Numbers of Truncations

[Hugh R. Tobin]

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