# [EM] IRV proponents figure out how to make IRV precinct-summable

Mon Mar 23 07:46:44 PDT 2009

```Dave Ketchum wrote:
> On Mar 22, 2009, at 4:24 PM, Kristofer Munsterhjelm wrote:
>
>> As stated, it's not summable. But note that the second round, which is
>> determined by the Plurality count, consists of a pairwise comparison.
>> Thus, one can make the method summable by simply storing the
>> information required to simulate any one-on-one runoff -- in other
>> words, by having a Condorcet matrix. Since Condorcet is not mutually
>> exclusive with summability, we know Condorcet matrices can be summed -
>> so that part is summable. We also know that Plurality counts are
>> summable - if A gets X votes in district 1 and Y votes in district 2,
>> A got X+Y votes in these districts combined.
>
> Agreed that Condorcet and Plurality, and even Approval, are summable.
>
> But suppose voters say A>B>C and A>C>B:
>      Condorcet will count A>B, A>B, A>C, A>C,  B>C, and C>B into the N*N
> matrix.
>      IRV will only see A>? and A>? until and unless A losing exposes
> what remains (B>? or C>?).
>
> True that ballot images could be forwarded, but that does not really
> make summable claimable.

Agreed (in turn) that forwarding ballot images doesn't make a method
summable, since otherwise, any method that doesn't care about the order
of the ballots would be "summable".

Also, IRV, in the general case, is not summable. However, what we're
talking about is the contingent vote, an "instant top-two runoff", which
is what the IRV proponents figured out how to make precinct summable (or
thought they had figured out how to make precinct summable). It agrees
with IRV if the number of candidates <= 3.

The contingent vote first counts plurality votes for the various
candidates, as top-two runoff does. Then, again as in top-two runoff,
the two "winners", Plurality wise, go to the next round. The difference
is that the contingent vote uses the same rank ballots for the second
round as for the first, only with all non-winners eliminated, whereas
true TTR has a separate second round.

Let's have a concrete example of how the contingent vote works, and why
my approach to it is summable.

District 1:     100: A > B > C
98: B > C > A
27: C > A > B

District 2:     104: C > B > A
121: C > A > B
50: A > B > C
25: B > A > C

Combined:       104: C > B > A
150: A > B > C
148: C > A > B
98: B > C > A
25: B > A > C

For the combined ballot, first do a Plurality count to see who advances.
Note that this Plurality count is what makes contingent vote equal to
IRV for number of candidates = 3, since preserving the top two is equal
to eliminating the last candidate.

Combined, plurality: 104 C, 150 A, 148 C, 98 B, 25 B
hence:               252 C, 150 A, 123 B

So C and A move to the second round.

Eliminating B, we get
104: C > A
150: A > C
148: C > A
98: C > A
25: A > C
summing up,
350: C > A
175: A > C

So C wins.

Let's look at the combined Condorcet matrix. It is

A    B    C    beats
A ---  227  350
B 298  ---  252
C 175  273  ---

Here you can see that the data we're looking for is "C beats A" and "A
beats C". Since there are only two candidates remaining from the first
round, the second round will be an one-on-one, which is the kind of
contest the Condorcet matrix stores information about. If C beats A more
often than A beats C, C is the winner. Incidentally, this shows that the
contingent vote passes Condorcet loser.

But let's do it again, with only the summable information about each
district (that is, the plurality count and the Condorcet matrix).

District 1:
Plurality count: 100 A, 98 B, 27 C

Condorcet matrix:    A    B    C    beats
A ---   98  125
B 127  ---   27
C 100  198  ---

District 2:
Plurality count: 50 A, 25 B, 225 C

Condorcet matrix:   A    B    C    beats
A ---  129  225
B 171  ---  225
C  75   75  ---

Let's sum this all up:
Plurality count: 150 A, 123 B, 252 C

Condorcet matrix:  A    B    C    beats
A ---  227  350
B 298  ---  252
C 175  273  ---

And run the election method again:
First "round": greatest two are A (150) and C (252)
So A and C go to the second "round".
Second "round": A>C by 175, C>A by 350, so C wins.

There you go, the contingent vote is summable.

>> I'm not sure about IRV - has anyone devised an STV variant that
>> handles equal rank? If not, then you're right - again, I'm not sure.
>
> Brian claims, and I cannot disprove, that IRV can be stretched to
> tolerate equal rank - questionable whether it would be worth the expense
> for real elections.

