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The probability that there's a cycle (there could only ever be at most one) longer than fifty turns out to be about 70%, and that corresponds to a prisoner needing to open more than fifty envelopes to get back to their number.
The solution is basically maximising the shareable information available to all the prisoners in a way which negates the "one prisoner loses, you all lose" rule by choosing a strategy which does that anyway.
What happens if one of the envelopes contains its own number (envelope #5 says go to envelope #5)?
That's just a cycle of length 1. That prisoner does less envelope-opening. ;) (But they may still die.)
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