You have two cards, neither of which are aces. After seeing 4 additional cards that aren’t aces?
you think your opponent has two aces with probability 1/3 and is bluffing (has no aces) with probability 2/3. The last
card that comes up (the “river”) is an ace. What is the probability that your opponent has two aces
- Anonymous1 decade agoBest Answer
Do you have any more information about HOW you know the probabilities in the question? - because very often this can affect how probabilities change based on new information. In this case the river card is the new information.
If, for example, you had NO information about your opponent's hand, then the initial probability that she holds two aces is (4C2)/(46C2) and this drops to (3C2)/(45C2) if the river comes up as an ace. (But these are obviously not your initial starting conditions as you have some additional information about her hand.)
Here is a scenario with the initial starting condition where the probability doesn't change at all: Suppose you have three opponents. Someone is standing behind them and sends you a signal that exactly one of them has two aces and the others have none, but doesn't tell you which. It's a contrived situation, I know, but the point is that each of them has a 1/3 chance of having two aces. If the river card comes up an ace, the probability DOESN'T change. Each one STILL has a one in three chance of having two aces. The reason is, the new information doesn't CHANGE the information you have already. So a lot depends on how you get the information.
If you're still still stuck, perhaps you can repost the question with more details? Also, does "bluffing" include the probability that she has only one ace?