There are two ways to handle equal rank, in theory, for a weighted
positional method. Plurality is just a weighted positional method with
the weights (1, 0, 0, ..., 0). The first is "whole", which means that if
you rank A = B > C, A and B has the same score, which is the same as A
in A > C. For plurality, that would turn it into Approval. The second is
"fractional", which means that the sum of the score for all ranked
candidates in a certain rank is the same, no matter how many you ranked.
For instance, for Plurality, ranking A = B > C would give half a point
to A and B (so that the sum is 1), and none to C, whereas ranking A > B
> C would give a full point to A and none to B or C.

Elimination would work the same way however equal rank would be treated.
If you vote A = B > C and A is eliminated, then for the next round, your
vote is B > C.

Assuming the voting machines can handle the input, where would the
expense lie in adding this support? It seems to be more a question of
whether the resulting system would be "IRV" or not... unless the expense
would be in "handling the input", but if you have a machine that can
handle A > B > C > D > E .. > Z, upgrading it to handle A = B = C > D >
E ... doesn't seem to be that expensive a change.

>> From what I've seen of voting equipment, most limitations seem to be
>> in the name of expediency. For instance, SF's RCV three-rank method
>> keeps voters from ranking more than three candidates - probably to
>> accomodate existing equipment.
>
> I am suspicious as to this relating to existing equipment, but:
>      Some ways of providing for more ranks significantly burden
> equipment design.
>      Providing for 2 ranks is essential to deserve claim as to ranks
> existing; 3 helps some; more than 3 helps real voters little.

One may then ask, how many ranks are required to break Duverger's Law?
Unfortunately, I don't know the answer.

>> What limitations may exist (such as your IRV example) may be handled
>> by having a voting machine that permits all ranking types (full,
>> truncated, equal rank), then having parameters that limit according to
>> what kind of voting system is being used in the back end (e.g no equal
>> rank).
>
> Sounds like building in expensive complications.
>
> Doing the specialization in software could be affordable.

There's a tradeoff at this point. Having a generalized machine lets you
build many that are all the same, so that you gain benefits of scale.
However, the generalized machine is more expensive because you can't cut
away what you don't need.

>> It is in theory possible to make it summable - see above. The method
>> they did use seems not to be, though - as far as I could see, they
>> checked, for all possible virtual runoffs (set by enforcing A and B as
>> winners in the first round), whether A or B won. Such a binary check
>> is summable only if the results are the same in both districts - but
>> when they're different, one runs into trouble. Consider this, for
>> instance:
>>
>> District 1      X>Y: 1000, Y>X:  990    X beats Y
>> District 2      X>Y:    1, Y>X:    2    Y beats X
>> -------------------------------------------------
>> Summed result                           X beats Y
>>
>> but also
>>
>> District 1      X>Y: 1000, Y>X:  990    X beats Y
>> District 2      X>Y: 1000, Y>X: 2000    Y beats X
>> -------------------------------------------------
>> Summed result                           Y beats X
>>
>> In both instances, X beats Y in the first district, and Y beats X in
>> the second district, but the summed result is different for the two
>> cases. Thus I think that they would have to store the entire Condorcet
>> matrix (numbers of voters, not just who won) in order to be summable.
>> If they did, then they're summable, but if they didn't, they aren't.
>
> Condorcet cares not as to number of voters - for it simply sum the
> matrices.

If nobody equal-ranks, then (A beats B) + (B beats A) = number of
voters. Apart from that, you're right, Condorcet doesn't care. What I
showed was that if they (the IRV proponents) tried to use only binary
arrays instead of integer arrays for their kinda-Condorcet matrices,
they would fail, because there's not enough information there. A
Condorcet matrix has to be integer (or even more fine grained, e.g for
CWP), even when that matrix is only to be used for determining the
winner of the contingent vote.

>> I mean freedom as a data format. A rated vote data format can emulate
>> a ranked vote format, as well as an approval-style data format.
>
> Saying freedom reminds me of something we sometimes ignore - how much
> complication do we burden voters with.

Voting is already irrational from a utilitarian point of view - your
chance of affecting the outcome is way too small for it to be worth
bothering to vote, let alone consider the issues to make an informed
decision. Yet we vote anyway.

That muddies the waters, because we can't use standard
utilitarian/economic theory to find out how much complication is too
much. Perhaps people wouldn't bother with anything more than Approval,
but that seems wrong (since people rate and rank things all the time).
So, how much is too much? I don't know.

Of course, the user interface should be good, but that's a separate
issue. I don't particularly like general purpose direct electronic
machines, so the "user interface" may be entirely transparent - put the
appropriate number in the box next to the candidate (for each
candidate), then at some later time, OCR reads off the numbers to parse
the ballot.

```