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Problem: At the end of a busy day $n$ fathers arrive at a kindergarten to pick up their kids. Each father picks a child to take home uniformly at random. . Let $X_{n}$ be the matching number among $n$ fathers and $n$ children. Find $EX_{n}$.

My Solution: Define, $x_{i}$ to be the $i$th father picking up his child. ( It should be a random variable)

Therefore: $X_{n}=x_{1}+x_{2}+\cdots+x_{n}$. Then $$EX_{n}=Ex_{1}+Ex_{2}+\cdots+Ex_{n}=n*\frac{1}{n}=1$$

Probably this is wrong, Can you tell me what I did wrong?

Rust32627
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  • I think the only thing you've done wrong is posting a question that has already been asked and answered on this website. – Gerry Myerson Oct 24 '22 at 05:08
  • I found two or three but they were a bit different solutions. – Rust32627 Oct 24 '22 at 05:14
  • Funny "coating" for this classical exercise that changes from the hat/peg version. A non-mathematical comment: "Each father picks a child at random". Is it the reason why it's better to ask the mothers to pick up their own child ? – Jean Marie Oct 24 '22 at 08:21

1 Answers1

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Once the first father picks a child, the 2nd only has $n-1$ children to choose from.

There's a $\frac{1}{n}$ probability that the second father's child was actually picked by the first father, and $\frac{n-1}{n}$ probability that it was not. So, the probability the second father picks his child is $\frac{n-1}{n} \times \frac{1}{n-1} = \frac{1}{n}$.

One can continue this inductively: $P($father $i+1$ picks his child$) = P($none of $1, \ldots, i$ pick this child) $ \cdot P($father $i+1$ selects his child, given that his child is still left$)$.

This can relatively easily be calculated, as $\left(\frac{n-1}{n} \cdot \frac{n-1}{n-2} \cdots \frac{n-i-1}{n-i}\right) \times \frac{1}{n-i-1}$ which simplifies to $\frac{1}{n}$.

So the answer is indeed $1$, as you say!

Fred T
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