We roll a fair die until a $5$ appears. What is the expected value of the minimum value rolled?

Assume that a $5$ is first seen on roll $n$.
$5$ is the lowest seen with $n-1$ $6$s then one $5$.
$4$ is the lowest seen with $n-1$ $4$s and $6$s, but not all $6$s then one $5$.
$3$ is the lowest seen with $n-1$ $3$s, $4$s, and $6$s, but not all $4$s and $6$s then one $5$.
$2$ is the lowest seen with $n-1$ $2$s, $3$s, $4$s, and $6$s, but not all $3$s, $4$s, and $6$s then one $5$.
$1$ is the lowest seen with $n-1$ $1$s, $2$s, $3$s, $4$s, and $6$s, but not all $2$s, $3$s, $4$s, and $6$s then one $5$.
$$ \begin{array}{c|l|l} \text{lowest}&\text{chance with $n$ rolls}&\text{sum over $n$}\\ \hline 5&\,\left(\frac16\right)^{n-1}\frac16&\frac15\\ 4&\,\left[\left(\frac26\right)^{n-1}-\left(\frac16\right)^{n-1}\right]\frac16&\frac1{20}\\ 3&\,\left[\left(\frac36\right)^{n-1}-\left(\frac26\right)^{n-1}\right]\frac16&\frac1{12}\\ 2&\,\left[\left(\frac46\right)^{n-1}-\left(\frac36\right)^{n-1}\right]\frac16&\frac16\\ 1&\,\left[\left(\frac56\right)^{n-1}-\left(\frac46\right)^{n-1}\right]\frac16&\frac12 \end{array} $$ Expected value $=5\cdot\frac15+4\cdot\frac1{20}+3\cdot\frac1{12}+2\cdot\frac16+1\cdot\frac12=\frac{137}{60}$

First, $X$ is not the minimum value rolled before obtaining a $5$, it is the minimum value rolled up to and including the first roll that comes up $5$, so that $X=5$ is possible.

The event $X=5$ means that $5$ comes up before any of $1$, $2$, $3$, or $4$ (we don't care about $6$). Since each of the five numbers is equally likely to come up first, $$P(X=5)=\frac15.$$

Now suppose $1\le x\le4$. Now the event $X=x$ means that, among the $x+1$ numbers $1,\dots,x,5$, the number $x$ comes up first, and $5$ second. Thus we have $$P(X=x)=\frac{(x-1)!}{(x+1)!}=\frac1{(x+1)x}\text{ for }1\le x\le4.$$