We have to start at the beginning and along the way we'll see why there are two ways to calculate the SD.

The total of the actual payoffs is a sum of independent random variables $\displaystyle \sum_i R_i.$ The standard deviation $\displaystyle S$ of this sum is the square root of the variance $\displaystyle V$ of the sum. The variance of a sum of independent random variables is the sum of the variances $\displaystyle V_i$ of each $\displaystyle R_i ,$ so

$\displaystyle V = \sum_i V_i .$

There are two ways to calculate the $\displaystyle V_i .$ The most accurate is to use the known pot sizes $\displaystyle P_i$ and the win probabilities $\displaystyle w_i$ to calculate the actual variance. Since $\displaystyle w_i P_i$ is the mean of the payoff $\displaystyle R_i ,$ the actual variance is

$\displaystyle V_i = w_i (P_i - w_i P_i)^2 + (1-w_i)(0 - w_i P_i)^2 .$

Then the standard deviation of $\displaystyle \sum_i R_i$ is

$\displaystyle S = \sqrt{V} = \sqrt{\sum V_i} .$

The 95% confidence interval around the expected total payoff $\displaystyle E = \sum{w_i P_i}$ is

$\displaystyle [ E- 1.96 S, E+ 1.96 S ].$

There is no multiplication by $\displaystyle \sqrt{n}$ of the sample size. Where did that go? Well, there is a second way to get the variance $\displaystyle V$ and that is to estimate it ignoring the $\displaystyle w_i$. We assume the random variables $\displaystyle R_i$ have the same variance $\displaystyle v'$ and estimate that as

$\displaystyle v' = \sum_i (R_i - M)^2 / (n-1)$

where $\displaystyle M = \sum_i R_i /n$ is the mean actual payoff.

Then the variance $\displaystyle V$ is estimated as

$\displaystyle V' = \sum_i v' = n v' .$

$\displaystyle V'$ is an estimate of the actual variance $\displaystyle V$ which is not exact because it treats the variation of $\displaystyle w_i$ as unknown.

The estimated standard deviation of $\displaystyle \sum R_i$ is

$\displaystyle S' = \sqrt{V'} = \sqrt{n v'} = \sqrt{v'}\sqrt{n} = s' \sqrt{n},$

where

$\displaystyle s' = \sqrt{v'} = \sqrt{\sum_i (R_i - M)^2 / (n-1)}$

is the estimated common standard deviation of the $\displaystyle R_i .$

Then the 95% confidence interval is

$\displaystyle [ E- 1.96 S', E+ 1.96 S' ] = [ E- 1.96 s' \sqrt{n}, E+ 1.96 s' \sqrt{n} ], $

which is the familiar one.