> The arrangement of the prisoners' numbers in the boxes is a permutation of 100 numbers. Every permutation consists of a number of cycles

That's impressive!,

... but I suppose you had former training in permutation groups and cycle notation?

Like in algebra lessons at university? :)

> don't want to claim this strategy would be optimal,

Well no approach can be better than 0.5, so a > 0.3 chance is already damn good.

> but it looks like chances of getting free are around 0.31 with this approach.

Yes, no matter how many prisoners, the chances are never worse than 1 - ln 2 = 0.3068

Cheers Rolf
(addicted to the Perl Programming Language :)
see Wikisyntax for the Monastery

In reply to Re^2: Proving Veritasiums riddle using Perl by LanX
in thread Proving Veritasiums riddle using Perl by cavac

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