[EM] counts of possible rank-order votes

[Warren Smith]

On 2/4/10, robert bristow-johnson <rbj at audioimagination.com> wrote:
>
> On Feb 3, 2010, at 4:31 PM, Warren Smith wrote:
>
>> as RBJ (and I earlier, and probably many others earlier) proved,
>> the number of possible rank-order votes
>> among C candidates is
>> C!
>> if rank-equality forbidden and all candidates must be ranked.
>
> actually Warren, i never attempted to prove that.

--I meant: that bundled with the next paragraph!

>and i didn't get
> at all how you came to say that until Kristofer told us that you were
> assuming "Australia rules" that requires every candidate to be ranked
> (a rule that didn't occur to me since that's not how IRV works
> here).  but given that rule and the no-equal-rank rule, i can quickly
> see that C! is correct.

--yah, I often consider what are mathematically simpler but maybe not
as nice in practice, rule variants, when doing such calculations.
There are many IRV rule variants.

>> ...The underlying reason for that is basically that the unranked
>> pseudolevel can be regarded as just another ranking level...
>
> well, i think that was the earlier argument i was having with Kathy
> and Abd ul.  maybe not, maybe it's something different.

--well, let's see if I can solidify that a bit.

If I cast a rank order (equal ranks allowed, and leaving people
unranked allowed)
ballot, with r ranks, then I'm effectively using r+1 ranks including
the unranked pseudolevel.   Summing over all possible values of r and
r+1, this is
the SAME count as if leaving people unranked is NOT allowed (!),
EXCEPT:
   *    I can also rank everybody and not-rank zero (doubling the count)
   *    The ranking of all equal (none unranked0 is the only one
without a "duplicate"

So G = 2*F - 1

where G=#ballot types where equalities and unranks both allowed,
F=#where equalities allowed but all must be ranked.

The way I was defining G, I had in mind that voters had to rank at
least one candidate.
Note G(1)=1 and F(1)=1.

To explain it in the next case G(2)=5 and F(2)=3:

F(2)=3 is because of the 3 ballot types A>B, B>A, A=B.

G(2)=5=2*3-1 is because of the 5 ballot types
     A>B, B>A, A=B *,
     A (B unranked), B (A unranked).
Note the *d ballot has no duplicate, but the others do get duplicated e.g.
"A>B" gets duplicated as "A(with B unranked)."

F(3)=13 is from
A>B>C and 5 others of this kind making 3!=6 in all,
A>B=C and 2 others (3 in all)
A=B>C and 2 others (3 in all)
A=B=C
so 6+3+3+1=13.

G(3)=25=2*13-1 then arises by duplicating everything (making the
last candidate now unranked) too, except A=B=C has no duplicate.

-- 
Warren D. Smith
http://RangeVoting.org  <-- add your endorsement (by clicking
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and
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