Consider the following problem:
Suppose a house with $n$ apartments is on fire. Every person living there can call the firefighters. Suppose the call costs $1 \$ $. If at least one person calls firefighters the fire will be extinguished. If noone calls them, then the house will burn down and everyone living there will lose $m \$ $ of property ($m > 1$). The people living in this house do not talk to each other and decide to call or not to call completely independently. For what $n$ the firefighters are less likely to be called?
The solution is the following one:
People living in the house not talking to each other means that we need to find a symmetric mixed Nash equilibrium. Suppose $\alpha$ is the probability of any given person calling the firefighters. Then anyones expected gain (in comparison to the situation when everything burns down) is
$$(1 - \alpha)^{n-1} \alpha (m - 1) + (1 - (1 - \alpha))^{n - 1} (\alpha (m - 1) + (1 - \alpha) m) = m - \alpha - (1 - \alpha)^n m$$
Let’s find for which $\alpha$ it is maximised.
$$\frac{\partial (m - \alpha - (1 - \alpha)^n m)}{\partial \alpha} = -1 + n (1 - \alpha)^{n-1} m$$
Thus the point of extremum is $$\alpha = 1 - (\frac{1}{m n})^{\frac{1}{n-1}} = \frac{\ln (n)}{n} + \frac{\ln(m)}{n} + O(\frac{\ln^2(n)}{n^2})$$
This indeed is what we need: the total value at this point is greater than at both ends of $[0; 1]$.
Now, when we know $\alpha$ let’s calculate the probability of firefighters being called. It is $1 - (1 - \alpha)^{n} = 1 - (\frac{1}{m n})^{1 + \frac{1}{n - 1}}$. It is not hard to see, that it is $1$ (it’s maximal possible value) both when $n = 1$ and $n \to \infty$. Thus the minimum is reached in the internal infimum.
$$\frac{\partial (1 - (\frac{1}{m n})^{1 + \frac{1}{n - 1}})}{\partial n} = - (\frac{1}{m n})^{1 + \frac{1}{n - 1}} \frac{1 + \ln(m) - \ln(n) - n}{n - 1}$$
Thus the minimum is reached when $m = n e^{n - 1}$. So the answer is $n = W(e m) = \ln(m) - \ln(\ln(m)) + o(1)$
Now, let’s make the problem a bit more complex:
Suppose a house with $n$ apartments is on fire. Every person living there can call the firefighters. Suppose the call costs $1 \$ $. If anyone calls firefighters the fire will be extinguished. If no-one calls them, then the house will burn down and $i$-th person living there will lose $X_i \$ $ of property ($X_i > 1$). Everyone knows only their value of $X_i$ and consider the corresponding values of everyone else to be i.i.d. random values, uniformly distributed on $[1;m]$ ($m > 1$). The people living in this house do not talk to each other and decide to call or not to call completely independently. For what $n$ the firefighters are less likely to be called?
I attempted to use a similar method: look for a symmetric Bayes-Nash equilibrium and see what is happening there, however the problem appears to be much trickier. Now, the strategy of each player is defined not as a single value $\alpha$ (probability of calling the firefighters), but as a function $\alpha: [1; +\infty) \to [0; 1]$, where $\alpha(x)$ is the probability of any given player calling the firefighters if they know they lose $x$ dollars if the fire will not be extinguished. Then for a fixed $X_i = x$ the expected gain of the $i$-th player will be $$x - \alpha(x) - x(1 - \alpha(x)) E (\Pi_{j \leq n, j \neq i}(1 - \alpha(X_j))) = x - \alpha(x) - x (1 - \alpha(x)) \big(\frac{\int_1^m (1 - \alpha(t))dt}{m - 1}\big)^{n-1}$$
And here I am stuck because I do not know how to find the function $\alpha$ (dependent on $m$) for which $x - \alpha(x) - x (1 - \alpha(x)) \big(\frac{\int_1^m (1 - \alpha(t))dt}{m - 1}\big)^{n-1}$ is maximal for every $x \in [1; m]$.
Any help will be appreciated.
Let $A'(x)=a(x)$ where $a\equiv\alpha$. We have $$\Bbb E[{\sf gain}]=x-a(x)-Mx(1-a(x))$$ where $M=(m-A(m)+A(1))^{n-1}/(m-1)^{n-1}$. Now $$\Bbb E[{\sf gain}]_x=1-a'(x)-M(1-a(x)-xa'(x))=0$$ for optimality which yields $(Mx-1)a'(x)=-Ma(x)-1+M$ so $$a(x)=\frac{C}{Mx-1}-\frac 1M+1$$ for some constant $C$. Since $a(m)=1-1/(mn)^{1/(n-1)}$ we obtain $$C=(Mm-1)\left(\frac1M-\frac1{(mn)^{1/(n-1)}}\right).$$ Integrating again, we obtain $$A(x)=\frac CM\log(Mx-1)+\frac{Mx}{M-1}$$ where we set the constant term to be zero for convenience. The values of $A(m)$ and $A(1)$ are determined through \begin{align}A(m)&=\frac CM\log(Mm-1)+\frac{Mm}{M-1}\\A(1)&=\frac CM\log(M-1)+\frac{M}{M-1}\end{align} but I don't think a closed form exists due to the complexity of the equations.