Poker odds according to Sklansky

[MrBeeks]

In Hold 'em, lets say you hold two cards pre-flop, and the flop comes so that you are drawing to an open ended straight flush. (ex. you hold 8h 9h, with a flop of Th Jh 2c). OK, so you quickly calculate that you have 15 outs to win the hand ( 9 hearts, 3 7's and 3 Q's, not counting the 7h and Qh twice). Therefore, you think that the odds of making either a straight or flush to be: 15/47 + 15/46 = 30/46.5 or roughly 64.5%. However, according to Sklansky in Appendix A of his book "Hold 'em Poker for Advanced Players", he says the correct odds are 54.1%. He doesn't explain how he arrived at this number. Can you explain?

[MickeyHoldem]

it's not 15/47 + 15/46

its (15/47) + (32/47)*(15/46) = 0.541165587

the second term must be multiplied by the chance you miss the turn

[MrBeeks]

ok.. thanks.. that makes sense. However, it seems that your answer explains the correct odds of making your hand by drawing to one of the 15 cards you need exactly once ( either on the turn or the river, but not both). In the situation where your considering an all-in bet post flop, and you want to know the odds of hitting one of your outs with either or both the turn and river cards, wouldn't you have 30 outs total (15 outs twice)?

[reubenf]

[ QUOTE ]
However, it seems that your answer explains the correct odds of making your hand by drawing to one of the 15 cards you need exactly once

[/ QUOTE ]

No, I think that would be (15/47)*(32/46) + (32/47)*(15/46). The first term is the chance you hit then miss, the second the chance that you miss then hit.

There are three ways you can hit at least one: Hit then miss, miss then hit, or hit then hit. Both "HM" and "HH" are accounted for by the first term Mickey gave, and "MH" is accounted for by the second term.

[MrBeeks]

thanks... i see it now..

[olavfo]

The chance of missing both turn and river is

(47-15)/47 * (46-15)/46 = 0.459

and so the chance of hitting on either turn or river is

1 - 0.459 = 0.541 = 54.1%

olavfo