Since you have no info on Player 1's hands, his hand is irrelevant. Player 2 has 2 random cards out of the 52 (not 50, since you don't know what cards to eliminate). Player 2's cards can be any of the 52 cards with equal probability. There are 4 x 48 = 192 combinations of 2-card hands with 1 ace. There are 52C2 = 26x51 = 1326 ways to deal those two cards. The probability that Player 2 has 1 ace is 192/1326 = 32/221 = 0.14479638.
Look at a more extreme case, Suppose you deal out 25 hands of 2 cards each and the last two cards to Player 2. You have no info on the 1st 25 hands. What's the probability that Player 2 has an ace? See how it's still the same? The fact that you've dealt out other hands doesn't mean anything unless you have info about them. Player 2 could have the last hand or the first hand--he still has an equal chance with every other hand of having any particular combination.
You can get that going the long way by assuming Player 1 has 0, 1, or 2 aces, calculating the probabilities for each case that Player 2 has 1 ace, and adding them up. You will get the same result. That would be a good exercise to go through if you are still not convinced. Edit: Actually, David's answer does it the hard way for you, verifying what I said about Player 2's hand being random.
· 6 years ago