[EM] IRV ballot pile count (proof of closed form)

[Abd ul-Rahman Lomax]

At 08:19 PM 2/4/2010, robert bristow-johnson wrote:

>On Feb 4, 2010, at 7:51 PM, Kathy Dopp wrote:
>
>>The general formula for the number of possible rankings (for strict
>>ordering, without allowing equal rankings) for N candidates when
>>partial rankings are allowed and voters may rank up to R candidates
>>(N=R if voters are allowed to rank all candidates) on a ballot is
>>given on p. 6 of this doc:
>>
>>http://electionmathematics.org/ucvAnalysis/US/RCV-IRV/ 
>>InstantRunoffVotingFlaws.pdf
>
>the only issue, Kathy, is whether the lower limit is i=0 or i=1.  you
>have to defend your use of i=0 for the case illustrated below (using
>the rules in Burlington VT and Cambridge MA).

i represents the number of blank positions, which when full ranking 
is allowed, and i = 0, represents explicit full rankings. Dopp states 
that for three candidates, there are 15 totals. There are actually 
two more, the empty rankings, i.e., ballots with no identifiable or 
legal vote. These are separately reported, divided into two 
categories, blanks and spoiled. Under some rules, blanks and spoiled 
ballots have different consequences, but I've never seen this apply 
to public elections. It does apply to preferential voting as 
described by Robert's Rules.

The core of the issue here is whether or not A>B>C is equivalent to 
A>B, and whether it is or is not equivalent depends on the definition 
of "majority." Is an A>B>C vote a vote for C? If write-ins are 
allowed, it certainly is! It may be moot in *most* situations, but 
not all. Kathy's formula is accurate for the general case, but misses 
the "spoiled ballot" issue. Note that ballots with a vote on them for 
anyone, including write-ins, may be considered in determining if the 
first round terminates having a majority. For practical purposes, 
write-ins may be reported as if they were a single candidate, and 
then batch-eliminated. But it could happen that the sum of all 
write-ins could create an ambiguity and this would require going back 
to the precincts for more detailed reporting. Not fun. Basically, a 
complete recount would be necessary. With more complicated reporting.

So, at the least, write-in ballots must be reported. Now, what 
happens if a write-in is in second place? What if every voter writes 
in the same candidate in second place, and nobody writes in this 
candidate in first place? The candidate is not in the list of 
eliminated candidates; I've never seen IRV rules that address this 
issue. By the intentions of IRV, though, I'd have to interpret that 
in order to proceed to the second round, all candidates not listed in 
the first round counts are eliminated, just on that basis. Yet this 
points out how IRV is definitely not equivalent to runoff voting 
where write-ins are allowed in subsequent rounds.... (as in the 
California general case).

>>In the US, R=3 in most IRV elections.
>
>in Burlington it was 5 in both 2006 and 2009.  N was also 5 (not
>counting any write-in).

But write-ins must be considered and reported in the first round. 
Routine practice is to report them as a single total, and normally 
that total then shows that, even if all write-ins were a single 
candidate, the candidate would be eliminated. So I've always seen the 
write-in total reported. Write-ins cannot be eliminated from the 
report. So in the Burlington election with 5 candidates on the 
ballot, there are really 6 that must be reported. Again and again and 
again, RBJ neglects and ignores this, and he then also imagines that 
the minor candidates can be similarly ignored, with reporting for the 
three major candidates as if it could be known that the minor 
candidates are irrelevant. For the ballot and election commission, 
there are no "minor candidates," except that write-ins are routinely 
treated as truly minor. All candidates on the ballot who receive even 
a single vote are reported in the full election results (which with 
IRV is reported as first round totals.)


>but Kathy, suppose N=R=3 and it's the "regular-old" IRV rules that do
>not require any minimum number of candidates ranked and do not allow
>ties.  to be clear, i need to also point out that only *relative*
>ranking is salient (at least in Burlington).  if a voter only ranks
>two candidates and mistakenly marks the ballot 1 and 3, the IRV
>tabulation software will close up the gaps and treat that precisely
>as if it was marked 1 and 2.

Kathy's formula does not separately report A>.>B and A>B. The 
clarification is correct but not in any way a criticism of Kathy's 
formula. The difference between the two is of interest in studying 
error rates, but not for determining the winner with IRV. With 
Bucklin, the two votes are quite different, but, of course, Bucklin 
is precinct summable, very simply. In the two votes that IRV 
collapses, B will be counted in the third round in the first case, 
and in the second round in the second. It's important to Bucklin, as 
it should be.

