Quit while you're ahead

In general, your expectation value for a strategy where you quit after rolling $n$ times is given by:

$E_n = n \times (\frac{5}{6})^n \times \$100$

This is maximized for $n = 5$ and $n = 6$ .

$5 + 6 = \boxed{11}$

We have the function $E(n) = 100\$ \cdot n \cdot {(\frac{5}{6}})^n$ . There are 3 ways to find out when this function is maximized.

Way 1: Plug in the numbers 1,2,3,4... until you found the maxima

Way 2: Find the derivative $E'(n) = ({\frac{5}{6}})^n \cdot (ln( \frac{5}{6} ) + 1)$ and solve for E'(n)=0. This will yield a number between 5 and 6 which means you have to check if E(5) or E(6) is or if they are equal.

Way 3: From the function for E(n) we can find the recursive definition $E(n) = E(n-1) \cdot \frac{5}{6} \cdot \frac{n}{n-1}$ we then solve for $\frac{5}{6} \cdot \frac{n}{n-1} = 1$ because that's the value for n where E(n) is just about to decrease again. Solving will yield $n=6$ which means that $E(5) = E(6)$ are the maxima of the function E (for natural n).

In my opinion way 3 > way 2 > way 1.