Subject: Re: [Bitcoin-development] Even simpler minimum fee calculation formula: f > bounty*fork_rate/average_blocksize

From: Michael Gronager <[email-masked]>
Date: Fri, 15 Nov 2013 12:58:14 +0100

>> Q = Total pool size (fraction of all mining power) q = My mining
>> power (do.) e = fraction of block fee that pool reserves
>> 
> 
> Unfortunately the math doesn't work that way. For any Q, a bigger
> Q gives you a higher return. Remember that the way I setup those
> equations in section 3.2 is such that I'm actually modeling two
> pools, one with Q hashing power and one with (1-Q) hashing power.
> Or maybe more accurately, it's irrelevant if the (1-Q) hashing
> power is or isn't a unified pool.

My Q and q are meant differently, I agree to your Q vs Q-1 argument,
but the q is "me as a miner" participating in "a pool" Q. If I
participate in a pool I pay the pool owner a fraction, e, but at the
same time I become part of an economy of scale (well actually a math
of scale...) and that can end up paying for the lost e. The question
is what is the ratio q/Q where I should rather mine on my own ? This
question is interesting as it will make bigger miners break away from
pools into solo mining, but I also agree that from pure math the most
advantageous scenario is the 100% mining rig.

> The equations give an incentive to centralize all the way up to 1
> miner with 100% hashing power.
> 
> Of course, if that one pool were p2pool, that might be ok!

Ha, yes, and then the math for p2pool starts... a math where we have
much more stales...