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They could certainly do even worse than this. If they all agreed in advance to open envelopes 1-50, then fifty of them would be guaranteed never to find their number and they'd all die. But could they come up with a strategy in advance which would improve their survival chances?
What makes this my favourite combinatorics problem is that there *is* a strategy which increases their likelihood of survival from 0.00000000000000000000000000008% to about 30%. So, what is the strategy, and why does it work?
Intuition before I actually do the math on it… They have one method of passing information during the trial: the fact that they haven’t yet all been killed. This indicates the range assigned to all the previous prisoners contained their respective numbers.
No, that turns out to be not usable information. ;)
Coud the 2-3 first guys that will open the envelope sort them from 1 to 100? (or by pile of 10)
No, that would constitute communication, so it's forbidden. :)
Each prisoner can open 50, so the first has a 50% chance of killing everyone. Can the first leave the envelopes sorted in some way? Can the prisoners decide the order in which they go?
First sentence: correct. Not permitted to sort the envelopes, as that would be considered communication. The prisoners can decide which order they go in, sure, but it doesn't really help.
Must they necessarily open exactly 50 envelopes? Perhaps a prisoner finds their number, then continues to open envelopes until they reach 50 or until they find their number + 1. This would indicate to the person next to them where their number was.
I should have been clearer that the envelopes are all returned to the exact same initial state for each prisoner. All closed, each in the same physical state and position, each containing the same prisoner number. They can keep opening more, but they can't use that information.
With this strategy will ALL the prisoners be alive at the end?
Yes, the probabilities are for everyone being alive. The only two outcomes are "everyone dies" and "nobody dies" anyway.
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