My good friend Keith asked me a math question. It goes like this:
Persons A, B, C, D receive individual gifts. But when the gifts are wrapped up, they look identical so there's no way to tell them apart. What is the probability that only one gift is given correctly?
I did the calculations for probability and got 1/3. I also listed every outcome and it was also 1/3. But Keith insists that it must be 1/4. I'm pretty sure he's wrong but I don't know if I'm incorrect either.
Can someone confirm my answer?
Maybe Keith was thinking of the probability that any particular person is a fixed point (gets the right gift), which is indeed $1/4$. For the probability of exactly one fixed point, condition on which person is fixed and then find the probability of a derangement on the remaining three people. Explicitly, let $X$ be the number of fixed points and compute: \begin{align} \mathbb{P}(X=1) &= \sum_{i=1}^4 \mathbb{P}(X=1 \mid \text{person $i$ fixed})\mathbb{P}(\text{person $i$ fixed}) \\ &= \sum_{i=1}^4 \frac{d_3}{3!} \cdot \frac{1}{4} \\ &= \sum_{i=1}^4 \frac{2}{6} \cdot \frac{1}{4} \\ &= 4 \cdot \frac{2}{6} \cdot \frac{1}{4} \\ &= \frac{1}{3} \end{align} More generally, for $n$ people, the same approach shows that the probability of exactly one fixed point is $d_{n-1}/(n-1)!$.