Idiot Question

[Freerollin`]

Hi folks,

I became a lawyer so I didn't have to do any serious math. My inability to find this answer in the search doesn't say much for my lawyering abilities...

In hold 'em, am I twice as likely to be dealt a given offsuit hand, say 83o, than a given pocket pair, such as KK?

I'm doing some review of my hand distributions in Poker Tracker (no I don't expect to get AA a certain amount of times in the next 25K hands, though it would be nice).

Thanks for the help.

[mrgold]

Yes. 6 combos of each pocket pair for each 12 combos of an offsuit hand for each 4 combos of a suited hand.

[AaronBrown]

You can get a pair in six ways. The first card can be any of the four suits, the second can be any of the remaining three; you divide by 2 because K/images/graemlins/spade.gif K/images/graemlins/heart.gif is the same hand as K/images/graemlins/heart.gif K/images/graemlins/spade.gif. 4 x 3 / 2 = 6.

You can get a non-pair in 16 ways. The first card can be any of four suits, so can the second. You don't divide by 2 because K/images/graemlins/spade.gif 6/images/graemlins/heart.gif is not the same hand as K/images/graemlins/heart.gif 6/images/graemlins/spade.gif. 4 x 4 = 16.

Those 16 non-pairs can be suited in any of four suits (4 ways). The remaining 12 ways (16 - 4) are unsuited.

So, yes, this is a long-winded way of saying you are twice as likely to get a given unsuited, unpaired hand than a given pair.

Since there are only 13 pairs but 13*12/2 = 78 unpaired hands, you are much more likely to an unpaired, unsuited hand than a pair. Since there are six times as many unpaired hands as paired hands, and each unsuited, unpaired combination has twice as many combinations as each pair, you will get 12 unsuited, unpaired hands for every pair.