Given (per player):
1 deck of 30 cards
15 different cards, so there are always 2 identical cards (let's name them ONE, TWO, ..., FIFTEEN)
1 hand of 5 cards (drawn from the deck)
I was wondering the following:
At one point, I know that there is a probability p (this is given, don't try to compute it) that my opponent has at least one of the two cards ONE in his hand. I cheat and try to look at his hand, and I can only see one card, which is a ONE. What is the probability that he has the second ONE in hand too at this moment?
Is it possible to solve this problem? If yes, how and what is the solution? If no, what information is missing?
Thank you in advance for your help,
Valentin
The opponent has $4$ more cards out of $29$ possible cards.
The probability that he has the second ONE-card is $\frac{4}{29}$. This can be calculated by $$\binom{28}{3}/\binom{29}{4}$$ ($\binom{29}{4}$ possible distributions and $\binom{28}{3}$ possible distributions , if the opponent has both ONE-cards of his deck).