The hundred numbered boxes

1. Key: A prisoner hits if and only if the loop it belongs to has length $\le 50$.
2. Since two cycles >50 cannot coexist in 100 elements, the failure event is the disjoint union: P(failure)=sum_ k=51 ^ 100 frac1k.
3. That ~31% drastically outperforms the random strategy (2^ -100 for collective success).

Answer: YES there is a strategy with ~31% probability of success.

Cycle strategy:
Each prisoner first opens his own box and then follows the chain indicated by the numbers found:

1. Open box $i$ (your number).
2. If there is $j$ inside, open box $j$.
3. Repeat until you find your number or exhaust 50 openings.

The roles form a permutation of $\{1,\dots,100\}$, and therefore a decomposition into disjoint cycles.
Key: a prisoner hits if and only if the cycle it belongs to has length $\le 50$.
Then everyone guesses correctly if and only if there is no cycle of length greater than 50.
Since two $>50$ cycles cannot coexist in 100 elements, the failure event is the disjoint union:

$$ P(\text{fracaso})=\sum_{k=51}^{100}\frac1k. $$

Therefore,

$$ P(\text{success})=1-\sum_{k=51}^{100}\frac1k\approx 0.31. $$

That ~31% drastically outperforms the random strategy ($2^{-100}$ for collective success).