But the way to mitigate the attack is similar to what you described (I'm not sure why you described it as the square root of the difficulties?), but just assign a blockchain score of:

block=1{summation}block=n of (PoW Difficulty in Block 1 * (PoS Difficulty in Block 1)^-1) + ... + (PoW Difficulty in Block n * (PoS Difficulty in Block n)^-1)

I don't get this. This is the weight for the total blockchain right?

PoS difficulty is the price of a ticket in terms of % of total coins available. The price is adjusted to sell a constant # of tickets. If the PoS Difficulty is low (tickets are cheap), then only a small fraction of total coins are participating in the lottery. The chain should be penalized for this. Your formula rewards the attacker for lowering PoS Difficulty, making the problem worse not better.

Again I suggest you use:

block=1{summation}block=n of (PoW Difficulty in Block 1)^0.5 * (PoS Difficulty in Block 1)^0.5 + ... + (PoW Difficulty in Block n)^0.5 * (PoS Difficulty in Block n)^0.5

The only time you can get absolute certainty that a double spend can succeed is with 51% PoW and 51% PoS. But you suspect there is a flaw in this logic...

Nothing to lose sleep over. Even though your logic is flawed, it does not reflect a fundamental flaw in your design. One chain will always be either better, equal to, or worse than another chain. The algorithm evaluates proof chains to decide this.

One group's chain controls a fraction x PoW and a fraction y PoS. The other group then has corresponding fractions 1-x PoW and 1-y PoS. The groups use their resources to construct long proof chains. Let f be a function evaluating the proof in a chain constructed using a PoW and PoS pair over a long time span. Of course PoW outcomes are random, but let's just ignore that to keep things simple.

We must have f (x, y) > f (1-x, 1-y) or f (x, y) = f (1-x, 1-y) or f (x, y) < f (1-x, 1-y). That is the first group wins, it is a tie, or the first group loses. The winning group gets complete control over which txns enter PoW blocks. The losing group becomes completely irrelevant.

f (x, y) = f (1-x, 1-y) describes a boundary on 0 <=x <=1 and 0 <=y <=1. If one group (good or bad) is on this boundary and adds on just a little bit of hashing power or a little bit of stake, then this group wins.

Bitcoin uses the boundary x = 0.5

My ideal boundary is y = 0.5 (that's what you approach as the number of signatories per block increases)

Your boundary needs to simulated because of the two difficulties.

But if we ignore that (only matters over long intervals anyway) and make an additional assumption, we can calculate it.

Here's the assumption:

You are including invalidated empty blocks in the block chain. If these empty blocks contribute to proof just like normal blocks, a 51% PoW attacker can select the winning chain, even with 0% stake. This makes PoS close to pointless. To avoid this, I'm going to assume that the empty blocks do not contribute to proof at all.

Then:

A PoW block contributes to proof iff it has at least 3 stake votes.

Proof Creation Rate = (Rate at Which Group Creates PoW Blocks) * (Probability Group Controls 3, 4, or 5 Stake Votes)

Probability Group Controls 3, 4, or 5 Stake Votes is a binomial distribution with Pr (X>=3) for a binomial distribution with parameters n=5 and p=y for group 1 and p=1-y for group 2.

http://en.wikipedia.org/wiki/Binomial_distributionPr (X>=3) = Pr (X=3)+Pr(X=4)+Pr(X=5)= 10y³(1-y)²+5y⁴(1-y)+y⁵ for group 1

Rate at which Group 1 creates PoW blocks is x*(total hash rate of both groups / PoW difficulty).

So we have,

Group 1 Proof Creation Rate = f(x,y) = (10y³(1-y)²+5y⁴(1-y)+y⁵) * (x) * (total hash rate of both groups / PoW difficulty)

Group 2 Proof Creation Rate = f(1-x,1-y) = (10(1-y)³(y)²+5(1-y)⁴(y)+(1-y)⁵) * (1-x) * (total hash rate of both groups / PoW difficulty)

The two groups are tied if

(10y³(1-y)²+5y⁴(1-y)+y⁵) * (x) * (total hash rate of both groups / PoW difficulty) = (10(1-y)³(y)²+5(1-y)⁴(y)+(1-y)⁵) * (1-x) * (total hash rate of both groups / PoW difficulty)

Solving for x gives us the boundary

x = -6y⁵+15y⁴-10y³+1 on 0<=y<=1

In table form:

x y

1 0

0.99144 0.1

0.94208 0.2

0.83692 0.3

0.68256 0.4

0.5 0.5

0.31744 0.6

0.16308 0.7

0.05792 0.8

0.00856 0.9

0 1

So yes, you probably weren't aware that you could win with 90% stake and 0.9% work. But why is that a problem?