Tossing the coin $2n$ times, win if tossed heads $>n$ times We have a coin with probability of heads $p = 0.48$, we toss it $2n$ times and win, if coin landed heads more than $n$ times. We can choose $n$. What $n$ should we pick?
 A: With $n=0$, our winning probability is $0$. With $n=1$ it is $p^2$. With $n=2$, it is $p^4+4p^3(1-p)=p^3(4-3p)$, etc. You can go on computing explicitly, but when can you stop? 
Note that the expected value is $E[X]=0.96n$, the variance is $V[X]=0.4992n$, and you need at least $n+1$ heads. Now use Chebyshev's inequality to obtain a bound.
A: For what $n$ is this a max?  Somebody who is sharper than I am will have to tell you that...
$$\sum\limits_{i=n+1}^{2n}{{2n}\choose{i}}.48^{i}.52^{2n-i}$$
A: $n$, in fact, seems to be around $11$ or $12$ or $13$. So for the lack of anything else, I decided to use De Moivre - Laplace integral theorem. This gives me that integral: $$\frac{1}{2\pi} \int_{0.0566139 \sqrt{n}}^{1.57359 \sqrt{n}}e^{-x^2/2} dx$$
But estimated error for $n$ with values around $12$ is pretty high — about $0.3$ or so. But from what Wolfram|Alpha tells, this function of $n$ seems convex, which means we can just manually check for the answer, inserting integer values in exact formula, which is $$\sum_{i=n+1}^{2n} C_{2n}^{i} 0.48^i 0.52^{2n-i}$$ and see, at which points it gives us the biggest answer, which gives us answer of 12 or 13. This is my answer, I'm not sure it's right.
A: The optimum $n$ is not unique: for $n = 12$ and $n = 13$, a maximum probability is attained:  $$P(n) = \sum_{k=n}^{2n} \binom{2n}{k} (0.48)^k (1-0.48)^{2n-k}$$ has value $$\frac{1222235987447480383482148518100992}{3552713678800500929355621337890625}$$ for both these cases, as found by computer search.  It is easy to see that $\displaystyle\lim_{n \to \infty} P(n) = 0$, so there is at least one maximum value for some finite $n$.
