Re: Odds of your oppenent holding AA against your KK (10 handed)
There are 4 Aces in the deck, so there are 6 possible 2-card combinations of pocket Aces. (4*3 / 2*1)
Taking out your own hand leaves 50 other cards in the deck, which can be combined in 2-card hands some 1,225 different ways (50*49 / 2*1).
Since you're not holding any Ace, that leaves those 6 combinations of pocket Aces among the total possible 1,225 combinations. So, the probability of your Heads-up opponent picking up (being dealt) one such pocket-Aces combination is 6/1225 = 0.00489795 or 0.489795%, which is also the probability of your Heads-up opponent picking up a pocket pair from a specific rank when you're holding no card from that rank.
You are playing now against 9 opponents, but you should not multiply that by 9 to find the answer for a 10-handed game : it is impossible for your opponents to simultaneously "pick" cards from those 1,225 combinations. (When one of them picks up for instance A[img]/forums/images/icons/diamond.gif[/img]5[img]/forums/images/icons/club.gif[/img] , all the combinations containing the A[img]/forums/images/icons/diamond.gif[/img] "disappear".)
This is a problem that only a qualified engineer can solve.
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