a poker-type question
A standard 52 card deck is shuffled and you and your opponent are dealt two 'hole' cards each. now the next three cards from the deck are exposed (the flop) and it contains two of the deck's four aces.
Does the fact that two aces were exposed make it less likely, more likely, or not important to the question 'what is the probability that my opponent was dealt an ace?'
My gut leads me to think it shouldn't matter - that the cards that were dealt on the flop don't affect what your opponent was dealt in the hole, but I suspect that's not correct and that the presence of the two aces decreases the probability that my opponent has an ace in the hole. Not sure by how much.
It does indeed decrease the probability that your opponent has an ace. Before the flop all you know is that there is a decreased chance of your opponent having any card you have, there are still 4 aces unaccounted for. If there's two aces on the flop then there are 47 cards left, and only two aces.
Originally Posted by demere
"that the cards that were dealt on the flop don't affect what your opponent was dealt in the hole"
Of course they don't literally affect what your opponent was dealt, but it strongly affects your knowledge of what your opponent might have. For example, if 3 aces came up on the flop, and the 4th ace came on the turn then you would know with 100% certainty that your opponent did not have an ace.