TITLE: You roll a die until you get a $5$, what is the expected value of the minimum value rolled?
QUESTION [7 upvotes]: I am struggling to work out a simple way to answer this, and a rationale behind this approach using tail sum (as I do not understand):
$$E\left( x\right) =\sum ^{5}_{k=1}P\left( x\geq k\right) =\dfrac {1}{6}\sum ^{5}_{k=1}\left( \sum ^{\infty }_{i=0}\left( \dfrac {k}{6}\right) ^{i}\right)  = \frac{137}{60}.$$
Does this always hold?
$$E\left( x\right)=\sum ^{n}_{k=1}kP\left( x= k\right)  =\sum ^{n}_{k=1}P\left( x\geq k\right)$$
I have never seen this formula, but working through it I understand.
EDIT: I can get the right answer with a long winded method, calculating each probability separately, which I believe the Tails sum speeds up:
This is my long winded approach.
$$E\left( X_{\min }\right) = 5P\left( x= 5\right) +\ldots +1P(x=1)$$
$$=5\left( \dfrac {1}{6}\sum ^{\infty }_{i=0}\left( \dfrac {1}{6}\right) ^{i}\right) +4\left( \dfrac {1}{6}\sum ^{\infty }_{i=1}\left( \dfrac {2}{6}\right) ^{i}- \dfrac {1}{6}\sum ^{\infty }_{i=1}\left( \dfrac {1}{6}\right) ^{i}\right)+\ldots$$ $$+1\left( \dfrac {1}{6}\sum ^{\infty }_{i=1}\left( \dfrac {5}{6}\right) ^{i}- \dfrac {1}{6}\sum ^{\infty }_{i=1}\left( \dfrac {4}{6}\right) ^{i}\right),$$
where $\dfrac {1}{6}$ represents getting a $5$, so the first sum is all the possibilities of getting repeated $6$s and then a $5$, or just rolling a $5$. The next sum is all the possibilities of getting a $4$ or a $6$, then a $5$, minus all the possibilities of just getting a $6$ then a $5$. So it represents all the strings of just $4$s or $6$s before getting a $5$, so all the possibilities of minimum value being a $4$.

REPLY [2 votes]: I wanted to point out that we don't need infinite series or fancy formulas to solve this problem.
Any infinite sequence of die rolls yields a permutation of $\{1,2,3,4,5,6\}$ by simply retaining all of the first appearances of each number. By symmetry, all $6!$ permutations are equally likely. Therefore to calculate the desired expectation, one can simply enumerate the $6!$ permutations, find the minimum values before the $5$ in each case, and average the results.