There are $3n$ people in the anonymous gift society. They're preparing the presents for one another. It's known that exactly $n$ people would like to have a tie as a gift, $n$ people would like to have a pair of socks as a gift, and $n$ would like to have a hand dinosaur. Everyone randomly picks and buys a present that they would not want themselves (e.g. if one wants a pair of socks, then he goes to the shop and buys either a tie or a dinosaur).
After a party was arranged, all members of the society put the gifts in the pile-up, and then picked it up randomly.
Alice and Bob do belong to the anonymous gift society. Alice would like to have a hand dinosaur, and Bob would like to have a pair of socks. Find the probability that neither Alice nor Bob will get the present they initially wanted.
ATTEMPT
Re-formulate the problem: there are 3 types of boxes, $n$ boxes of each type.
Type $T$ is a box donated by a person who would like to have a tie, so it contains either a dinosaur or a pair of socks, with probability $P = \frac{1}{2}$.
Type $D$ is a box donated by a person who would like to have a dinosaour and contains either a tie or a pair of socks.
Type $S$ is either a tie or a dinosaur.
Alice and Bob pick a box at random. What is the probability of both the boxes containing an unwanted item?
Denote $A_i$ the event that Alice got the box of type $i$, $A$ the event that Alice didn't get the present she wanted. (same for Bob, $B$) $$\sum_{i,j}P(A,B\space|\space A_i, B_j)P(A_i, B_j)$$
There are only 9 terms in this summation and each term is easy to compute, for example: $$P(A, B \space | \space A_D, B_T)P(A_D, B_T) = 1 \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \bigg(\frac{n}{3n-1} \bigg)$$ $$P(A, B\space | \space A_T, B_T)P(A_T,B_T)=\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \bigg(\frac{n}{3n - 1}\bigg)$$ ... and so on. Putting it all together, one obtains $$\frac{16n - 5}{36n - 12}$$
- Is this correct?
- Is there a more elegant way to tackle this problem?
As has already been stated in the comments, your approach is correct.
I don’t know whether you would consider this more elegant; perhaps marginally so:
The probability of Alice and Bob getting the same present is $\frac{n-1}{3n-1}\cdot\frac12+\frac{2n}{3n-1}\cdot\frac14=\frac{2n-1}{2(3n-1)}$. If they do get the same present, the probability that it’s unwanted for both of them is $\frac13$. That covers $1$ of the $2\cdot2=4$ out of $3\cdot3=9$ cases in which they both get unwanted presents, so $3$ out of the remaining $6$ cases where they get different presents are doubly unwanted, so if they get different presents the probability that they’re doubly unwanted is $\frac12$. Thus the overall probability is
$$ \frac{2n-1}{2(3n-1)}\cdot\frac13+\left(1-\frac{2n-1}{2(3n-1)}\right)\cdot\frac12=\frac{2(2n-1)+3(4n-1)}{12(3n-1)}=\frac{16n-5}{12(3n-1)}\;. $$