jmc

For $n$ dice, there are $\left(\genfrac{}{}{0ex}{}{n}{2}\right)=\frac{n(n-1)}{2}$ ways to choose two of them. For each of these ways, there is a ${\left(\frac{5}{6}\right)}^2{\left(\frac{1}{6}\right)}^{n-2}$ chance that all but the chosen two dice will roll a 1. Therefore, we need to find the value of $n$ for which $\frac{25n(n-1)}{2\cdot 6^n}=\frac{25}{216}$, or $108n(n-1)=6^n$. Plugging in values for $n$, we can see that $n=\boxed{4}$ works and no value of $n$ less than 4 works. Now we just need to prove that no values of $n$ greater than 4 work. Note that if $n\ge 5$, then $n<3^{n-3}$ and $n-1<2\cdot 2^{n-3}$. We can multiply these inequalities to get that when $n\ge 5$, we have $n(n-1)<2\cdot 6^{n-3}$, or $108n(n-1)<6^n$.