[vote] Update OpenWrt rules

[David Woodhouse]

On Tue, 2025-11-04 at 15:14 +0000, Daniel Golle wrote:
> 
> Example:
> 
> 9 votes for (A)
> 4 votes for (B)
> 8 votes neutral
> 
> In the currently proposed model, this would result in
> 
> (A) 9 + 0.5 * 8 = 13  (61.9%)
> (B) 4 + 0.5 * 8 = 8   (38.1%)
> 
> While the ratio of the non-neutral votes was actually
> 
> 69% (A) vs. 31% (B)
> 
> If this had been a decission which required a 2/3rd majority, then
> the neutral votes would have changed the outcome in this case.
> 
> What we can do is simply not count them but assume quorum is
> reached once enough members indicated their mere awareness of the
> voting process.
> 
> What Adrian probably means is that a "neutral" vote should be
> counted as 0.69 for (A) and 0.31 for (B) rather than 0.5 for each.
> In this way it would not change the outcome at all.

Yes. But I don't think we can use (A) and (B); it isn't multiple
choice. I think you meant *yes* where you said (A) and *no* where you
said (B). The two-thirds threshold is important in the yes/no model.

We should *discount* the neutral votes and count that the 9 yes votes
are 9/13 == 69.2% of the total yes/no votes.

The more confusing way of doing it is to keep the neutral votes, and
count them as 2/3 of a vote. So it's 9 + (8 * ⅔) == 14⅓ out of 21,
which is 68.3%.

You don't have to count the neutral vote as *precisely* the same as the
ratio between yes/no, because you don't have to avoid affecting the
*numeric* result. You just have to avoid taking it across the 2/3
threshold in either direction, so a neutral vote should mathematically
be 2/3.

Congratulations, you found an even *more* confusing way to do it :)

We should just discount the neutral votes when evaluating the yes/no
against the 2/3 threshold :)

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