[EM] The Equal Vote Coalition and robla

[Kristofer Munsterhjelm]

On 2025-04-24 18:41, Richard wrote:
> On 4/24/25 02:20, Kristofer Munsterhjelm wrote:
> 
>> What would happen if there were three such voters, all voting B=D? To be
>> fair, it would seem you would have to place one and a half voter in B's
>> line and the other in D's. But your constraints say you can't divide the
>> voters into fractions.
> 
> If the conference-hall simulation matched actual election counting 
> rules, the third voter would not vote in that round.
> 
> If a fourth voter also has the B=D equal preference, then that's two 
> pairs of voters cooperating.
> 
> If decimal numbers were used, the counts would be rounded down to the 
> nearest integer.  This correctly simulates the pairing approach. 
> Specifically, with three voters having the B=D preference, the count of 
> three would be divided by the number of candidates involved, which is 2 
> (B and D), that yields 1.5 and that's rounded down to 1, which is the 
> number of votes going to candidate B and the number of votes going to 
> candidate D.

This seems problematic, though. In your original post, you said "Here's 
how to correctly count overvotes in governmental elections, without 
departing from the principle of one voter supporting only one candidate 
in each round of voting."

My (implied) point is that the conference-hall method, and its 
simulation, leaves something to be desired because additional voters 
with an equal preference may be deprived of a say in the outcome. And if 
it does leave something to be desired, it cannot unconditionally be 
called "the correct" way to count. If "the principle of one voter 
supporting only one candidate" is meant to be a constraint, it might be 
(given the constraint), but then it remains to be shown that this 
condition has value in itself.

As an example of uneven voting power, say there's a near tie election 
with the following first preference counts so far:

A: 1700
B: 1500
C: 1300
D: 1202
E: 1200

and suppose that three ballots that rank A=E are counted next, then two 
ballots ranking A=B=E, and finally three ballots ranking A=B=C=E.

By the conference-hall method, the three A=E ballots count for 1 point 
for each:

A: 1701
B: 1500
C: 1300
D: 1202
E: 1201

and the two A=B=E ballots count for zero (rounding down), as do the 
three A=B=C=E ballots. Hence, as far as I understand, E is eliminated.

By fully fractional counting, instead the three A=E ballots in total 
count as 1 + 1/2 for each candidate. To mitigate roundoff problems, I'll 
keep a table of how many of each fraction has been counted:

    wholes  /2   /3   /4   /5
A: 1701    1    0    0    0
B: 1500    0    0    0    0
C: 1300    0    0    0    0
D: 1202    0    0    0    0
E: 1201    1    0    0    0

The two A=B=E ballots count, in total, as 2/3 for each:

    wholes  /2   /3   /4   /5
A: 1701    1    2    0    0
B: 1500    0    2    0    0
C: 1300    0    0    0    0
D: 1202    0    0    0    0
E: 1201    1    2    0    0

and the three A=B=C=E ballots count as 3/4 for each:

    wholes  /2   /3   /4   /5
A: 1701    1    2    3    0
B: 1500    0    2    3    0
C: 1300    0    0    3    0
D: 1202    0    0    0    0
E: 1201    1    2    3    0

Then D is eliminated instead of E because 1201 + 1/2 + 2/3 + 3/4 > 1202.

> The difference between what I'm suggesting and what you're suggesting is 
> when division occurs.  I'm suggesting it happens once, after counting 
> the ballots that share the same pattern.  You're suggesting that a 
> ballot can be split into 0.5 votes for candidate B and 0.5 votes for 
> candidate D without waiting to see if there are an odd number of voters 
> with that ranking pattern.  This means your approach does division at 
> each ballot, instead of just once at the end of a counting round.

If by "division" you mean the splitting of ballots into fractions, 
that's correct. But if you mean the mathematical operation of division, 
then not necessarily. Consider the tabular form above, and let's look at 
D and E here:

    wholes  /2   /3   /4   /5
D: 1202    0    0    0    0
E: 1201    1    2    3    0

Until this point, we haven't done any division, we've just been counting 
fractions. Now we can do a division at the very end and find out that 
1201 + 1/2 + 2/3 + 3/4 > 1202.

This way of counting lends itself to easy reduction while the ballots 
are counted, because once a candidate's count of halves (/2 fractions) 
equals 2, then you can set it back to zero and increment the 
wholes/units place. Same with thirds (once you get three, just increment 
a unit and set the thirds' count to zero).

This "increment the units" strategy has a similar effect as your 
approach: two A=B voters contribute one point to each, in this case 
because each adds a 1/2 fraction to both candidates, and then once you 
get 2/2, you can increment the unit. In addition, you can keep a proper 
count without having to keep track of how many e.g. B=C ballots you've 
seen before when you encounter another one. Instead of thinking "I last 
gave a point to C when I saw a B=C ballot, next one must be to B", you 
can just add one /2 point to each, every time, and then resolve into 
full votes once enough fractions accumulate.

The difference is that this tabular method keeps the residual fractions 
until the end of the process so ties can be broken if necessary.

If the full vote values are too far apart, then you don't need to do a 
division at all. E.g. if the result had been:

    wholes  /2   /3   /4   /5
D: 1250    0    0    0    0
E: 1201    1    2    3    0

then since every fractions column contributes less than a full vote 
(otherwise, that full vote would have been drawn into the units place), 
the value for E can't be greater than its unit count plus the number of 
fractions columns, here 1201 + 4. Since 1205 < 1250, there's no need to 
check the division: E would be the loser anyway.

This might be more complex, but it doesn't seem like the conference-hall 
method is unambiguously "the correct way" to do things.

-km