100-sided die probability

If the expected value of this game is $a$, then at a die roll of $X$ you have the choice of either collecting $X$ or paying a dollar and restart, which gives you an expected value of $a-1$. To maximize the expected value, you should take $X$ if $X> a-1$ and start over if $X\le a-1$ (it does not really matter what we do when $X=a-1$). We obtain therefore $$ a = \frac1{100}\left(\lfloor a-1\rfloor\cdot a+\sum_{k=\lfloor a-1\rfloor+1}^{100}k\right) =\frac1{100}\left(\lfloor a-1\rfloor\cdot a+\frac{100\cdot101}{2}-\frac{\lfloor a-1\rfloor \cdot\lfloor a\rfloor}{2}\right). $$ I find numerically (didn't do much code checking, but the results are somewhat plausible) $$a\approx87.3571 $$ which seems to be exactly (and of course the true result must be rational) $$a=87\frac{5}{14}.$$ But I'm sure you can do the justification after the fact, i.e. show that the strategy that consists in continuing until you roll at least $87$ gives you $87\frac{5}{14}$ as expected value.

For your convenience, here is the PARI one-liner:

solve(a=1,100,sum(k=1,100,max(a-1,k))/100-a)

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If an extra roll costs two dollars instead of one, the result would be $$a=82\frac12$$ instead, and with a cost of only $0.1$ dollars it would be $$a=96\frac1{10}.$$

Hint:

If your value now is $X_t$, what is the marginal value of the roll? If you roll $R \sim \mathcal{U}[1,100]$, then if $R > X_t+1$ you gained and if $R \leq X_t+1$, you either lost or became indifferent. So what is the marginal value?