Numbers of Truncations
Hugh R. Tobin
htobin at ccom.net
Fri Nov 22 20:55:36 PST 1996
Lowell Bruce Anderson wrote:
>
> Don't allow only ties for last, i.e., truncations; instead, allow a voter
> to have ties at any position on his or her ballot.
>
> Then, in the voting system, each voter casts a ballot by ranking the
> candidates from that voter's first choice to that voter's last choice with
> ties allowed throughout. In particular, not explicitly ranking (i.e.,
> truncating) k of the candidates is equivalent to explicitly ranking those
> k candidates as being tied with each other behind every other candidate.
> For n > 0, let R(n) = the number of different ballots that a voter could
> cast, including the 1 "trivial" ballot on which all of the candidates are
> tied with each other, when there are n candidates. Let R(0) = 0. Then
> R(n) = the sum from m=1 to m=n of (n!/(m!(n-m)!))R(n-m)
> = the sum from m=0 to m=(n-1) of (n!/(m!(n-m)!))R(m)
> for n > 0.
>
> n=1: n!= 1, R(n)= 1, R(n)/nR(n-1)=1.000000
> n=2: n!= 2, R(n)= 3, R(n)/nR(n-1)=1.500000
> n=3: n!= 6, R(n)= 13, R(n)/nR(n-1)=1.444444
> n=4: n!= 24, R(n)= 75, R(n)/nR(n-1)=1.442308
> n=5: n!= 120, R(n)= 541, R(n)/nR(n-1)=1.442666
> n=6: n!= 720, R(n)= 4683, R(n)/nR(n-1)=1.442699
> n=7: n!= 5040, R(n)= 47293, R(n)/nR(n-1)=1.442695
> n=8: n!=40320, R(n)=545835, R(n)/nR(n-1)=1.442695
>
> I still do not know if 1.442695... has an intrinsic interpretation.
>
> Bruce
Perhaps one should even allow the voter to rank a candidate, C, as tied
with a range of ranked candidates. When that range is all other
candidates, it means that C is not ranked above or below any candidate,
and is equivalent, in Donald's Condorcet ballot, to abstaining in each
pairwise race involving C. For example, (A>B)=C. This might be rational
for a voter who has absolutely no information about C, as frequently
happens in elections for minor offices. Is there any reason not to allow
such a ballot?
-- Hugh Tobin
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