Hold 'em prob. calculations

[BogusPomp]

I am trying to teach myself how to do some of these calculations.... and still not sure I'm doing them properly

Would someone/anyone mind checking these answers for me. Much obliged.

First-- Chances of flopping 2nd nut flush only to have nut flush out:

Dealt Kx suited= 4*12/(52c2)
If holding Kx suited, flopping flush= (11c3)/(50c3)
Any one of 9 opponents is holding Ax of this suit = 9 * (7/[47c2])

This event will occur to any one person on any particular hand approx 1 in 56300 (assuming no folds)


Next---Chances dealt AA and flopping full-house, only to have quads out:

Dealt AA = (4c2)/(52c2)
Flop comes either remaining Ace plus any pair = 2 * [12 * (4c2)]/ ( 50 c 3)
Any one of 9 opponents holds quads with flopped pair = 9/(47c2)

This event will occur to any one person on any particular hand approx 1 in 3 613 008 (assuming no one folds)

thank you

[AaronBrown]

Not quite. First off, when you do the Kx suited, you have to multiply by 11, not 12, because x cannot be the A.

But a deeper problem is you have to start with the flop. It will be suited with probability 4*C(13,3)/C(52,3) = 22/425.

You must have the second highest outstanding suited card (not necessarily the K), plus one of the 8 other suited cards that is not the highest outstanding. That's 8/C(49,2) = 1/147.

One of the other 9 players must have the highest outstanding suited card, plus one of the 7 remaining. That's 63/C(47,2) = 63/1,081.

Multiply these together and get 22*1*63/(425*147*1,081) = 66/3,215,975. That's about 1 in 48,700 rather than 56,300.

The second problem is perfect.

[BruceZ]

[ QUOTE ]
I am trying to teach myself how to do some of these calculations.... and still not sure I'm doing them properly

Would someone/anyone mind checking these answers for me. Much obliged.

First-- Chances of flopping 2nd nut flush only to have nut flush out:

Dealt Kx suited= 4*12/(52c2)
If holding Kx suited, flopping flush= (11c3)/(50c3)
Any one of 9 opponents is holding Ax of this suit = 9 * (7/[47c2])

[/ QUOTE ]

Your Kx suited calculation includes x = A which would give you the nuts if you flop a flush (barring any str8 flushes). The Kxs can also be the nuts if the A flops.

It is also possible to flop a 2nd nut flush without holding the K if the A or K flops, or if a str8 flush is possible.


[ QUOTE ]
Next---Chances dealt AA and flopping full-house, only to have quads out:

Dealt AA = (4c2)/(52c2)
Flop comes either remaining Ace plus any pair = 2 * [12 * (4c2)]/ ( 50 c 3)
Any one of 9 opponents holds quads with flopped pair = 9/(47c2)

This event will occur to any one person on any particular hand approx 1 in 3 613 008 (assuming no one folds)

[/ QUOTE ]

This is fine. It can also happen if you flop 3 cards of the same rank, and someone holds the 4th.