The binomial formula for the probability of x heads on n flips with a probability of 0.5 is:$$\displaystyle \frac{\frac{n!}{x!(n-x)!)}}{2^{n}}$$
For the probability of getting exactly n/2 heads, this becomes: $$\displaystyle\frac{\frac{n!}{(\frac{n}{2})!(\frac{n}{2})!}}{2^{n}}$$
I don't know how to evaluate this in the limit that n approaches infinity, but I used a spreadsheet's binomdist function (x,n,0.5,false) to calculate the probability of exactly 50% heads on n flips (n = 2x), where x ranges from ten to a billion, with steps increasing by powers of 10. Up to 1 billion heads on 2 billion flips, the probability of the 50% heads decreases per the table below.
| Heads | Flips | Probability |
|---|---|---|
| 10 | 20 | 0.176197052 |
| 100 | 200 | 0.05634847901 |
| 1000 | 2000 | 0.01783901115 |
| 10000 | 20000 | 0.005641825312 |
| 100000 | 200000 | 0.001784121886 |
| 1000000 | 2000000 | 0.000564189513 |
| 10000000 | 20000000 | 0.0001784124094 |
| 100000000 | 200000000 | 0.00005641895828 |
| 1000000000 | 2000000000 | 0.00001784124116 |
It seems that the probability of an outcome with exactly 50% heads decreases as the number of flips increases. This makes sense to me because the number of categories of outcomes that are near (but not equal to) 50% heads increases with increasing numbers of flips, so some of the most central probability would be reapportioned to the near-neighbor categories. I also know that the likelihood of getting extreme results decreases as n increases, so that more of the probability is centralized around the 50% heads rather than the tails.
However, from another point of view, it seems surprising. If the trend continues in the limit that n approaches infinity, then the probability of getting exactly 50% heads is zero when the coin is flipped an infinite number of times. This contradicts how I had thought expectation values work. I had thought that by flipping a coin an infinite number of times, the limiting-case behavior equates to the expectation value. The expectation value of heads for the coin is 0.5, but the binomial function suggests that we never reach it.
I recognize that the probability distribution becomes increasingly centralized about 50% in a symmetrical way. However it seems that it never actually lands there, even after an infinite number of flips, which seems paradoxical.
Am I understanding this correctly?
Stirling's approximation can be used to show that we have
$$\frac{ {2k \choose k} }{2^{2k}} \sim \frac{1}{\sqrt{\pi k}}$$
so you're correct that the probability of getting exactly 50% heads decays as $n \to \infty$ to zero but slowly.
How do we reconcile this with the intuition that as $n \to \infty$ we expect the fraction of heads to converge to 50%? In other words, does this contradict the law of large numbers? The answer is that it does not, because to ask for exactly 50% heads as $n \to \infty$ is much too strict (and that's not what the law of large numbers says). Here's an example of what we can prove: if we ask for the probability of getting between 49.9% heads and 50.1% heads, that probability converges to $1$ (very rapidly). More generally, for any $\varepsilon > 0$, the probability that the fraction of heads is between $\frac{1}{2} - \varepsilon$ and $\frac{1}{2} + \varepsilon$ converges to $1$. This is the weak law of large numbers.
To get even more specific than this, the central limit theorem characterizes the probability that the fraction of heads is between $\frac{1}{2} - \frac{x}{\sqrt{n}}$ and $\frac{1}{2} + \frac{x}{\sqrt{n}}$; the limit turns out to be a function of $x$ which grows towards $1$, so that e.g. we can find a specific value of $x$ such that this probability converges to more than 99%. Informally, the central limit theorem allows us to say that the fraction of heads is within some small multiple of $\frac{1}{\sqrt{n}}$ away from $\frac{1}{2}$ with high probability, which is pretty good for large $n$; e.g. when $n = 10^6$ this is about $\frac{1}{10^3}$.