Value of using a better normal distribution I tried to derive this on my own, but my stats education proved too far back…
(This is a problem in Bayesian decision theory – if that makes you uncomfortable, feel free to reformulate it)
Let's say I have options $o_i$ with payout $u_i \sim \mathcal N(\mu_i, \sigma_i)$. How much would I profit from reducing the variances to $\sigma_i ' = a_i \sigma_i$. For simplicity, let's assume the values of $a_i$ are given.
Example: I have 5 tasks that I can do. Either I do the one that I assume has the best return on investment straight away, or I investigate first which one is the best. Investigating has some cost attached and offers only a better estimate of the return on investment, not the exact value. 
 A: (I'm the original requester: I continued looking into this while waiting for an answer. Maybe this helps someone who finds this via search).
Expected Loss
The way to look at this: You get $U_i$ if you pick option $o_i$. If you had perfect information, you would get $\max_j U_j$. So by picking any option $o_i$, your loss is $\mathcal{L}(u) = U_i - \max_j U_j$.
As a simplification, we can assume the $\mu_i$ ordered, so that $\max \mu_i = \mu_1$. We can further transform the options so that $\mu_1 = \sigma_1 = 0$ by subtracting $U_1$ from the payouts (effectively increasing the standard deviation of the other options).
This leaves us with $\mathcal{L}(U) = - \max_j U_j$. The expected loss is now
\begin{align}
 \mathbb{E}\left(\mathcal{L}(\boldsymbol{U}) \right) &= \int \mathbb{P}(\boldsymbol{U}=\boldsymbol{u})\mathcal{L}(\boldsymbol{u}) d\boldsymbol{u} \\
&= - \int \mathbb{P}(\boldsymbol{U}=\boldsymbol{u}) \max_j(u_j) d\boldsymbol{u}
\end{align}
Case $N=2$
For the case of $N>2$ things get a bit messy, so let's consider the case $N=2$ first, where we only get a second option $o_2$ with estimated payout $U_2$:
\begin{align}
 \mathbb{E}\left(\mathcal{L}(U_2) \right) &= \underbrace{\mathbb{E}(0 | U_2 \leq 0)}_{=0} + \mathbb{E}(U_2\, |\, U_2 > 0) \\
&= \frac{1}{2} \operatorname{erfc}{\left (\frac{\mu_2}{\sqrt{2}  \sigma_2} \right )}
\end{align}
The value of information is then the difference in expected loss (which is now easy to calculate):
\begin{align}
VoI(b) &= \mathbb{E}\left(\mathcal{L}(U')\right) - \mathbb{E}\left(\mathcal{L}(U)\right) \\
&= \frac{1}{2} \operatorname{erfc}{\left (\frac{\mu_2}{\sqrt{2} b \sigma_2} \right )} - \frac{1}{2} \operatorname{erfc}{\left (\frac{\mu_2}{\sqrt{2} \sigma_2} \right )}
\end{align}
Maximum of multiple random variables
The maximum of multiple random variables can be best understood in its cummulative distribution function:
\begin{align}
\mathbb{P}(\max_j U_j \leq x) &= \prod \mathbb{P}(U_j \leq x)
\end{align}
(read as: the probability of the max being smaller than $x$ is the probability of all elements being smaller than $x$.)
The expected value is in terms of the density however, which means we need to take the derivative further down.
Multiple options
I haven't been able to work this out properly. The loss is between the maximum individual loss and the sum of individual losses. This is a good enough approximation for me.
