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I've been trying to work this out for a while but I'm really struggling.

I know the odds for a single player getting a hand from a range is simply:

Combinations of hands in range / Total possible combinations

The problem is when let's say you are on the BTN and you want to work out the odds that at least one of the opponents has a hand from a range.

For example for a range specifically of aces (given that I don't have an ace in my hand) I tried:

[4C2 x 48C2] / 50C4 = 36/1225

Unfortunately this seems to bloat the odds (4+48 > 50). Yet I can't use 46C2 either because it doesn't matter if an ace is in one hand as long as the other has two aces.

What should I do? Is a solution possible with Combinatorics?

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There's a couple possible ways using just combinations to figure out the chance that either of two remaining opponents holds pocket aces:

  1. 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.

  1. 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.

  • Good answer for solving for aces range. I'm guessing then doing this for say top 20% of hands would be an absolute nightmare because of all the possible combinations of cards that create hands you don;t want. I'll have a look for a program to do this or write one myself. Thanks for showing me how to do this! – user4555 Aug 30 '16 at 15:26
  • I realised for a range of hands you have to use permutations. Permutations and computing don't mix unfortunately. Really don't think I can solve this sadly. – user4555 Sep 5 '16 at 17:34

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