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shagjohn
01-29-2005, 01:04 AM
Hope this isn't a newb q, but I looked through the other pages of posts and didn't find anything.

In the hopes of understanding the odds better, just wanted to verify I'm doing this right...
On pg. 107 of Hold'em, Sklansky has a table of the probability of hitting your outs (assuming you only need one) on the turn or the river.

Is this math right? Or harder than it needs to be?

x = (47 - outs) Choose 2

# ways to see turn and river (not in order) = 47 Choose 2, which is 1,081

prob of hitting 1 of the outs = (1,081 - x) / 1,081
?

Thanks!

kodonnell
01-29-2005, 01:52 AM
Straining to remember statistics from a bazillion years ago, that looks right. However, it is harder than it needs to be. The way I do it is:
outs x 4 after the flop
outs x 2 after the turn
After 10 outs it starts to skew a bit but it gets you in the ballpark.