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%?
1) Interpreting the revised question as asking how many tosses $n$ of a fair coin are required to have the probability that $|H-T| \lt 0.001 n $ greater than or equal to $0.99$, where $H$ is the number of heads and $T$ the number of tails with $H+T=n$, this is equivalent to asking for $n$ such that $\mathbb{P}\left(-0.0005n \lt H-\frac12{n} \lt 0.0005n\right) \ge 0.99$ since if $H=\frac12{n}+0.0005n$ then $T=\frac12{n}-0.0005n$ and $H-T= 0.001n$.

It is typical to use a normal approximation to the binomial distribution for questions like this.  $H$ has a mean of $\frac12{n}$, a variance of $\frac14n$ and a standard deviation of $\frac12\sqrt{n}$.  So using $\Phi^{-1}$ as the inverse cumulative distribution function of a standard normal random variable, we have $\mathbb{P}\left(-\Phi^{-1}(0.995) \lt \dfrac{H-\frac12{n}}{\frac12\sqrt{n}} \lt \Phi^{-1}(0.995)\right) \gtrsim  0.99$ or  $\mathbb{P}\left(-\Phi^{-1}(0.995)\frac12\sqrt{n} \lt H-\frac12{n} \lt \Phi^{-1}(0.995)\frac12\sqrt{n}\right) \gtrsim 0.99$.

So we need to solve  $0.0005n \gtrsim \Phi^{-1}(0.995)\frac12\sqrt{n}$ which gives  $n \gtrsim (1000\Phi^{-1}(0.995))^2$ and since $\Phi^{-1}(0.995)\approx 2.575829$ this gives  $n \gtrsim 6634897$.

But this is only an approximation. We could use computer calculations of the binomial distribution to get exact answers, but there is an issue that due to rounding every thousand, the probability is not a monotonic function of $n$.  I believe that the smallest $\mathbb{P}\left(\left|\frac{H-T}{n}\right| \lt \frac1{1000}\right) \approx 0.9900006$ for $n=6633001$ but about $0.989995$ for $n=6636000$, with any greater $n$ giving over $0.99$.  So the normal approximation gives a reasonable answer.

2) The original question is actually slightly different, asking "In how many flips on average will the delta between percentage of heads and percentage of tails will be less than $0.1\%$?" Interpreting this as asking for the expected cumulative number of times in the $n$ tosses it is true that $\left|\frac{H-T}{H+T}\right| \lt \frac1{1000}$, it is possible to use linearity of expectation and to add up the probabilities of being in this interval.  For small values of $H+T$, it makes sense to use the binomial distribution as rounding matters: for example with $n \le 1000$ the probability is zero for odd $n$.

As I said in a comment while the question was closed, my calculations suggest that for $1$ flip, the expectation is $0$; for $10$ flips the expectation is ${2 \choose 1}\frac1{2^{2}}+ {4 \choose 2}\frac1{2^{4}}+ {6 \choose 3}\frac1{2^{6}}+ {8 \choose 4}\frac1{2^{8}}+ {10 \choose 5}\frac1{2^{10}}\approx   1.71$; for $100$ flips about $7.04$; for $1000$ flips about $24.25$; for $10^4$ flips about $540.19$; for $10^5$ flips about $16663.55$; for $10^6$ flips about $483950.29$.  For small $n$ the expected number is much smaller than $n$, but for very large $n$ most cases will fall in the relative small interval around the mean.
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.