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.
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.