There's a couple possible ways using just combinations to figure out the chance that either of two remaining opponents holds pocket aces:
- Find out how many ways the first player (SB) can have aces. Find out how many ways the first player can have one ace and then given that he holds one ace, get the odds of the BB having AA. Find out how many ways there are that the first player has zero aces and given that he holds zero aces, get the odds of the BB having AA. Then add together the odds for each of the three scenarios that result in somebody having pocket aces.
How many possible hands dealt to the first player? 50C2 = 1225
How many possible hands dealt to the second player after removing cards dealt to the first player? 48C2 = 1128
How many ways can the first player get aces? 4C2 = 6
How many ways can the first player get exactly one ace? 4C1 X 46C1 = 184
How many ways can the second player get aces if we know exactly one ace is missing? 3C2 = 3
How many ways can the first player get zero aces? 46C2 = 1035
How many ways can the second player get aces if we know that no aces have been used? 4C2 = 6
So we end up with [(6/1225)(1128/1128)] + [(184/1225)(3/1128)] + [(1035/1225)*(6/1128)] = 0.0097916, or ~102:1 odds.
- This is a simpler way. We know 4 cards must be dealt to the two players. Of all the combinations dealt, how many result in either AAxx or xxAA, where any of the x could also be an A?
How many ways can the 4 cards be dealt? 50C4 = 230300
How many ways could the 4 cards be 4 aces? 4C4 = 1
How many ways could there be exactly 3 aces in the 4 cards? 4C3*46C1 = 184 (note that at least one of the players definitely has pocket aces in this situation)
How many ways could there be exactly 2 aces in the 4 cards? 4C2*46C2 = 6*1035 ...but wait. There's 6 ways for the 2 aces to be distributed, but not each distribution guarantees AA for one of the players. In fact, there's only two ways--AAxx and xxAA where x is not an ace. So really we'll use 2*1035 to get 2070.
Now we have (1 + 184 + 2070)/230300 = 0.0097916, or ~102:1 odds, as above.