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Wisco Kidd
03-12-2004, 02:33 PM
So I just started reading Winning Low Limit Holdem and one of the odds he puts out there is if you hold AK in the pocket you have a 2.1:1 of hitting either an A or K on the flop. Math is not my strong suit, so can someone please expain how this is the case. I understand that, there are 3 Aces left and 3 Kings left out of 50 possible cards. So the odds on the first card to hit the flop is 44:6 (is that much correct?) and the second card on the flop will be 43:6 and the third card is 42:6. How do those work out to be 2.1:1? Thanks for any help you can offer.

pudley4
03-12-2004, 02:58 PM
44:6 = 6/50
43:6 = 6/49
42:6 = 6/48

If you want the probability of exactly one A or one K on the flop it's:

(6/50 * 44/49 * 43/48) * 3 = 2.45:1

(The probability of hitting on the first and then missing on the next two. We don't care what order it comes in, so we multiply by three - Hit, Miss, Miss; MHM;, MMH)

If you want the probability of at least one A or K it's:

1 - (44/50 * 43/49 * 42/48) = 2.1:1

(This is one minus the probability of missing all three times)

Wisco Kidd
03-12-2004, 03:14 PM
I'm looking at the numbers, but I just don't get it.
Your over my head, can you dumb it down a little /images/graemlins/blush.gif

Wisco Kidd
03-12-2004, 04:12 PM
never mind, I figured it out. Thanks for the help.