[ARIN-consult] Consultation on Implementing Single Transferrable Voting for ARIN Elections

[Richard Laager]

On 2/10/22 18:58, William Herrin wrote:
> A majority ranking a
> candidate last offers only one credible inference: that the majority
> disapproves of the candidate.
That's not correct, at least not for a definition of "disapprove" I 
would use.

Example: "Where should we go for lunch: A, B, or C?"

Imagine I'm genuinely fine with any of those. That is to say, I approve 
of all of them / I don't disapprove of any. We do a quick approval vote. 
It turns out that everyone feels like me: all choices are acceptable to 
everyone.

We still need a decision of which one place to go. Being voting system 
enthusiasts, we then quickly run an STV election (which for a single 
winner election is the same as IRV/RCV). This forces everyone to rank 
their choices from most to least preferred. The results might look like 
your example.

If you looked at just the voting results in isolation, you would declare 
that the majority disapproves of the last-placed choice. But here we 
know that's false; there is unanimous approval of all options.

It's true that the loser is the "least preferred", but that's different 
from "disapprove[d] of".

Put differently, I'd define "disapproved of" as "would rather leave the 
seat empty than have that candidate" or "would rank lower than 'None of 
the Above'".


I think what you might really care about is that STV is violating the 
Condorcet winner criterion. In your particular example, where STV 
elected 3 & 4, I agree that it did. I finally took the time to do the 
pairwise comparisons by hand. I showed my work below.

Unless I made a mistake, the Condorcet winners are 1 & 3. That is, 1 
wins every pairwise election and 3 wins every pairwise election other 
than with 1. Further, the Condorcet loser is 4; i.e. 4 loses every 
pairwise election. Also, from what I can see, this election has a strict 
Condorcet ranking: 1 > 3 > 2 > 5 > 4.

So maybe we want to pick a system that upholds the Condorcet winner 
criterion. What are our choices? There are various Condorcet methods 
that produce single-winners. As far as I can find, there are two 
Condorcet methods that can be used in multiple seat elections: CPO-STV 
and Schulze STV. The ballots are the same between traditional STV and 
those methods, but the computations to find the winners are much more 
involved; in CPO-STV, you're calculating all possible pairs of election 
outcomes.

However, while those systems produce what I'd argue are better results 
(honoring the Condorcet winner criterion), I think it's important that 
the voters understand how the system works so that they can trust it. 
When you get into systems that are so complicated that they need 
computers to calculate the winner, that might be the wrong trade-off in 
real life.

And to be clear, my personal view is that traditional STV is definitely 
better than FPTP.

----

 > Suppose you have five candidates for two positions. The votes are:
 > 30% 1, 3, 2, 5, 4
 > 20% 3, 1, 2, 5, 4
 > 19% 4, 1, 2, 5, 3
 > 10.5% 2, 3, 1, 5, 4
 > 10.5% 2, 4, 1, 5, 3
 > 5% 5, 4, 1, 2, 3
 > 5% 5, 3, 1, 2, 4

Calculate pairwise elections:
1 > 2: 30 + 20 + 19 +             + 5 + 5
1 > 3: 30 +      19 +        10.5 + 5
1 > 4: 30 + 20 +    + 10.5 +            5
1 > 5: 30 + 20 + 19 + 10.5 + 10.5
2 > 3:           19 + 10.5 + 10.5 + 5
2 > 4: 30 + 20 +      10.5 + 10.5 +     5
2 > 5: 30 + 20 + 19 + 10.5 + 10.5
3 > 4: 30 + 20 +      10.5 +            5
3 > 5: 30 + 20 +      10.5
4 > 5:           19 +        10.5

Sum:
1 > 2: 79
1 > 3: 64.5
1 > 4: 65.5
1 > 5: 90
2 > 3: 45
2 > 4: 76
2 > 5: 90
3 > 4: 65.5
3 > 5: 60.5
4 > 5: 29.5

Evaluate:
2 > 3 is false; flip it as 3 > 2.
4 > 5 is false; flip it as 5 > 4.

1 > 2
1 > 3
1 > 4
1 > 5
3 > 2
2 > 4
2 > 5
3 > 4
3 > 5
5 > 4

Simplify:
1 > 2345
3 > 2
2 > 45
3 > 45
5 > 4

Simplify:
1 > 2345
3 > 2 > 45
5 > 4

Simplify:
1 > 235 > 4
3 > 2 > 5 > 4

Simplify:
1 > 3 > 2 > 5 > 4

-- 
Richard