[binions]

jay

If 6 outs (ie 6 ways to make a pair) is worth about 4%, can we say that with AK, there is a ~4% chance AA or KK is out at a full table?

And that with AQ, there is 12 possible AK, 6, KK, 3 AA and 3 QQ such that there is a ~16% chance of being dominated (24/6 = 4 * 4%) at a full table?

Of course, these numbers are more subject to elimination effect, because if it is folded to you on the button with AQ, AA-QQ + AK is not going to be out as often as if someone raised or even limped UTG.