Probability of having 2 cards in hand knowing the probability of having 1

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?

Valentin

• barrycarter: you don't share the same deck with the opponent, you may have the exact same cards in it, but each player has his own deck with 30 cards in it. – Valentin Moullet Jan 10 '16 at 22:48
• Yes, I deleted my comment after I saw yours. This sounds suspiciously like the "chance of having two boys if it's known they have at least one boy" problem. – barrycarter Jan 10 '16 at 22:49

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).