Numbers of Truncations

Lowell Bruce Anderson landerso at
Wed Nov 20 08:10:27 PST 1996

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.


More information about the Election-Methods mailing list