Suppose I have a fair coin that I will flip 100 times. Obviously the expected outcome is 50 heads and 50 tails.
This is what I want to know:
How can I find the probability of this expected outcome happening. In other words, what is the probability that, given 100 flips, my outcome will be EXACTLY 50 heads and 50 tails.
$$\binom {100} {50}\left[\frac{1}{2}\right]^{50}\left[\frac{1}{2}\right]^{50}=\frac{\binom {100} {50}}{2^{100}}$$ You "choose" the 50 places, then probability of 50 heads, then probability of 50 tails.
A slightly different perspective:
Choose 50 numbers from $1,...,100$ without replacement. There are $(100)\cdots (51) = {100! \over 50!}$ ways of doing this. If we order the result, there are $50!$ equivalent selections, hence there are ${100! \over 50! 50!}$ ways of getting exactly 50 heads.
Since the total number of possibilities is $2^{100}$, we get the result ${1 \over 2^{100}} {100! \over 50! 50!}$.
There are $2^{100}$ possible sequences that can result from your $100$ flips, and all are equally likely. Of all these sequences, there are $\binom{100}{50}$ that have exactly $50$ Heads and $50$ Tails, so the probability of getting one of those is just $$\frac{\binom{100}{50}}{2^{100}}.$$
