of a weighted voting, where weights depend on subjective confidences, but not on individual performances. Members do not know their own — or their peers’ — performance scores, so there is no justification for assigning more or less weight to a particular member throughout the experiment. The only thing that matters is each member’s confidence in the present trial.

#### Results {#results-3 .unnumbered}

To calculate $P_{WCS}(c)$, we need to compute, given stimulus $c$, the probability of obtaining set $\vec{z}$ with a positive sum . Thus $$\begin{aligned}
P_{WCS}( c) &= \int_{x_1/\sigma_1 + \ldots + x_n/\sigma_n \geq 0}  \exp \left[ - \frac{(c+b_1-x_1)^2}{2\sigma_1^2}+\right.\\
&\left.-\ldots- \frac{(c+b_n-x_n)^2}{2\sigma_n^2}\right]\frac{dx_1\cdots dx_n}{(2 \pi)^{n/2} \sigma_1 \cdots \sigma_n}\nonumber\\
&= H\left[ \sqrt{2\pi} s_{WCS} \left( c + b_{WCS}\right)  \right],\end{aligned}$$ where the integration is based upon the fact that a sum of Gaussian random variables $z_i$ is a Gaussian random variable [@Piau2011]. The resulting parameters are: $$\begin{aligned}
s_{WCS} & = \sqrt{n} \times \frac{s_1+\ldots+s_n}{n}, \\
b_{WCS} & =\frac{s_1 b_1+\ldots+s_n b_n}{s_1+\ldots+s_n}.\end{aligned}$$

Again, note that the above result for $s_{WCS}$ is the lower boundary value for optimal Bayesian reasoning, exact only for $n=2$ (due to symmetry) and a group of participants with the same performance. By knowing the exact distribution of individual performances, we can obtain a 