Wikipedia writes For example, suppose a fair coin is flipped 100 times. Using the law of averages, one might predict that there will be 50 heads and 50 tails. While this is the single most likely outcome, there is only an 8% chance of it occurring.
Where can I get formula and explanation, how is it possible?
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As others have said, the formula is given by the binomial distribution.
As for an intuitive explanation, the reason why the probability is so low is that there are so many other outcomes that are not so unlikely either. $49/51$, $48/52$, and their symmetric, are all reasonably likely. If you set the probability of $50/50$ to $90\%$, there is no room for these other options since all probabilities have to sum up to $1$.
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Arnaud and Badjohn have addressed the math to calculate the exact answer, so I'll add another way of explaining intuitively.
8% seems small, but it's actually very large. There are 101 possible results for 100 coin flips (100 heads, 99 heads, 98 heads,... all the way to 0 heads). If each result was equally probable, the each event would have a probability of slightly less than 1%. So the actual probability for 50/50, 8%, is 8 times the "average" probability.
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When thinking about large numbers in probability, it's often useful to go back and examine how the same system works with smaller numbers. You can then intuitively extrapolate the qualitative behaviour more accurately.
For a single coin, there are only two outcomes (0 heads, 1 head), and only one "path" to achieving each.
For two coins, there are three outcomes (0 heads, 1 head, 2 heads). For the extreme outcomes, there's only one path to each. But for the median outcome, there are two paths, depending on whether the first or the second coin happened to be the one coming up heads. So it will occur 50% of the time. As there are a total of four (2^2) paths, that means the extreme outcomes each occur with 25% probability.
For three coins, there are four outcomes (0, 1, 2, and 3 heads). Again, there's only one path to each of the extrema, and these occur once every eight (2^3) trials. But now there's obviously three ways to get one head, and three ways to get one tail (two heads). The probability of each is three-eighths, or 37.5%. But since this is a dual mode, the pattern might not yet be entirely clear.
With four coins, there are five outcomes, and the extrema occur with 1/16th (6.25%) probability each. There are four ways to get one head, and four to get one tail, for 25% probability each. With ten paths of the sixteen (2^4) possible accounted for, that leaves six paths to the median, for 37.5% probability. In detail, those six paths are: HHTT, HTHT, HTTH, THHT, THTH, TTHH. So already the median and mode outcome occurs less than 50% of the time.
You may wish to continue the series by yourself, and confirm that the results you obtain by hand match the formulae given in other answers. Note that because the number of paths increases exponentially to the number of coins, you'll only be able to inspect exhaustively for small numbers of coins, but you could use a computer to go a bit further - not all the way to 100 though!
The formula is simple enough, you use the binomial distribution and you find it is ${100 \choose 50} 2^{-100}$. In general the probability of an exact 50/50 split on $2n$ flips of a fair coin is ${2n \choose n} 2^{-2n}$. Using asymptotics for factorials one can show that ${2n \choose n} 2^{-2n}$ behaves like $1/\sqrt{n}$, so it decays albeit slowly. This occurs because the distribution of the actual number of heads and tails broadens as you do more flips. The distribution of the fraction of heads vs. tails becomes narrower as you do more flips, but this is a different thing. For example, the probability of 48% to 52% heads includes the 50/50 split as well as the 48/52, 49/51, 51/49, and 52/48 splits, all of which are pretty likely as well.