> I'm curious to know how my other objection is handled. The order in a Perl hash is random, but fixed during a run. So this algorithm will be biased to some results.

OK, after a second coffee I finally got my had wrapped around it.

Definition/Interpretation matters:

The random pick $r between from 0..N-1 represents what the voter $r has chosen.

Hence order doesn't matter as long as it is fixed, since a 100 wide interval will always have the same probability of 100/550, no matter where it occurs.˛

But I don't see any intuitive way to invert this interpretation, because $r has not-voted to all other choices.

Assigning "inverted" votes like suggested in your other post might work, as long as they sum up to 550 again°. But the question of the weighting factor and distribution will come up again.

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

°) respectively N' is corrected.

update

˛) IOW no matter if voters John and Jim swapped their place, their probability to be chosen stays the same.

In reply to Re^4: Randomly choosing from a set of alternatives with varying popularity by LanX
in thread Randomly choosing from a set of alternatives with varying popularity by ibm1620

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