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.
I tried several strategies, and the best seems to be to reroll on $9$ or less and stop with $10$ or more, which gives an expected payout of $\frac{293}{27}=10.8518518518519$ just above $\$10.85$
Reroll on $8$ or less
Let's say we reroll on $8$ or less and stay with $9$ or greater. The probabilty of rolling $9$ or greater is
$$
\frac{6+6+6+5+4+3+2+1}{60}=\frac{33}{60}
$$
so on average, it takes $\frac{60}{33}$ rolls to get $9$ or greater paying on average $\frac{27}{33}$ dollars to get there. The expected payout of $9$ or greater is
$$
\frac{6\cdot9+6\cdot10+6\cdot11+5\cdot12+4\cdot13+3\cdot14+2\cdot15+1\cdot16}{6+6+6+5+4+3+2+1}=\frac{380}{33}
$$
Thus, the expected value of this strategy is $\frac{380}{33}-\frac{27}{39}=\frac{353}{33}=10.6969696969697$
Reroll on $9$ or less
Let's say we reroll on $9$ or less and stay with $10$ or greater. The probabilty of rolling $10$ or greater is
$$
\frac{6+6+5+4+3+2+1}{60}=\frac{27}{60}
$$
so on average, it takes $\frac{60}{27}$ rolls to get $10$ or greater paying on average $\frac{33}{27}$ dollars to get there. The expected payout of $10$ or greater is
$$
\frac{6\cdot10+6\cdot11+5\cdot12+4\cdot13+3\cdot14+2\cdot15+1\cdot16}{6+6+5+4+3+2+1}=\frac{326}{27}
$$
Thus, the expected value of this strategy is $\frac{326}{27}-\frac{33}{27}=\frac{293}{27}=10.8518518518519$
Reroll on $10$ or less
Let's say we reroll on $10$ or less and stay with $11$ or greater. The probabilty of rolling $11$ or greater is
$$
\frac{6+5+4+3+2+1}{60}=\frac{21}{60}
$$
so on average, it takes $\frac{60}{21}$ rolls to get $11$ or greater paying on average $\frac{39}{21}$ dollars to get there. The expected payout of $11$ or greater is
$$
\frac{6\cdot11+5\cdot12+4\cdot13+3\cdot14+2\cdot15+1\cdot16}{6+5+4+3+2+1}=\frac{266}{21}
$$
Thus, the expected value of this strategy is $\frac{266}{21}-\frac{39}{21}=\frac{227}{21}=10.8095238095238$
Best Answer
You have a $6/12$ chance of rolling under $7$, and then a $6/12$ to roll the average of $7,8,9,10,11$ and $12$. I.e. the expected result is $$7\frac{6}{12}+\frac{7+8+9+10+11+12}{6}\frac{6}{12}=8.25$$