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 $4$.
 A: I have found a different way to think about to this problem for those still a little stuck. Let m = the minimum number found. 
E(m) = 1*P(m=1) + 2*P(m=2) + ... + 5*P(m=5)
Notice we do not include the probability that m=6 as we know we have rolled a 5 (this is when the game stops) so we cannot have our smallest number to be 6. 
Lets have a look at P(m=1):
This is effectively the event that we get a 1 before we get a 5. Whats more likely, getting a 1 before a 5 or a 5 before a 1? Naturally they are both equally likely so P(m=1) = 0.5
Now lets look at P(m=2). We need two things here. To get a 2 before a 5 and to get a 5 before a 1. I.e we have 2 as our minimum but the game has stopped before we had a chance to change that minimum to 1. We now have 6 possible choices in the race of 1,2 and 5. We could have 512 or 125 or 251 or so on. Think of it as a race between 3 horses, all outcomes equally likely. (in essence we are also "racing horse 4 5 and 6" but we dont care how they perfom in the race just the relative positions of 1 2 and 5)  We have 3 choices for First place, 2 for second and 1 for third. 3! = 6. Of these 6 choices the ONLY way in which m=2 is if we have 251. 1 choice out of 6 thus P(m=2) = 1/6.
Now onto P(m=3) again we are looking at the positions of horses 1,2,3 and 5. With a total of 4! possibilities we must have either 3512 or 3521. P(m=3)=2/4! = 1/12.
P(m=4). we have 5 horses now racing: 1,2,3,4 and 5. with a total of 5!, 120 outcomes and m=4 only occuring on the outcomes: 45xyz. with x,y,z taking any choice in 123. there are 3! choices for xyz thus 3!/5! = 1/20 = P(m=4)
P(m=5). Just look the step above we are looking at the relative positions between horses 1,2,3,4 and 5. This time m=5 if and only if 5 wins the race! So successful outcomes are of the form 5xyzt. with xyzt in 1234. There are 4! ways to arrange horses 1234 thus P(m=5) = 4!/5! = 1/5. 
Can you spot a pattern for P(m=i)?
Putting this all in gives E(m) = 1*1/2 + ... + 5*1/5 = 137/60. 
:) Oskar 
A: 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 $\frac{137}{60}$
A: 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.
A: The formula
$$E\left( x\right) =\sum ^{n}_{k=1}P\left( x\geq k\right)$$
is a special case of a fairly well-known more general result:
\begin{align}
E(X) &=\int_0^\infty P(X>x)\ dx
\end{align}
for any random variable $\ X\ $ which assumes only non-negative values.  For a non-negative integer-valued random variable, this becomes
\begin{align}
E(X) &=\sum_{k=0}^\infty\int_k^{k+1} P(X>x)\ dx\\
&= \sum_{k=0}^\infty\int_k^{k+1} P(X>k)\ dx\ \ \text{ because }P(X>x)= P(X>k)\\
&\hspace{10.2em}\text{ for } k\le x < k+1\\
&= \sum_{k=0}^\infty P(X>k)\\
&= \sum_{k=0}^\infty P(X\ge k+1)\\
&= \sum_{k=1}^\infty P(X\ge k)\  ,
\end{align}
and for the case being considered here, the sum can be terminated at $\ k=5\ $, because $\ P(k\ge k) = 0\ $ for $\ k\ge 6\ $.
The identity
$$
\sum_{k=1}^5 P(X\ge k) = \frac{1}{6}\sum_{k=1}^5\sum_{i=0}^\infty \left(\frac{k}{6}\right)^i
$$
comes from the decomposition
\begin{align}
\left\{X\ge k\right\}&=\bigcup_{i=1}^\infty\left\{k\le T_j\le4\text{ or } T_j=6 \text{ for } 1\le j<i\right\} \cap \left\{T_i=5\right\}\ ,
\end{align}
giving
\begin{align}
\sum_{k=1}^5 P(X\ge k)&=\sum_{k=1}^5\sum_{i=0}^\infty \left(\frac{6-k}{6}\right)^i\frac{1}{6}\\
&= \sum_{k'=1}^5 \sum_{i=0}^\infty \left(\frac{k'}{6}\right)^i\frac{1}{6}\ ,
\end{align}
where the final sum is obtained by replacing the index $\ k\ $ in the previous one with $\ k'=6-k\ $.
