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Mr. Curious
09-30-2005, 04:02 PM
Three all-ins on a board of AQT:

Hand 1: AQ
Hand 2: AT
Hand 3: QT

For the flop to come up AQT, the probability is 2/46 * 2/45 * 2/44 = .000087834

How do I adjust the odds of the hands to figure out the total likelyhood of the board + the hands?

Is it ((4/52 * 4/51 AQ) * (3/50 * 4/49 QT) * (3/48 * 3/47 AT)) + (2/46 * 2/45 * 2/44)?

Thanks!

09-30-2005, 04:13 PM
Actually:
You don't care about the order of the cards on the flop or in the player's hands.
With 6 known cards, there are 46 choose 3 = 15180 possible flops, and you wanted to see one of 2*2*2=8 of those so the probability is 8/15180=.000527...

Now, assuming we don't care about the order of the hands, there are:
52 chose 2 * 50 chose 2 * 48 chose 2 *46 chose 3/6
possible ways to get 3 hands and a flop. And, there are
4*4*4 ways to get an AQT flop
3*3 ways to get an AQ hand provided the flop is AQT
2*3 ways to get a QT hand provided the flops is AQT and there is a AQ hand, and
2*2 ways to get a AT hand provided everyhing abouve is true.
So that's
13824 in 4635635004000

Mr. Curious
09-30-2005, 07:06 PM
[ QUOTE ]
13824 in 4635635004000

[/ QUOTE ]

Or pretty friggin' unlikely /images/graemlins/tongue.gif

Thanks for the info!