# Double urn probability and multiple instances

I've got a probability problem based on the following problem. I provided some scenarios which I need to solve but I'm thankful for any hints or links on how to solve just one of them or how to described this scenario (and/or solutions) in proper 'mathematical notaion'

The Problem:
An urn contains 10 balls numbered from 1 to 10, every number occurs excactly once. Person A picks balls from the urne and distributes the balls randomly among 4 other persons, say persons B, C, D and E (it is not neccessary for every person to get the same amount of balls). After the distribution process, person C picks $N$ ball(s) from its own urn, which holds another 10 balls, again numbered from 1 to 10. What is the probability that person C received one or more balls from person A which have the same number as the ball(s) person C just picked from its own urn?

Some scenarios:
Scenario 1: Person C received one ball with nr.7. What is the probability that person C picks the ball with the same number (7) from its own urn?
Scenario 2: Person C received two balls with nr.7 and nr.8. What is the probability that person C picks the same two balls (7 and 8) from its own urn?
Scenario 3: Person C received two balls with nr.7 and nr.8. What is the probability that person C picks one ball from its own urn which number is either 7 or 8?
Scenario 4: Person C received one ball with nr.7. What is the probability that person C picks two balls of its own urn out of which one is ball nr.7?
Scenario 5: Person C received one ball with nr.7 and person D received three other balls, say balls 4, 5 and 6. What is the probability that person C picks one ball from its own urn which number is either 4, 5, 6 or 7? (Person C and person D are merged, they count as one)

• If A picks all the balls and distributes each one independently and with equal probabilities between B,C,D and E, and then C picks one ball, the probability C's pick matches a ball already held by C is obviously $\frac14$. – Henry Apr 11 '15 at 14:16
• How many balls does C draw from their own urn in each of the scenarios? The main problem suggests only one, but that seems clearly contradicted by the intent of (e.g.) Scenario 2. – Brian Tung Apr 12 '15 at 5:58
• @Henry: Person C doesn't pick one of the balls it received from person A. C picks one or more balls from its own urn and compares the picked ball(s) with the balls received by A. – froua Apr 12 '15 at 9:04
• @BrianTung: You're right that was a bit miselading, I corrected it. Person C picks $N$ ball(s) from its own urn depending on the scenario – froua Apr 12 '15 at 9:07

For the general question, assuming picks are without replacement:

• for each numbered ball picked by C from their own urn, the probability C did not get that numbered ball from A is $\displaystyle \frac34$ (i.e. it went to B, D or E),

• so the probability of no matches is $\displaystyle \left(\frac34\right)^N$,

• the probability of at least one match is $\displaystyle 1-\left(\frac34\right)^N$ and

• the probability of $n$ matches is $\displaystyle {N \choose n}\frac{3^{N-n}}{4^N}$.

For the scenarios the numbered balls received by C from A are determined by the scenario, so:

• if C picks one from ten balls, the probability it is number $7$ is $\dfrac{1}{10}$

• if C picks two from ten balls, the probability they are numbers $7$ and $8$ is $\dfrac{1}{45}$

• if C picks two from ten balls, the probability exactly one is number $7$ or $8$ is $\dfrac{16}{45}$, and the probability at least one is number $7$ or $8$ is $\dfrac{17}{45}$

• if C picks two from ten balls, the probability one is number $7$ is $\dfrac{1}{5}$

• if C picks one from ten balls, the probability it is numbers $4$, $5$, $6$ or $7$ is $\dfrac{2}{5}$