Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

10 kids each place a jacket on a hangar. Later, each of them randomly grabs a jacket. What's the variance for the number of kids who end up grabbing their own jacket?

I know we should expect just a single kid to grab his/her own jacket, but how do we use that to find the variance?

share|cite|improve this question
I'm surprised TSA would let kids be running around an airport like that these days... – Emily Nov 8 '12 at 21:15
Updated for clarity. Thanks! – David Faux Nov 8 '12 at 21:40
If they are randomly distribute i.i.d then wouldn't the expectation be just $\frac{1}{10}$ and you can proceed in finding the variance? – diimension Nov 8 '12 at 21:40
up vote 5 down vote accepted

Let $X_i=1$ if the $i$-th kid grabs her/his jacket, and $0$ otherwise. Then the number $Y$ of kids who get their own jacket is given by $$Y=X_1+X_2+X_3+\cdots +X_n,$$ where $n$ is the number of kids, here $10$. One common way to compute the variance is to use the fact that the variance is $E(Y^2)-(E(Y))^2$. You know the second part, so we concentrate on $E(Y^2)$.

We have $E(Y^2)=E((X_1+X_2+\cdots+X_n)^2)$. Expand the square, and use the linearity of expectation. We get $$E(Y^2)=\sum_{i=1}^n E(X_i^2) +2\sum_{1\le i\lt j\le n}E(X_iX_j).$$

Calculate. Since $X_i$ is $0$ or $1$. $E(X_i^2)=\Pr(X_i=1)=\dfrac{1}{n}$.

For the other part, to find $E(X_iX_j)$, it is enough to find $\Pr(X_iX_j=1)$. The probability that $X_i$ is $1$ is $\dfrac{1}{n}$. Given that $X_i=1$, the probability that $X_j=1$ is $\dfrac{1}{n-1}$. Thus $E(X_iX_j)=\dfrac{1}{n(n-1)}$.

Finally, put things together. Things simplify very considerably.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.