Forty-four interns have been kidnapped by the diabolical François, who, in his great kindness, gives them a chance to escape his trap. He has them enter one by one into a large room containing 44 drawers, each of which contains the name of an intern (the names are assumed to be distinct). If by opening fewer than 22 drawers the intern finds his name, he moves to another room where he must wait for his friends. Otherwise, François kills everyone. Find a strategy that gives the interns at least a $30\%$ chance of survival.

## Sunday, August 22
We number the interns and the drawers. The system of drawers thus corresponds to a permutation. We denote $p(i)$ as the content of drawer $i$.

Mathematician $i$ opens drawer $i$. If he is not there, he opens drawer $p(i)$ and continues in this manner, so the $k^{\text{th}}$ drawer he opens is drawer $p^{(k)}(i)$. If the permutation has only cycles of length less than 22, the interns survive.

Let's find the probability that a permutation has a cycle of length greater than 22. There are exactly

$$
\binom{42}{k}(42-k)!(k-1)!
$$

permutations that have a cycle of length $k$ because there are $(k-1)!$ permutations of $k$ numbers.

And we have:

$$
\frac{1}{42!} \sum_{k=22}^{42}\binom{42}{k}(42-k)!(k-1)!=\sum_{k=22}^{42} \frac{1}{k}<0.68
$$

Therefore, the probability of the interns surviving is greater than $1-0.68=0.32$.

## Sunday, August 22