - approximately once in 8.5 times.

[ QUOTE ]
There are 19,600 flops.

You will flop a made flush 165 times . . . (11x10x9) / 3!.

You will flop a flush draw 2,145 times . . . (11x10x39) / 2!.

You'll flop a flush OR flush draw 11.47% of the time - approximately once in 8.5 times.

[/ QUOTE ]

Please can you give the details of the factorial equation you used here.

[ohnonotthat]

The flopped flush is the easier of the two.

There are 50 unseen cards, so . . . 50x49x48/3! = 19,600 (possible flops)

There are 11 cards of your suit; you need three to flop, so . . . 11x10x9/3! = 165


*

The figures for the flopped draw are done the same way but here we need 2 of your suit and one NOT of your suit (the inclusion-exclusion principle); if you don't do it this way you'll double count the flopped flushes. (In other words, you'll count them as both made hands AND flopped draws).

PM me if you want it explained in greater detail.

[ohnonotthat]

You got it.

[shutupndeal]

Thanx so much!