A reminder: guest Student Puzzle editor Stanislav Volkov set this problem in the previous issue. He explains the solution below.

Well done to student members Samprit Chakraborty (Indian Statistical Institute, Kolkata) and Aniv Mazumder (Indian Statistical Institute, Delhi), who both sent full solutions to Puzzle 60, above; and a good attempt was made by Rahul Vishwakarma (also from Indian Statistical Institute, Delhi) to solve the problems.

 
So, it is clear that if Alice decides to stop when the drawn value is $x$, she should stop if it is some $x’>x$. Let $L$ be the smallest threshold when Alice decides to stop playing, and let $\mu=\mu(L)$ be her expected payoff. Because there is no time discount, it is clear that at each draw she should use the same strategy. Hence, the number of times she plays $\eta$ has the geometric distribution,

$
{\mathbb{P}}(\eta=k)={\mathbb{P}}(\xi \lt L)^{k-1}{\mathbb{P}}(\xi\ge L),\quad k=1,2,\dots
$

and her expected payoff is

\begin{align}\label{eq}
\mu={\mathbb{E}}(\xi\mid\xi\ge L)-c {\mathbb{E}} (\eta-1)=c+\frac{{\mathbb{E}}(\xi 1_{\xi\ge L})-c }{{\mathbb{P}}(\xi\ge L)}.
\end{align} [Equation 1]

This immediately implies that if $c\ge {\mathbb{E}} \xi$ then for any $L>0$ we have $\mu(L)<{\mathbb{E}}\xi $ and hence it's not worth drawing again, so the optimal strategy is just to take the result of the first draw with $\mu_{\rm opt}={\mathbb{E}} \xi$. Hence, from now on, we assume that $\boxed{{\mathbb{E}} \xi>c}$.

In case $\xi$ has a continuous distribution with density $f$, ${\mathbb{P}}(\xi\ge L)=\int_L^\infty f(x)dx$ and
$$
\mu=c+\frac{\int_L^\infty x f(x)dx -c}{\int_L^\infty f(x)dx}=\frac{\int_L^\infty x f(x)dx -c\int_0^L f(x)dx}{\int_L^\infty f(x)dx},
$$
yielding
\begin{align*}
\frac{\partial\mu}{\partial L}&=\frac{-Lf(L)\int_L^\infty f(x)dx+f(L)\left(\int_L^\infty x f(x)dx -c\right)}{{\mathbb{P}}(\xi\ge L)^2}\\ &
=\frac{f(L)}{{\mathbb{P}}(\xi\ge L)^2}\left[\int_L^\infty (x-L) f(x)dx -c \right]
=\frac{f(L)}{{\mathbb{P}}(\xi\ge L)^2}\left[{\mathbb{E}} \max\left(\xi-L,0\right) -c \right].
\end{align*}
The expression in square brackets is positive when $L=0$ and is monotonically decreasing to $-c$ as $L\to\infty$. Hence, there is an $L^*$ such that $\frac{\partial \mu}{\partial L}$ is non-negative for $L \lt L^*$ and non-positive for $L \gt L^*$. Then her expected payoff will be
$$
\frac{\int_{L^*}^\infty x f(x)dx -c {\mathbb{P}}(\xi< L^*)} {{\mathbb{P}}(\xi\ge L^*)}. $$ If $\xi$ is Uniform$[0,1]$, then the expression in the square brackets equals $\frac{(1-L)^2}{2}-c$, hence the optimal strategy is to play until she gets $1-\sqrt{2c}$ or more (this formally also covers the case when $c\ge 1/2$). Under the optimal strategy, Alice's expected payoff equals $$ \mu_{\rm opt}=1-\frac{\sqrt{2c}}{2}-c\left(\frac{1}{\sqrt{2c}}-1\right) =1+c-\sqrt{2c} $$ when $c<1/2$, and $\mu_{\rm opt}=1/2$ otherwise. If $\xi$ has a uniform discrete distribution in $\{1,\dots,n\}$, assume that Alice stops when the result of a draw is at least $L=n-\ell+1$; as before, we assume that $c<{\mathbb{E}}\xi=\frac{n+1}2$. Equation [1] becomes $$ \mu=n + c + \frac{1}2-\frac12\left[\frac{2cn}\ell +\ell \right]. $$ Since $\frac{a}\ell+\ell$ (where $a>0$) is a convex function minimized at $\ell=\sqrt{a}$, $\mu$ is maximised when $\ell$ is one of the two integer closest to $\sqrt{2cn}$, i.e., $\lfloor\sqrt{2cn} \rfloor$ or $\lceil\sqrt{2cn} \rceil$ (for example, if $n=6$ and $c=1$, corresponding to a usual rolling of a die, then the best strategy for Alice is to stop whenever $L=L^*$ where we can take both $L^*=3$ and $L^*=4$).