[EM] RE: Lotteries undefeated "in median"

Simmons, Forest simmonfo at up.edu
Sat Jul 9 17:41:41 PDT 2005


Jobst,
 
I'm still digesting this.  I'm always interested in potentially good "lottery methods."
 
Here's my latest attempt:
 
Voters first submit approvals, and the candidates are listed in order of approval.
 
Each voter then submits a number between 1 and the number of candidates (to be used as explained below).
 
Let K be the median of these submitted numbers.
 
A random voter picks a candidate from the top K candidates in the approval list.
 
I have a way of automating this from ordinal ballots with approval cutoffs, but I don't have time to describe it here.
 
Example:
 
51 A>B>>C
49 C>B>>A
 
The approval order is B>A>C .
 
The 51 direct supporters of A submit the number 2, which is the median K..
 
The winner is chosen by random voter from the top two approval candidates B and A.
 
So the winning lottery is  51A+49B.
 
Even though B had 100 percent approval, the majority winner A gets the most probability.
 
Forest
 
 
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