Sum of the number shown on the die A die is repeatedly rolled and the number shown on the die is summed where 
$N$ is the sum. 
What can you say about $E(R_N)$, the expected number of rolls until the sum is 
greater than or equal to $N$?
I used Wald's identity to show that
$E(X)=21/6=3.5$ where $X$ is the number rolled by the die.
$E(R_N)=\dfrac{N}{E(X)}$
But I'm pretty sure the solution isn't so simple since there is a distinction 
between at least and exactly. How do I account for the amount of rolls that
account for a sum greater than $N$? This suggests to me that the expectation
may be non-finite, but I could use some help on the formalization to test
this hypothesis.
 A: This is easy enough to compute recursively. If $N>6$ then throwing the die one time reduces the problem to $N-i$ for $i\in\{1,2,3,4,5,6\}$.  Specifically, $$E[R_N]=1+\frac 16 \times \sum_{i=1}^6E[R_{N-i}]$$
Then we just need to compute it explicitly for $N=\{1,2,3,4,5,6\}$.  
$E[R_1]=1$
$E[R_2]=\frac 16(E[R_1]+1)+\frac 56 1=1.1666...$
$E[R_3]=\frac 16(E[R_2]+1)+\frac 16 (E[R_1]+1)+\frac 46 1=1.3611...$
$E[R_4]=\frac 16(E[R_3]+1)+\frac 16 (E[R_2]+1)+\frac 16 (E[R_1]+1)+\frac 36 1=1.58796...$
$E[R_5]=\frac 16(E[R_4]+1)+\frac 16(E[R_3]+1)+\frac 16 (E[R_2]+1)+\frac 16 (E[R_1]+1)+\frac 26 1=1.85262...$
$E[R_6]=\frac 16 (E[R_5]+1)+\frac 16(E[R_4]+1)+\frac 16(E[R_3]+1)+\frac 16 (E[R_2]+1)+\frac 16 (E[R_1]+1)+\frac 16 1=2.16139...$
Your approximate formula isn't exact, though it is decently close...for example, using the recursion we get $E[R_{100}]=29.0476$ while $\frac {100}{3.5}=28.571$.
The recursive is in standard form, a constant plus something linear with constant coefficients, so it can be solved in closed form but I have not attempted to do that.
