<s>[INST] You roll a die until you get a $5$, what is the expected value of the minimum value rolled? 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 $4$.
 [/INST] I seem to get $\dfrac{137}{60}$, slightly more than $2$, which seems as plausible to me as your similar answer
More precisely: $$1 \times \frac12 + 2 \times \frac16 + 3 \times \frac1{12}+4 \times \frac{1}{20}+5 \times \frac15$$ 
I think you are saying that if only $6$s are thrown before the first $5$ then the minimum is $6$ to calculate 
$$1 \times \frac12 + 2 \times \frac16 + 3 \times \frac1{12}+4 \times \frac{1}{20}+5 \times \frac16+6 \times \frac1{30}$$ 
I am saying the minimum is $5$, since a $5$ is thrown to stop the game.  This is what is causing the difference between $\frac{139}{60}$ and $\frac{137}{60}$
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