How to find probability with multiple pieces of evidence?

Question

I'm trying to find the probability of P5 having C1, C3, or C4. Here are the constraints of the problem.

C1 - C5: Are cards that can be held by 1 person. P1 - P5: Are players that each have 1 card. X means the person does not have the card. O means the person has the card ? means it's unknown who has the card.

 P1 P2 P3 P4 P5

C1 ?   X   X   ?   ?

C2 X  O   X   X   X

C3 ?   X   X   ?   ?

C4 ?.  X.  X   X.  ?

C5 X.  X. O.  X.  X

I've calculated by looking at each row:

C1 has a 1/3 chance of being in P5 C3 has a 1/3 chance of being in P5 C4 has a 1/2 chance of being in P5

I've calculated by looking at P5 column:

There is a 1/3 chance that they have C1, C3, or C4

I'm thinking there is a way to combine these probabilities to give me an answer as to the chance of P5 having C1, C3, or C4. Will you please show me the math and let me know what this is called in statistics?

Answer 1

You need another assumption. There are $4$ consistent ways to deal out C1, C3, and C4 to P1, P4, and P5 respectively: P1P4P5, P4P1P5, P4P5P1, and P5 P4P1. If you add the assumption that all consistent deals are equally likely, then the probability P5 gets C1 is $1/4$; C3, $1/4$; and C4, $1/2$. In many practical situations, the assumption that all consistent deals are equally likely is not correct because the hidden cards held have affected the information which has been revealed.

The technique you used to assign probabilities by looking at individual columns or rows is not correct. You will come up with positive probabilities for impossible assignments. For example, if there are only $2$ cards and $2$ players, and you have no direct restriction on the card held by P1, but you know P2 can't hold C1, then you can deduce that P2 has C2 so P1 has C1. Your method would assign probabilities of $1/2$ to the cards for P1, or for the locations of C2.