Suppose someone wants to convince the public that he possesses a superpower, that can affect outcomes of fair coin tosses. To demonstrate his abilities he is demanded to toss a fair coin, while maximising the fraction of heads. However, because the public does not know probability theory, they allow him to chose when to stop. What is the largest possible expected fraction of heads he can get?
Formally, the problem can be reworded the following way:
Suppose $X_1, X_2, ...$ are i.i.d. Bernoulli random variables with success probability $\frac{1}{2}$. The reward for stopping on $n$-th turn is $Y_n = \frac{\sum_{i=1}^n X_i}{n}$. What is the optimal stopping rule?
One of the possible strategies (though maybe not the best one) is a $1$-stage look-ahead rule, that tells us to stop on $n$-th step if $Y_n \geq E[Y_{n+1}|Y_n]$. Here this condition is equivalent to $Y_n \geq \frac{1}{2}$. I do not know exactly, what expected gain does this strategy yield, but it is clearly $\geq \frac{1}{2} + (1 - \frac{1}{2})\frac{1}{2} = \frac{3}{4}$.
However, there may still be some even better strategy...
Although it is very easy to implement the dynamic programming solution if the number of coin tosses is upper bounded, it seems like the general problem is not not solved and called the "Chow-Robbins game" [1]. For example, if the number of tosses is upper bounded by N=100, one finds that the optimal strategy has an expected fraction of heads equal to (1+0.5677)/2 > 3/4.
[1]: Häggström, O. and Wästlund, J., 2013. Rigorous Computer Analysis of the Chow–Robbins Game. The American Mathematical Monthly, 120(10), pp.893-900. https://arxiv.org/pdf/1201.0626.pdf :