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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

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  • $\begingroup$ 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. $\endgroup$ – Valentin Moullet Jan 10 '16 at 22:48
  • $\begingroup$ 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. $\endgroup$ – barrycarter Jan 10 '16 at 22:49
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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).

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  • $\begingroup$ Thank you. This seems correct without taking into account that you know there's a probability p of your opponent at least having 1 ONE-card. Here I was wondering if, for example, you cheat, and before that your opponent draw his 5 cards, you take both his ONE-cards and place them randomly in the first 15 (for example) cards. This will change this probability p depending on how you cheat, right? So my question is: knowing how you cheated (more formally, knowing the probability p of your opponent having at least 1 ONE-card in his hand), what is the probability of him having the second one? $\endgroup$ – Valentin Moullet Jan 10 '16 at 23:02
  • $\begingroup$ But if you see a 'ONE' card in the opponent's hand, isn't it more likely he has 2 'ONE' cards than having only 1 'ONE' card? $\endgroup$ – barrycarter Jan 10 '16 at 23:06
  • $\begingroup$ I interpreted the qeustion in the way that you see a ONE-card and ask for the probability that the opponent has the second ONE-card as well. But the intent seems to be : If the opponent has at least one ONE-card, what is the probabilty that he has both, right ? This can be calculated with the theorem of Bayes. $\endgroup$ – Peter Jan 10 '16 at 23:09
  • $\begingroup$ Peter: yes, that is how you have to interpret it. I "just" added one thing: the opponent deck is biased, and you know how, which change the probability of your opponent having drawn certain cards (the ONE-cards in this case). That's why I say you "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". Is it still possible to solve? $\endgroup$ – Valentin Moullet Jan 10 '16 at 23:13
  • $\begingroup$ OK, now I got it. I do not think that it is possible. It should depend on the way the probabilities are manipulated. $\endgroup$ – Peter Jan 10 '16 at 23:21

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