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Dave H.
11-15-2004, 09:55 AM
I know there are 169 starting hands in holdem. Can someone tell me how to compute that (in elementary terms)?

Is AKo the same as AKs, i.e. is that considered ONE of the 169 possible hands or TWO of the possible starting hands?

Thanx much!

Lost Wages
11-15-2004, 10:13 AM
There are 13 ranks, hence 13 unique pairs. A non-pair hand can have one of the 13 ranks for it's first card and one of the 12 remaining non-pairing ranks for its second card so there are (13*12)/2 = 78 non-pair hands. We divide by two to eliminate permutations since we are only interested in unique combinations, e.g. JT is the same hand as TJ. A non-pair hand can either be suited or unsuited so there is a total of 13 pairs + 78 suited non-pairs + 78 unsuited non-pairs = 169 unique hands.

Lost Wages

Dave H.
11-15-2004, 11:06 AM
Thanks so much!

Dave H.

DoubleupDinan
11-15-2004, 02:21 PM
Good question. I was wondering the same thing.

This has probably been discussed before(so I apologize in advance) but I was also wondering if there are any really good books to read up on to learn about probability?