Paying for extra uniform draws to raise the max

You receive one free draw from $\text{Unif}(0,1)$. After seeing all draws so far, you may buy another independent draw for $0.02, as many times as you like. When you stop you are paid the largest value drawn. Under optimal play, what is your expected net payout, i.e. the largest value minus all fees paid?

$0.82$. Buying one more draw when your best is $m$ gains $E[(U-m)^+]=\tfrac{(1-m)^2}{2}$, which is worth the $0.02$ fee iff $m<0.8$, so stop at the first value at least $0.8$. That value averages $0.9$; the number of paid draws averages $\tfrac{1}{0.2}-1=4$, costing $0.08$, so the net is $0.82$.