[EM] Puzzle example X 1000

MIKE OSSIPOFF nkklrp at hotmail.com
Sat Mar 10 14:58:58 PST 2001

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[regarding complete abstention]

>We can't rule it out, but without some basis for assigning a probability to
>this
>
>event, how can we factor it in? Maybe in the real world you could look at
>historical patterns. For an isolated example, there's nothing to go on.
>Might
>as well solve the problem for the case where everybody is certain to vote
>for at least one candidate.

That seems best.

>likelihood of these scenarios, I think it would be impossible to improve
>on the 50-50 assumption in any foolproof way.

That sounds right. I wasn't comfortable about assuming that each way
of voting was equally likely. The assumtpion that the voter is
equally likely to vote for 1 or 2, and that if it's 2, the 2nd one
is equally likely to be either one of the remaining 2--that seems
best. Of course the voter's 1st choice is equally likely to be any of
the 3.

>
> > I've meant to add up your benefit by voting for B in addition to A,
> > with all 729
>
>I'm not sure about that number. If you add the no-vote possibility, but
>not the ABC vote, each person can vote one of 7 ways, so it becomes
>7x7x7 or 343. Did I miss something?

I agree about assuming that no one abstains completely. So I figure that
these are the ways of voting:

A, B, C, AB, AC, BA, BC, CA, CB

9 ways for someone to vote, so 729 ways for the 3 voters to vote.

But, as I said, I didn't feel comfortable assuming that each way of
voting is equally likely for a voter. The 50% assumption makes sense.
So a particular voter is equally likely to vote for any of the 3 as
favorite, and is equally likely to have either of the remaining 2 as
his 2nd choice, and is 50% likely to vote for his 2nd choice.

As you suggested, 6 equally likely preference orderings, and 50%
chance of voting for one's 2nd choice.

Mike Ossipoff

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