How many times do you have to toss a coin so that the probability that #heads = #tails is 0.01?

Question

I thought it would be something like: $Binomial(n, \frac{1}{2})$. So $0.01 = \binom{n}{\frac{n}{2}}(\frac{1}{2})^{\frac{n}{2}}(\frac{1}{2})^{\frac{n}{2}}$, because half of $n$ should be heads and the other half should be tails.

Answer 1

Your suggestion is correct.

For even $n$, an approximation to the central binomial coefficient is $\displaystyle {n \choose n/2} \sim 2^n\sqrt{\frac{2}{\pi n}}\text{ as }n\rightarrow\infty$ so we would be looking for $\sqrt{\dfrac{2}{\pi n}} \approx 0.01$ or $n \approx \dfrac{2}{ 0.01^2 \pi} \approx 6366.2.$

$6366$ turns out to be the best even integer answer.

 n   C(n,n/2)/2^n
==== ============
6364 0.01000133
6366 0.00999976
6368 0.00999819