Re: [Bitcoin-development] User vote in blocksize through fees

[Matt Whitlock <bip@mattwhitlock.name>]

On Friday, 12 June 2015, at 7:44 pm, Peter Todd wrote:
> On Fri, Jun 12, 2015 at 02:36:31PM -0400, Matt Whitlock wrote:
> > On Friday, 12 June 2015, at 7:34 pm, Peter Todd wrote:
> > > On Fri, Jun 12, 2015 at 02:22:36PM -0400, Matt Whitlock wrote:
> > > > Why should miners only be able to vote for "double the limit" or "halve" the limit? If you're going to use bits, I think you need to use two bits:
> > > > 
> > > > 	0 0 = no preference ("wildcard" vote)
> > > > 	0 1 = vote for the limit to remain the same
> > > > 	1 0 = vote for the limit to be halved
> > > > 	1 1 = vote for the limit to be doubled
> > > > 
> > > > User transactions would follow the same usage. In particular, a user vote of "0 0" (no preference) could be included in a block casting any vote, but a block voting "0 0" (no preference) could only contain transactions voting "0 0" as well.
> > > 
> > > Sounds like a good encoding to me. Taking the median of the three
> > > options, and throwing away "don't care" votes entirely, makes sense.
> > 
> > I hope you mean the *plurality* of the three options after throwing away the "don't cares," not the *median*.
> 
> Median ensures that voting "no change" is meaningful. If "double" + "no
> change" = 66%-1, you'd expect the result to be "no change", not "halve""
> With a plurality vote you'd end up with a halving that was supported by
> a minority.

I'm very confused.

Let's say, out of the 12,000 blocks in a voting period:
• 1000 blocks express no preference,
• 2000 blocks vote to halve the limit,
• 3500 blocks vote to double the limit, and
• 5500 blocks vote to keep the limit the same.

 The plurality vote is to keep the limit the same. The median vote is… well, I'm not sure, since there are four choices, and an average of the two "middle" choices doesn't seem to make sense.