Flipping fair coin - Examining an event. Let's say we flip a fair coin many times. Let's say 1.000.000 times. We know that even though each flip is not connected, influenced by past or future flips, in the long run each side heads or tails will end up approximately around  500.000 (50/50 event - theory of equilibrium) 
My question is: What price ranges can the difference take between heads and tails throughout the 1.000.000 flips? 
For example, if we could see the results at 1892 flips we may have witnessed a "70" difference (981 Heads over 911 Tails). 
Or maybe if we could see the results at 15250 flips we may have witnessed a "6" difference (7622 Heads over 7628 Tails) 
So... What could be considered a normal difference range? $\pm 70$? $\pm50$? 
And what could be considered an EXTREME RARE difference range somewhere in the flips? $\pm500$ $\pm2000$? $\pm10.000$?
 A: Given a fair coin, the number of heads (in $1000000$ flips) has a binomial distribution with a mean of $500000$, but it is very close to being normally distributed with that same mean, and with a standard deviation of
$$
\sigma = \sqrt{1000000 \left(\frac{1}{2}\right) \left(\frac{1}{2}\right)} = 500
$$
Using the usual properties of the normal distribution, we find that the number of heads should be within $500$ (one sigma) of $500000$ about $68$ percent of the time, within $1000$ (two sigmas) about $95$ percent of the time, and so on.
ETA: (Semi-)fun fact—The probability that we are outside one sigma is approximately $\frac{1}{\pi}$; the probability that we are outside two sigmas is about $\frac{1}{7\pi}$; the probability that we are outside three sigmas is about $\frac{1}{16e^\pi}$.
A: This is given by the standard deviation $\sigma$. In the case of a coin flipped N times, if we want to know the number of heads, the standard deviation, from the average of $N/2$ is $\sigma = \sqrt{.25N} = \sqrt{N}/2$.
Within 2 standard deviations is considered normal. Anything more than 2 standard deviations from the mean is usually considered an outlier.
A: Over many coin flips, the number of "Heads" is approximately normally distributed. (See: normal approximation to binomial.)
For $n$ coin flips, the distribution of the proportion of heads is approximately:
$$N\left(\frac{1}{2}n,\sqrt{\frac{1}{4}n}\right)$$
Construction a $95\%$ probability interval for this distribution, we can say that 95% of the time, the number of heads in $n$ coin flips will be within:
$$\left(\frac{n}{2}-(1.96)*\sqrt{\frac{1}{4}n},\frac{n}{2}+(1.96)*\sqrt{\frac{1}{4}n}\right)$$
For $n=1000000$ this implies that $95%$ of the time, the number of heads will be within the interval $$\left(499500,500500\right)$$
A: Note: This situation deserves some additional considerations which  put the problem in a different light.

The following is from chapter III: Fluctuations in Coin Tossing and Random Walks of the classic An Introduction to Probability Theory and Its Applications, Vol. I by W. Feller.
(W. Feller): For example, in various applications it is assumed that observations on an individual coin-tossing game during a long time interval will yield the same statistical characteristics as the observation of the results of a huge number of independent games at one given instant. This is not so.

He continues with

According to widespread beliefs a so-called law of averages should ensure that in a long coin-tossing game each player will be on the winning side for about half the time, and the lead will pass not infrequently from one player to the other.

But, in fact this is wrong and contrary to the usual belief the following holds:

With probability $\frac{1}{2}$ no equalization occurred in the second half of the game regardless of the length of the game. Furthermore, the probabilities near the end point are greatest.

The reasoning is based upon the Arcsine law for last visits (see e.g. Vol 1, ch.3, section 4, Theorem 1 in W. Feller's book): The   probability that up  to  and  including epoch $2n$ the last   visit to the origin occurs at      epoch     $2k$ is     given by
\begin{align*}
\alpha_{2k,2n}=\frac{1}{4^n}\binom{2k}{k}\binom{2n-2k}{n-k}
\end{align*}
Since according to Stirling's  formula
\begin{align*}
\binom{2k}{k}\sim    \frac{1}{\sqrt{\pi    k}}
\end{align*}
it     can  be shown that for fixed $0<x<1$ and $n$ sufficiently large
\begin{align*}
\sum_{k<xn}\alpha_{2k,2n}\approx  \frac{2}{\pi}\arcsin \sqrt{x}
\end{align*}

A consequence of the Arc since law are the following examples
Suppose that a great many coin-tossing games are conducted simultaneously at the rate of one per second, day and night, for a whole year.
  
  
*
  
*On the average, in one out of ten games the last equalization will occur before $9$ days have passed, and the lead will not change during the following 356 days.
  
*In one out of twenty cases the last equalization takes place within $2\frac{1}{2}$ days,
  
*and in one out of a hundred cases it
  occurs within the first $2$ hours and $10$ minutes.

(W. Feller): Anyhow, it stands to reason that if even the simple coin-tossing game leads to paradoxical results that contradict our intuition, the latter cannot serve as a reliable guide in more complicated situations.
