Numbers of Truncations

Mike Ossipoff dfb at
Fri Nov 22 20:29:06 PST 1996

Hugh R. Tobin writes:

[Mike writes: Just before I mailed this, I checked the header,and
there was no copy to EM, just a copy going to Tobin. So I added
EM at that time. So apparently the listserver is again failing
to sent replies using the "reply" option to the EM list, but is
instead just sending them to the person who wrote the letter that
you're replying to. So I'd suggest editing the header to include
election-methods-list at, or else using the 
"group reply" option, so the reply will go wherever the original
went. That 2nd solution is obviously the easier of the 2]

> 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?

Yes. It complicates the ballot, in order to add an option that
a voter would be unlikely to want. If you don't know anything
about someone, do you really want to rank hir equal to lyour
favorite? :-)

This, & the other added options that you proposed in an
earlier letter are things that I have no objection to on
principle. But they, separately or in combination, make
for a horrendously complicated ballot, and a horrendously
difficult job of trying to explain it & justify it. People
will reject the reform as too complciated, &/or they'll
not bother voting.

The option to vote a "bottom ranking", of one's _last_ choices
sounds ok until you realize that you might accidentally leave
out someone (candidates you don't know about might be pretty
bad--do you want to count on knowing every candidate & write-in?).

Not desirable, from the voter's standpoint. Still, as I said,
I wouldn't object to it on principle, even though it's a bad
idea to use it, except for the way it would complicate the
ballot & the explation of the reform.

But I absolutely oppose your suggestion to maybe count only
bottom rankings. As an option, bottom rankings aren't desirable.
As the only ranking, they're entirely unacceptable.


> -- Hugh Tobin
> .-


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