One hundred boxes

Idea. The strategy makes the prisoners' successes depend only on the cycle structure of the permutation. They all succeed or fail together, depending on whether there is a cycle longer than $50$, and that has a simple probability.

Step 1: what the strategy does. Starting at box $k$ and following the numbers walks around the cycle containing $k$; the number $k$ is the last one found on that cycle. So prisoner $k$ succeeds if and only if their cycle has length at most $50$. All succeed if and only if every cycle has length at most $50$.

Step 2: a long cycle. A permutation of $100$ elements has at most one cycle longer than $50$. The number of permutations with a cycle of length exactly $\ell >50$ is $$\left(\genfrac{}{}{0ex}{}{100}{\ell }\right)(\ell -1)!(100-\ell )!=\frac{100!}{\ell }:$$ choose the elements of the cycle, arrange them in a cycle, and arrange the rest freely. So that cycle exists with probability $\frac{1}{\ell }$.

Step 3. The events for different $\ell >50$ are disjoint, so $$\mathbb{P}(\text{all succeed})=1-\sum _{\ell =51}^{100}\frac{1}{\ell }\approx \mathrm{0.3118.}$$

Why this is surprising. If each prisoner opened $50$ random boxes, the probability would be $2^{-100}$. The cycle strategy makes their successes perfectly correlated, and the answer stays above $1-\ln 2\approx 0.307$ for any number of prisoners.