A hundred-sided die, one dollar per reroll

Idea. The game has no memory: every reroll puts you back at the start, one dollar poorer. So the best strategy accepts any roll of at least some fixed $k$. Compute the value of each $k$ and pick the best one.

Step 1: the value of a threshold. Accept any roll of $k$ or more. Each roll is accepted with probability $\frac{101-k}{100}$, so the number of rolls is geometric with mean $\frac{100}{101-k}$, and you pay 1 dollar for every roll after the first. The accepted roll is uniform on $k,\dots ,100$, with mean $\frac{k+100}{2}$. So $$f(k)=\frac{k+100}{2}-\frac{100}{101-k}+1.$$

Step 2: the best threshold. $$\begin{array}{rl}f(86)& =94-\frac{100}{15}\approx 87.333,\\ f(87)& =94.5-\frac{100}{14}\approx 87.357,\\ f(88)& =95-\frac{100}{13}\approx \mathrm{87.308.}\end{array}$$ Raising $k$ by one adds $\frac{1}{2}$ to the average accepted roll but also raises the expected fees, and the fees grow faster and faster. The gain stops at $k=87$.

Step 3: the value. $$f(87)=\frac{187}{2}-\frac{50}{7}+1=\frac{1223}{14}\approx \mathrm{87.36.}$$

Check. With value $V$, rerolling is worth $V-1\approx 86.36$ dollars, so you should accept a roll exactly when it is at least $87$, which is the threshold used. The value is the solution of $$V=\mathbb{E}[\max (X,V-1)].$$