Numbers of Truncations
Lowell Bruce Anderson
landerso at ida.org
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.
Bruce
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