Good looking out: obviously not optimal. I tried to figure out for myself what was going wrong.

First, I simplified the code itself a little further:

That does basically the same thing, except calls rand() once less per comparison.

Staring at it didn't do much good for me so I came up with a little test script of my own:

use strict;
use warnings;

sub compare {
        my ($a, $b) = @_;
        my $rand = rand();
        print "$a: $rand; $b: 0.5\n";
        return $rand <=> 0.5;
}

print join "", sort { compare($a, $b) } split //, "abcde";
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The output of which was something like:

$ perl test.pl
a: 0.876941476789401; b: 0.5
c: 0.0768385185833438; d: 0.5
b: 0.203632482365844; c: 0.5
c: 0.234952540459695; a: 0.5
a: 0.746254431212112; d: 0.5
b: 0.648884088019244; e: 0.5
ebcda
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After running this a few times and looking at the results, it hit me. This is going to be a rough explanation, but I'll give it a shot. After comparing the first four letters to one another it selects the "smallest" one (the one with the smallest rand() anyway) and compares it to the final letter: 'e' in this case. Because 'e' is the last to be compared (in the initial go, at least) and because the rand() number is regenerated at each comparison, it always has roughly a 50% chance of being the "smallest" since it has to only pass a single test to be the "smallest". The letters in front of it must not only pass their first test, but will always have to be compared to the letters following it, thus reducing their chances of being the "smallest".

In this way, my "solution" favors the letters towards the end and is more likely to stick them in front. That's why the distribution is so uneven. If you follow sort()'s algorithm ( quicksort, I believe (C's qsort(), I think ) ) through, it'll make sense.

I hope that makes some sense ...