Following statman's lead, perhaps I can help resurrect this forum by starting threads, beginning with this easy one motivated by skimming the old threads on this topic.

You're heads-up on the turn, you get all-in and one of you is drawing to 9 outs.

a) If the variance of net profit when running it once is v, what's the variance when running it twice?

b) Does the answer change if, instead of dealing two separate rivers without replacement (as is typical), you reshuffle and redeal the river?

This thread flopped, so I'll just give the solution.

For simplicity, let's say the final pot is 1. And since the money is already in, net and gross profit are equal.

One way to calculate the variance is directly. The EV is 9/44, and when running it once that's also the expected squared profit, so the variance running it once is 9/44 - (9/44)² ≈ .16271
Running it twice, we know the EV is the same, so the variance is E(profit²)-E(profit)² = (9/44)(8/43) + 2(9/44)(35/43)(1/2)² - (9/44)² ≈ .07946
The reduction factor is thus 0.48837, or 21/43 to be exact.

Another approach is to set it up as a function of two random variables. Suppose X and Y are the profit from the 1st and 2nd run respectively if that run counted for the entire pot. Note that X and Y are identically distributed.

Since each run counts for half the pot, the variance when running it twice is:

Var(X/2 + Y/2) = Var(X/2) + Var(Y/2) + 2*Cov(X/2, Y/2) = Var(X)/4 + Var(Y)/4 + Cov(X,Y)/2

Cov(X,Y) = E(XY) - E(X)E(Y)

When X and Y are independent, E(XY)=E(X)E(Y) and so the covariance is 0. This is the case if the river is reshuffled.
Independent or not, Var(Y)=Var(X), so with reshuffling the variance is Var(X)/2, exactly half that of running it once.
Without reshuffling, X and Y aren't independent and their covariance is negative because hitting/missing the first run decreases/increases your chance of hitting the next one. So the variance running it twice must be less than half that of running it once.

E(XY)=(9/44)(8/43) and E(Y)=E(X), so Cov(X,Y) ≈ -.003784
Var(X/2 + Y/2) = Var(X)/2 + Cov(X,Y)/2 = (.16271 - .003784)/2 = .07946

Now let's generalize to running it n times with k outs and d unknown cards. This is where the random variables approach shines. Since each random variable has equal variance and each pair of RV's has equal covariance, the properties of variance tell us:

Var(Σ X_j / n) = [Var(X₁) + (n-1)*Cov(X₁ , X₂)]/n

We're interested in that divided by Var(X₁), meaning: [k/d - n(k/d)² + k(k-1)(n-1)/(d(d-1))]/n / (k/d - (k/d)²)

Reducing eliminates k and leaves us with (d-n)/(dn-n)

If each run involved reshuffling, the variance reduction would be 1/n, but instead it's 1/n times (d-n)/(d-1)