>now, given those parameters, are you telling us that the 9 tallies
>shown on Warren's page:
>  http://rangevoting.org/Burlington.html (i have very similar
>numbers, no more different than 4), are not sufficient to apply the
>IRV rules and resolve the election?
>
>     1332  M>K>W
>      767  M>W>K
>      455  M
>     2043  K>M>W
>      371  K>W>M
>      568  K
>     1513  W>M>K
>      495  W>K>M
>     1289  W

They are not sufficient as to what the precincts must report, which 
is the point here. First of all, what is the number of ballots 
containing a vote in the first round? That is the basis for the first 
round majority, and the first round terminates the tabulation if a 
majority is found. In order to limit the reporting to the patterns 
above, we must assume that none of the minor candidates or write-ins 
got, overall, as many votes as the least of M, K, or W. In order to 
know that, the totals from all precincts must be known. The election 
officials cannot assume that *any* candidate is "minor," even if that 
candidate only gets 1 vote in the precinct, because the candidate 
might get sufficient votes from other precincts.

Practically speaking, I'd assume, the precincts would be provided 
with a spreadsheet showing the possible combinations, and they would 
report the combinations using the spreadsheet, transmitting it. So 
some cells would be blank or zero. With 5 candidates on the ballot, 
the spreadsheet has gotten large, but it's still doable. What happens 
if preferential voting encourages more candidates to file, as it 
tends to do? 23 candidates in San Francisco? Even with three-rank 
RCV, it gets hairy.

>is there any reason those 9 tallies could not have been summed from
>subtotals coming from all 7 wards of Burlington?

Yes. I stated it. Those are not the first round sums, they are the 
second round, after batch elimination of the two minor candidates and 
the write-ins. Elimination cannot be done at the precinct level, for 
the reason stated above, it must be done centrally. RBJ has skipped a 
step, based on assumptions about write-ins and minor candidates. It 
is considered allowed to report all write-ins as a single tally, but 
the votes for all candidates on the ballot, as well as a write-in 
total, as well as a spoiled ballot total, must be reported.

Then, for each minor candidate, treating "write-in" as another minor 
candidate, the subsequent rankings must be reported separately. RBJ 
has lumped them all together. If it should turn out that "write-in" 
would not be eliminated, the precincts would have to then recount 
write-ins to categorize them to determine if there is a viable 
write-in candidate, at least to advance to the next round.

>please tell us why those 9 piles are not enough, given the parameters
>stated above?

In Warren's analysis, the first round of batch eliminations have been 
done; more complete ballot totals should have been reported, 
actually, for a complete study. (It would be acceptable for Warren's 
purposes if all minor candidates and write-ins had been lumped as a 
single "minor" category, but this is legitimate only for later 
analysis. The precincts cannot report that way, because they can't 
know which candidates are minor until the first overall canvass is 
done. Reporting write-ins as a single category is a compromise that 
*usually* works, which is why it is done. But that *cannot* be done 
with candidates on the ballot. It's rude not to tell a candidate how 
many votes they got. At least first preference votes! I have never 
seen an election report that did not report totals for all explicit 
candidates, with a sum for write-ins.)

For this analysis, M>K>W is not a complete ranking, as the first 
example. It could be simply that vote, or it could be write-in>M>K>W, 
or minor candidate>M>K>W. There are thus three other possibilities 
that must be reported if they exist, as there were two minor 
candidates. In addition, the spoiled ballot total must be reported, 
but that's only a single total overall. As I recall, undervotes are 
also reported, i.e., blank, also a single total.

This is true for all the combinations shown, so the number of 
combinations that are mandatory to report is four times that shown by 
RBJ (neglecting the spoiled and blank ballot reports).

If RBJ wants to challenge this, he should provide a means for the 
precincts to know what totals to report, that works in the general 
case. He's been asked for this before, but he's ignored it, and has 
continued arguing on a strange mixture of pure math and unstated assumptions.

There is a way to avoid such massive reporting, which is to report 
interactively, which is what is done in Australia. Only one set of 
totals is reported from a precinct at a time, the totals for the 
current round. (which can be just uncovered votes due to eliminations 
that have been reported to the precinct from central tabulation.)

However, the problem with this is that a single error in a precinct 
can require, then, all precincts to have to retabulate.