Let $v$ denote the expected value of the game. If you roll some $x\in\{1,\ldots,100\}$, you have two options:
- Keep the $x$ dollars.
- Pay the \$$1$ continuation fee and spin the dice once again. The expected value of the next roll is $v$. Thus, the net expected value of this option turns out to be $v-1$ dollars.
You choose one of these two options based on whichever provides you with a higher gain. Therefore, if you spun $x$, your payoff is $\max\{x,v-1\}$.
Now, the expected value of the game, $v$, is given as the expected value of these payoffs:
\begin{align*}
v=\frac{1}{100}\sum_{x=1}^{100}\max\{x,v-1\}\tag{$\star$},
\end{align*}
since each $x$ has a probability of $1/100$ and given a roll of $x$, your payoff is exactly $\max\{x,v-1\}$. This equation is not straightforward to solve. The right-hand side sums up those $x$ values for which $x>v-1$, and for all such values of $x$ that $x\leq v-1$, you add $v-1$ to the sum. This pair of summations gives you $v$. The problem is that you don't know where to separate the two summations, since the threshold value based on $v-1$ is exactly what you need to compute. This threshold value can be guessed using a numerical computation, based on which one can confirm the value of $v$ rigorously. This turns out to be $v=87\frac{5}{14}$.
Incidentally, this solution also reveals that you should keep rolling the dice for a $1 fee as long as you roll 86 or less, and accept any amount 87 or more.
ADDED$\phantom{-}$In response to a comment, let me add further details on the computation. Solving for the equation ($\star$) is complicated by the possibility that the solution may not be an integer (indeed, ultimately it is not). As explained above, however, ($\star$) can be rewritten in the following way:
\begin{align*}
v=\frac{1}{100}\left[\sum_{x=1}^{\lfloor v\rfloor-1}(v-1)+\sum_{x=\lfloor v\rfloor}^{100}x\right],\tag{$\star\star$}
\end{align*}
where $\lfloor\cdot\rfloor$ is the floor function (rounding down to the nearest integer; for example: $\lfloor1\mathord.356\rfloor=1$; $\lfloor23\mathord.999\rfloor=23$; $\lfloor24\rfloor=24$). Now let’s pretend for a moment that $v$ is an integer, so that we can obtain the following equation:
\begin{align*}
v=\frac{1}{100}\left[\sum_{x=1}^{v-1}(v-1)+\sum_{x=v}^{100}x\right].
\end{align*}
It is algebraically tedious yet conceptually not difficult to show that this is a quadratic equation with roots
\begin{align*}
v\in\left\{\frac{203\pm3\sqrt{89}}{2}\right\}.
\end{align*}
The larger root exceeds $100$, so we can disregard it, and the smaller root is approximately $87\mathord.349$. Of course, this is not a solution to ($\star\star$) (remember, we pretended that the solution was an integer, and the result of $87\mathord.349$ does not conform to that assumption), but this should give us a pretty good idea about the approximate value of $v$. In particular, this helps us formulate the conjecture that $\lfloor v\rfloor=87$. Upon substituting this conjectured value of $\lfloor v\rfloor$ back into ($\star\star$), we now have the exact solution $v=87\frac{5}{14}$, which also confirms that our heuristic conjecture that $\lfloor v\rfloor=87$ was correct.
Both approaches look correct! The $5+{}$in the first equation does account for the winnings you get to keep.
The series you found in the second case is
$$
\frac52 \sum_{k=1}^\infty \bigg( \frac12 \bigg)^k k.
$$
This is indeed a notch more complicated than a simple geometric series, but we can still handle it! Starting from
$$
\sum_{k=0}^\infty x^k = \frac1{1-x},
$$
valid for $|x|<1$, we can take the derivative of both sides to get
$$
\sum_{k=0}^\infty kx^{k-1} = \frac1{(1-x)^2},
$$
which implies
$$
c\sum_{k=1}^\infty kx^k = \frac{cx}{(1-x)^2}.
$$
This is exactly the series you want, with $c=\frac52$ and $x=\frac12$, and therefore your series equals
$$
\frac{(5/2)(1/2)}{(1-1/2)^2} = 5,
$$
which is the correct expected value.
Best Answer
Both games: you roll some number of 1,2,3s before rolling a 4,5,6. The expected number of rolls is 1 ( the sum of 1/2, 1/4, 1/8 ...). On average you'll have a single dollar when 4,5,6 comes up.
Game 1: if you roll 4,5 you get a fresh start - as if you just started playing with no accumulated winnings. If you roll 6 you can expect to keep 1. E = 1
Game 2: you roll 6 1/3 of the time and win 0. You will roll 4,5 2/3 of the time an expect to keep 1. E = 2/3