Optimal stopping in coin tossing with finite horizon There's a classic coin toss problem that asks about optimal stopping. The setup is you keep flipping a coin until you decide to stop, and when you stop you get paid $H/n%$ where $H$ is the number of heads you flipped and $n$ is the number of times you flipped. I believe that this problem is unsolved for an infinite horizon.
If we limit the horizon to allow $t$ tosses at maximum, what's the optimal stopping strategy?
 A: I think the correct strategy is as follows: Let's say you've made $n$ flips with $n<t$, and you have $x$ heads so far.  (1) If $\frac{x}{n}<\frac{1}{2}$ you should flip again; (2) If $\frac{x}{n}>\frac{1}{2}$ you should stop; and (3) If $\frac{x}{n}=\frac{1}{2}$ you are indifferent to stopping or not stopping.
Here's why.  For (1), You can quit now and receive $\frac{x}{n}<\frac{1}{2}$.  But if you flip again, your expected ratio after the next flip is $\frac{1}{2}(\frac{x}{n+1}+\frac{x+1}{n+1})=\frac{2x+1}{2n+2}$. Because $\frac{x}{n}<\frac{1}{2}$, we have $\frac{x}{n}<\frac{2x+1}{2n+2}<\frac{1}{2}$ (you can check these inequalities by cross multiplying).  Thus by flipping again, you expect a better result.  
Similar arguments show (2) and(3) as well.  
A: The coin-tossing problem is a fascinating problem. The finite-horizon case can easily be solved by dynamic programming. The infinite-horizon case has still unanswered questions. A nice survey of the problem can be found in
http://personal.vu.nl/h.c.tijms/APMNarticle.pdf
I found it surprising that in the infinite-horizon case the simple heuristic of stopping as soon the proportion of heads is more than 0.5 has an expected ratio of $\pi/4=0.785398...$, being very close to the conjectured value of the minimal expected ratio.
A: this is just a guess, but if t is sufficiently large than the amount paid should be close to expected value, well we notice that for any t you stop at the expected value is $E(\frac{H}{t})=\frac{1}{t}E(H)=\frac{1}{t}\frac{t}{2}=\frac{1}{2}$ so for any t you have the same expected value but notice that variance for a given t is $Var(\frac{H}{t})=\frac{1}{t^{2}}Var(H)=\frac{1}{t^{2}}\frac{t}{4}=\frac{1}{4t}$ so the higher t you pick the less "risky" of a game it becomes and closer chances you will get expected value. So personally Im a pretty risk adverse person so I like I would choose t=n since possible outcomes of $\frac{H}{n}$ are less volatile than $t<n$. But thats just my personal choice, if you want to have more of a chance on getting a higher pay than just $\frac{1}{2}$ than you would want a more volatile outcomes so you may want to choose a $t<n$
Note: after reviewing my strategy this works more for a game if you have to choose $t$ before you start, but if you can choose at any time during game than this strategy may not be the best, because obviously if you flipped a heads first you would want to stop right then
