Then when you say that rest of the network has 1% of coin-confirmations it means that each person has 0.001% of coin-confirmations. You need to plug this value into hash-target calculation, not the sum of coin-confirmations they have.
Sorry, you are right and I screwed up again. Not much sleep recently, I apologize.
As you point out, my numbers applied to a single legit miner and the short-term reorg situation is different when miners are numerous. For short-term reorgs, the correct reference is to a representative, average miner like you say. However, when the number of miners increases, the average number of coin-confirmations in a successfully mined block tends would tend to increase in a similar proportion. Thus, it is not obvious that it actually becomes progressively easier to do reorgs as mining becomes more atomistic. We need to analyze the entire system to get an answer here.
Take the 50-50 weighting. For simplicity normalize total coins and hashing power to 1.
Suppose there is one legit miner who owns c 0
Excluding periodic reorgs, the legit miner will consecutively mine every single block with c coin-confirmations. The legit miner's voting power is xc. The attacker will need to exceed this for 6 consecutive blocks to do a reorg. Let w be the number of blocks the attacker waits between attacks. The attacker's coin-confirmations are w(1-c). The attacker is tied with the legit miner if w*(1-c)*(1-x)= 6xc. Solving for w indicates how long the attacker need to wait between attacks. w=6xc/[(1-c)(1-x)]
Thus, if the attacker has 5% of coins (c=0.95) and 5% of hashing power (x=0.95), he can attack every 2166 blocks. That is about once every 15 days. Attacking generates 6 blocks. Legit mining would generate 0.05*2166=108 blocks. There is a sacrifice of 102 blocks to facilitate attacks. Each attack pays off if it facilitates a theft equivalent in value to 102 block rewards.
What if mining is more atomized than this. Suppose there are n identical legit miners, who own nc coins in total, 0
Excluding periodic reorgs, each legit miner will mine once every n blocks with nc coin coin-confirmations. An individual legit miner's mining power will be nxc. Collectively, the n individual miners have mining power of n^2xc. The attacker will need to exceed this collective mining power for 6 consecutive blocks to do a reorg. Let w be the number of blocks the attacker waits between attacks. The attacker's coin-confirmations are w(1-nc). The attacker is tied with the legit miners if w*(1-nc)*(1-nx)= 6n^2xc. Solving for w indicates how long the attacker need to wait between attacks. w=6(nx)(xc))/[(1-nc)(1-nx)].
Thus, if the attacker has 5% of coins (nc=0.95) and 5% of hashing power (nx=0.95), again he can attack every 2166 blocks. Atomizing mining doesn't change anything.
Suppose we decide that this is to frequent and want to enforce a threshold of once a year (every 52560 blocks) in this scenario (nc=0.95) (nx=0.95). This would mean that each 6 block reorg would need to pay off 2622 blocks rewards in theft value to payoff. Remember the attacker is forgoing 2622 mined blocks in order to prepare each double spend. Seems like a lot.
To achieve this, the stake weighting needs to be lowered. We started out with 50-50, we would achieve this target if we reduce stake to 41 and raise work to 59. Perhaps 40 stake-60 work is a good split.
Notes:
We must find g such that 52560=2166^g
g=ln 52560/ln2166
g=1.41520944787
(1-p)/p=g (where p is the stake weighting and 1-p is the work weighting)
p=1/2.41520944787=0.41