A player either calls or raises the big blind. I assume there is a 50% chance he has an ace. I also have an ace. The flop comes and includes an ace. How should I now assess the probability that he has an ace. A problem in Bayesian statistics I believe.

  • Would it not be as simple as saying that if with 3 aces remaining before the flop he has a 50% chance of having one... Now that the flop reveals one that means there are only 2 aces remaining. It's 33% harder for him to have an ace now, so 0.66 * 0.5 = 0.33333 or 33.3%
    – pingu2k4
    Commented Jul 31, 2015 at 0:27
  • Thank you, but there are only 2 aces possibly remaining. Assuming he has one (or at least I think there is a 50% chance he has one) and I obviously know I have one, the flop of an ace should lower, after the fact, my judgment that he has a 50% chance of holding an ace.
    – coldeye
    Commented Jul 31, 2015 at 1:53
  • Thats what I mean. Preflop you beleive there is a 50% chance, when the only known ace is the one in your hand (IE 3 remaining out there either in his hand or in deck). After you see an ace on the flop, known aces = 2, so you lose one third of unknown aces. That 50% chance * 0.66 (for 2/3 remaining unknown aces) would equal 0.33 or 33.33%
    – pingu2k4
    Commented Jul 31, 2015 at 7:21
  • Again Matthew, thank you, but I would say 2 aces are know, 100% sure I have an ace an 50% sure he has an ace, Then the third ace comes on the flop. I could be wrong or misguided that's why I posted.
    – coldeye
    Commented Aug 1, 2015 at 2:36
  • I understand totally what Matthew explains, but an old player said to me once that it's always 50/50, either he has it or not :) this even works with outs :)
    – disco beat
    Commented Aug 1, 2015 at 22:21

1 Answer 1


Given the fact that he's voluntarily participating in the pot (VPIP) pre-flop, what's the probability he has the ace?

P(Ace | VPIP) = (Prob(VPIP| Ace) * P(Ace)) / Prob(VPIP)

So, in order to solve the problem, you need to know not only the probability of being dealt an ace given that you've seen two (it's about 9.5% percent, let's round to 10 for easy math), but his average pre-flop raising percentage, a little about his range, etc.

If he plays about 20% of hands, and he plays every ace to the flop (this would be fairly loose), the probability that he has an ace given his preflop actions would be (1 * .10) / .2 = 50%. This is a very ace-heavy player: 15% of hands have a ace in them to begin with, so he's playing almost all pairs (5.8%), all aces suited or unsuited (14.8%), and not much else.

A more reasonable player might have the same 20% range, but only playing a third of the aces (A-K -> A-10), and mixing in more suited connectors, etc.

Now there's only a 16% chance he has an ace.

Another reasonable loose player might play 25% of his hands, and play half the aces (A-8 or better), giving about 20%.

A tight player who over values unpaired aces might play 15% of hands and play 75% of aces (say, all but ace-6 through A-8), giving 55%.

I'd say anyone who's got the ace more than 1/3rd of the time is probably playing too transparently.

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