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WiredCrabs
11-19-2004, 02:16 AM
Situation, I am holding K10 offsuit...

Playing at a 10 handed table, 4 handed after the flop of 2-4-5 rainbow...having over cards, and knowing there are 47 unseen cards in the deck, to find out the prob of hitting a K or 10...would I take the 3 remaining 10's + 3 remaining Kings and divide that by 47, multiply by 2 for the turn and the river? so (3+3/47*2)= .255 = 25%??

schroedy
11-19-2004, 02:27 AM
Not exactly. Sklansky has a section on how to do this in one of his books. Better to just look it up in tables somewhere.

The chances that you will hit at least once = 1 minus the chances that you miss both times. 1 - (41/47 (odds first card misses you) * 40/46 (odds second card misses you after first card missed you)) = approximately what you said but not exactly.

(Warning: I use the table/memory method)

I know this is blasphemy around here, but Ken Warren's very basic Winning Texas Hold Em has good tables of this sort.