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08-23-2002, 06:07 PM
Question I should know the answer to, but here it is anyway. What is the proceedure for figuring out drawing odds with two cards to come?

For some reason I thought that the probabilities were determined by combining each street and dividing by 2, but this is not coming out correctly. Thanks for a clarification.

08-23-2002, 10:34 PM
Example 10 outs:47 unseen cards. With 2 to come there are 47*46/2*1 combinations of cards=1081. Chance of hitting your card twice = 10*9/2*1=45 ways. Chances of hitting your card once( hit/miss combo)is 10 hits * 37 miss(47cards-10 hits)=370. Add 370 hit/miss combos + 45 Hit/Hit combos =415. Odds =1081 Total-415 (combinations which include at least one of your outs):415 or 1081-451:415=666:415 = 1.6:1

08-24-2002, 03:29 AM
Thanks Steve, I knew I was missing a step.

08-24-2002, 09:50 AM
If we use Steve's reasoning for n outs instead of 10 outs we get:

(C[n, 2] + n*(47 - n))/C[47, 2]

= (93 - n)*n/2162.

Which is petty close to 4% per out.