If the current maximum is $x$, your expected maximum if you draw is $$ x^2 + (1-x) \Bigl( \frac{x+1}{2} \Bigr) $$ so the expected improvement is $$ \left( x^2 + (1-x) \Bigl( \frac{x+1}{2} \Bigr) \right) - x $$ which exceeds ${\large{\frac{1}{50}}}$ (the cost of the draw) when $x < {\large{\frac{4}{5}}}$.

Hence the optimal strategy is to draw if $x < {\large{\frac{4}{5}}}$, otherwise stop.

Let $e$ be the expected value of the game assuming the above stopping rule.

Then we get $$ e = \Bigl(\frac{1}{5}\Bigr) \Bigl(\frac{9}{10}\Bigr) + \Bigl(\frac{4}{5}\Bigr) \Bigl(e-\frac{1}{50}\Bigr) $$ which yields $e={\large{\frac{41}{50}}}=.82$.

If the current maximum is $x$, the expected maximum if you draw is $$ x^2 + (1-x) \Bigl( \frac{x+1}{2} \Bigr) $$ so the expected improvement is $$ \left( x^2 + (1-x) \Bigl( \frac{x+1}{2} \Bigr) \right) - x $$ which exceeds ${\large{\frac{1}{50}}}$ (the cost of the draw) when $x < {\large{\frac{4}{5}}}$.

Hence the optimal strategy is to draw if $x < {\large{\frac{4}{5}}}$, otherwise stop.

Let $e$ be the expected value of the game assuming the above stopping rule.

Then we get $$ e = \Bigl(\frac{1}{5}\Bigr) \Bigl(\frac{9}{10}\Bigr) + \Bigl(\frac{4}{5}\Bigr) \Bigl(e-\frac{1}{50}\Bigr) $$ which yields $e={\large{\frac{41}{50}}}=.82$.

If the current maximum is $x$, your expected maximum if you draw is $$ x^2 + (1-x) \Bigl( \frac{x+1}{2} \Bigr) -\frac{1}{50} $$ which exceeds $x$ when $x < {\large{\frac{4}{5}}}$.

Thus the optimal strategy is to draw if $x < {\large{\frac{4}{5}}}$, otherwise stop.

Let $e$ be the expected value of the game assuming the above stopping rule.

Then we get $$ e = \Bigl(\frac{1}{5}\Bigr) \Bigl(\frac{9}{10}\Bigr) + \Bigl(\frac{4}{5}\Bigr) \Bigl(e-\frac{1}{50}\Bigr) $$ which yields $e={\large{\frac{41}{50}}}=.82$.

If the current maximum is $x$, your expected maximum if you draw is $$ x^2 + (1-x) \Bigl( \frac{x+1}{2} \Bigr) $$ so the expected improvement is $$ \left( x^2 + (1-x) \Bigl( \frac{x+1}{2} \Bigr) \right) - x $$ which exceeds ${\large{\frac{1}{50}}}$ (the cost of the draw) when $x < {\large{\frac{4}{5}}}$.

Hence the optimal strategy is to draw if $x < {\large{\frac{4}{5}}}$, otherwise stop.

Let $e$ be the expected value of the game assuming the above stopping rule.

Then we get $$ e = \Bigl(\frac{1}{5}\Bigr) \Bigl(\frac{9}{10}\Bigr) + \Bigl(\frac{4}{5}\Bigr) \Bigl(e-\frac{1}{50}\Bigr) $$ which yields $e={\large{\frac{41}{50}}}=.82$.