OK, I get your point. I don't think the error in my approximation is quite as horrendous as you do, but whatever.
In any case, I finally crunched the numbers, and I worked out the proper solution using the multinomial approach, which properly accounts for the covariance. Here it is, for the benefit of the OP:
As explained above, we want to find Var(G3 - G2).
By the properties of variance
, we know that for two dependent random variables G3 and G2:
Var(G3 - G2) = Var(G3) + Var(G2) - 2*Cov(G2,G3).
The variance calculations are just like those from the binomial formula, which OP knows:
Var(G2) = (.21*(1-.21))/100 = 0.001659
Var(G3) = (.12*(1-.12))/100 = 0.001056
Now for the covariance, using the multinomial formula we get:
Cov(G2,G3) = (-1*p1*p2)/n = -1*(.21*.12)/100 = -0.000252
Var(G3) + Var(G2) - 2*Cov(G2,G3) = 0.001659 + 0.001056 + 2*0.000252 = 0.003219.
Now to get SE(G3 - G2), I just take the square root = sqrt(0.003219) = 0.05673623.
Finally, I use the normal approximation to get the Pr((G3-G2) < -.095). Using the calculator here
, I plug in a mean of 0, an SD of 0.05673623, and find the area below -0.095 = 4.75%.
Voila, that's the way to do it using the proper (multinomial) approach.
I've double-checked all this with my own simulations.
So... Hopefully this will be BruceZ approved