Flipping coins- percentages of heads vs tails If I flip a coin multiple times and count the number of time it fell on heads and the number of times it fell on tails and keep a track of them. In how many flips on average will the delta between percentage of heads and percentage of tails will be less than 0.1%?
 A: 

*If I flip a coin "$2$" times... On average the delta between percentage of heads and percentage of tails will be less than $0.1 \%$. Thank you.

As It can be $\{HH \}$ or $ \{TT \}$ or $\{HT\}$ or $\{TH \}$
I went through a lot of sleeplessness because of this question randomly popping up somewhere I don't even remember now.
A: Great question! So let's look at the binomial distribution. We can represent the standard deviation with the equation $\mu_x = \sqrt{npq}$ and $p$ in our case is the probability of heads. This is $.5$. Same with tails. $n$ is the number of tosses. So $\mu_x = \sqrt{.5 * .5 * n} = .5*\sqrt{n}$
Our mean here is zero and our standard deviation is $.5\sqrt{n}$. Because we can apply the same rules as the normal distribution, if we had, let's say $10,000$ tosses, 68% will fall within 1 standard deviation, so $.5\sqrt{10,000}$ which is $50$. For 99% confidence, we would do 3 standard deviations, so for this would be $150$. So we can be 99% confident that the difference is equal to or less than 150 tosses. You can apply this logic to your problem to get the answer. 
Note: I'm not an expert in this field so please do feel free to edit anything that isn't proper terminology.
A: 
I want to check if a coin is fair(lands 50% of the times on each
  side. I assume that delta of 0.1% between them is fair). How many
  flips do i need in order to be 99% confident that the coin is fair?

For large n you can apply moivre laplace. So you approximate the binomial distribution by the normal distribution. 
Let $p=\frac{X}{n}$, where $X\sim Bin(n,0.5)$. The equation becomes
$P(|\frac{X}n-0.5|\leq 0.001)=2\Phi\left(\frac{0.001}{ \frac{0.25}{\sqrt n}} \right)-1=0.99$
$\Phi\left(\frac{0.001}{ \frac{0.25}{\sqrt n}} \right)=0.995$
$\frac{0.001}{ \frac{0.25}{\sqrt n}} =\Phi^{-1}\left(0.995\right)$
$\frac{0.001}{ \frac{0.25}{\sqrt n}} =2.576$
$\sqrt n=2.576\cdot \frac{0.25}{0.001}$
$n=414,736$
