Jeff,

is there a hand that AAAKK can beat that AAAQQ can not beat?
remember if you hold AAAQQ it is not possible that another player holds AAAKK since there are only 4 aces in the deck!

I claim that their strength as poker hands is identical.

Nir, that is true. AAAQQ can beat any hand that AAAKK can beat.

I understand the problem.

But above you say that sixes to nines are the best to have to be assured of not being beaten. Why? AAA with any pair whatsoever assures me of the best possible full house. Doesn't it?

This math stuff worries me. Maybe I should stop playing poker without cheating.

Jeff my friend,

I just claimed (as did others - please read Scott's or Tomas' posts) that AAA66 is stronger than AAAKK because it blocks more possible straight flushes for your opponents.

The advantage is however so minute that it doesn't really matter...
Nir

I read those other posts. But I was drunk, so I didn't catch on.

So I just went back and read Scott's post. I think I've got it now.

The two worst beats I ever took were when I was holding a full house, so I figure the math isn't going to help me anyway. I get blind and goofy and can't think straight, especially if others are holding good hands and throwing chips into the pot. One time I was actually convinced I had wiped everyone at the table out and STOOD UP TO GO HOME! (This is a true story.) Then one of the guys showed me four jacks and I seriously considered eating my hat.

Was it hold'em?
in hold'em if the board is paired and you got a full house (no other way) it can very well be that others hold 4 of a kind.
happened to me as well one time.

Nir, it was indeed Hold 'em. I KNEW he was holding a jack, and I played accordingly. I played the hand beautifully.

Except he was holding not one jack but two.

Maybe I should eat my hat right now. It makes me feel stupid again just telling the story.

Here's a different tack on this problem for anyone who wants to try it out.

With the answer I gave above, 16 out of 40 possible SFs are blocked (leaving 24 SFs that can beat yours), and 2 out of 13 possible FOAKs (Four-of-a-kinds) blocked (leaving 11 FOAKs that can beat you). With the answer above, there are only 35 hands that can beat you.

Let's say that, in the situation above, I've decided that I choose my Full House to be Tens over Fives.

This seems to be much better than the previous answer, doesn't it? The Tens stop anyone else from getting 3 RFs, 3 King-high SFs, 3 Queen-high SFs, 3 Jack-high SFs, 3 Ten-high SFs, and the Fives stop anyone from getting 2 Five-high SFs, 2 Six-highs, 2 Seven-high SFs, 2 Eight-high SFs, 2 Nine-high SFs.

This gives us a total of 25 SFs blocked! That leaves only 15 SFs that can beat you! Again, there are still only 11 FOAKs that can beat you.

Since there are 3 Tens in the hand, any Full House with 3 Jacks, Queens, Kings, or Aces will beat it, but that's only 4 additional Full Houses that can beat you.

11 FOAKs + 4 Full Houses + 15 SFs gives a total of only 30 hands that can beat you! That's fewer than the 35 hands that could beat you with Aces over Sixes through Nines!

So, have I discovered a previously unknown and better answer to this question? Or am I just trying to pull the wool over your eyes? If you think I'm trying to pull a fast one, prove it by explaining the fallacy in my argument.

"Since there are 3 Tens in the hand, any Full House with 3 Jacks, Queens, Kings, or Aces will beat it, but that's only 4 additional Full Houses that can beat you."

There are many more than 4 Full Houses that consists of those cards.

/Tomas