I have been thinking of the following game and the best strategy to follow.
Let us assume I have a biased coin which gives head with probability $0.8$ and tails with probability $0.2$. I have been offered two options:
- Toss the coin $10$ times. If the number of heads is greater than $8$, I win; if it is less than $8$, I lose; if it is $8$, we play again.
- Toss the coin $20$ times. If the number of heads is greater than $16$, I win; if it is less than $16$, I lose; if it is $16$, we play again.
What would be the best strategy? Also, what would be the optimal number of tosses should I select have I given the option?
Let's assume you want to maximize your expected value. The first strategy is a random variable $$X_1= \begin{cases} 1 & \textrm{if } k > 8,n=10 \\ 0 & \textrm{if } k < 8,n=10 \\ Y & \textrm{if } k = 8,n=10 \\ \end{cases} $$ where $X_1 \sim Y_1$. So $$\mathbb{E}[X_1]=(p_{9,10}+p_{10,10})+p_{8,10}\mathbb{E}[Y_1]$$ $$\mathbb{E}[X_1]=\frac{p_{9,10}+p_{10,10}}{1-p_{8,10}}$$ Similarly the second strategy has expected value $$\mathbb{E}[X_2]=\frac{p_{17,20}+p_{18,20}+p_{19,20}+p_{20,20}}{1-p_{16,20}}$$ Those $p_{k,n}$ are all binomial probabilities. Turns out that $\mathbb{E}[X_1]>\mathbb{E}[X_2]$