<s>[INST] We roll a fair die until a $5$ appears. What is the expected value of the minimum value rolled? 
Question: Given a fair dice, we roll until we get a $5.$ What is the expected value of the minimum value rolled?

Answer is $\frac{137}{60}.$
There is a similar question asked in MSE but I do not understand the method used by Henry. 
In particular, if we let $X$ be the minimum value rolled up to and including $5$, then 
$$E(X) = \sum_{x=1}^5 xP(X=x) = 1 \times \frac12 + 2 \times \frac16 + 3 \times \frac1{12}+4 \times \frac{1}{20}+5 \times \frac15 = \frac{137}{60}.$$
It seems that we are using the fact that 
$$P(X=x) = \frac{1}{x(x+1)}.$$

I do not understand how to obtain the equation above.

 [/INST] The probability that the minimum is $X\in\{1,2,3,4,5\}$ can be found as follows:
The probability of rolling a sequence of length $k$ using numbers from the set $\{X,X+1,\dots,6\}-\{5\}$ is $\left(\frac{6-X}{6}\right)^k$.  Similarly, the probability of getting a sequence with values in $\{X+1,\dots,6\}-\{5\}$ is $\left(\frac{5-X}{6}\right)^k$. So the probability of a sequence of length $k$ with all elements in $\{X,\dots,6\}-\{5\}$ and at least one instance of $X$ is $\left(\frac{6-X}{6}\right)^k-\left(\frac{5-X}{6}\right)^k$.  The probability of rolling a $5$ right after this sequence is $1/6$.  
We sum over $k$ to find the probability of getting a sequence of any length with minimum $X$.  This works out to $$P(X)=\frac{1}{6}\sum_{k=1}^\infty \left(\frac{6-X}{6}\right)^k-\left(\frac{5-X}{6}\right)^k = 
\left\{
\begin{array}{ll}
        \frac{1}{X} - \frac{1}{1+X} = \frac{1}{X(X+1)} & \text{for } X=1,2,3,4\\
        \frac{1}{X} = \frac{1}{5} & \text{for } X=5\\
        \end{array}\right.
$$ Which is what we wanted. 
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