repeated toss a coin and you can stop anytime and payoff is just #times you got head divided by total number of throws, how do you maximize your pay.
Does anyone have a clever strategy for this? This was changed from another problem I solved involving flipping over cards from a deck and being paid the fraction of red. Also never saw a satisfying solution
On
Suppose the probability of head is $p$. Suppose you have thrown $n$ tosses and landed $a$ heads. And suppose you contemplate playing $m$ more tosses. The expected payoff of this is $$ \frac{a+pm}{n+m}. $$ This is greater than or equal to $\frac{a}{n}$ iff $$ (a+pm)n-(n+m)a=m(np-a)\geq 0. $$ So if $a>pn$, you shouldn't continue. If $a\leq np$, then you should continue.
So I think that you should stop whenever the ratio of heads to flips is greater than 1/2. Like obviously since there is no maximal number of flips you can "eventually" get this as the ratio so stoping when your ratio is lower is silly but you can't guarantee you will "eventually" get a higher ratio so if you ever do get that (example sequences H or THH) then you should stop.