Is the coin fair or unfair?

We assume, that both sides of the coin appear with the same probability $p = 1 - p = \frac{1}{2}$ . The probability $P_n(k)$ of having exactly $k$ times tails (or heads) for an total number $n = 1000$ of throws is given by a binomial distribution

$P_n(k) = {n \choose k} p^k (1 - p)^{n - k} = \frac{n!}{k! (n - k)!} \frac{1}{2^n}$

The mean value $\mu$ and the root mean square $\sigma$ (RMS) results to $\begin{aligned} \mu &= \langle k \rangle = \frac{1}{2} n = 500\\ \sigma &= \sqrt{\langle (k - \langle k \rangle)^2 \rangle} = \frac{1}{2} \sqrt{n} \approx 15.8 \end{aligned}$ The result $k = 440 \approx \mu - 3.8 \cdot \sigma$ is almost 4 standard deviations away from the mean value and therefore extremly unlikely. To be more precise, we calculate the probability $P_n(k \leq 440)$ , that $k$ assumes a value in the interval $[0, 440]$ . We use the fact, that the discrete binomial distribution can be approximated by the continuous normal distribution with the same mean value and RMS for $n \gg 1$ (central limit theorem)

$\begin{aligned} P_n(k \leq 440) &= \sum_{k = 0}^{440} {n \choose k} p^k (1 - p)^{n - k} \\ &\approx \int_{0}^{440} \frac{1}{\sqrt{2 \pi} \sigma} \exp \left(- \frac{(x - \mu)^2}{2 \sigma^2} \right) dx\\ &= \frac{1}{2} \left[ 1 + \text{erf} \left(\frac{440 - \mu}{\sqrt{2} \sigma } \right) \right]\\ &\approx 7 \cdot 10^{-5} = 0.007 \,\% \end{aligned}$ where $\text{erf}(x)$ denotes the error function. Thus, such a significant difference between "heads" and "tails" occurs only in one out of every 14,000 cases for a fair coin. With a probability bordering to certainty, we can say, that this coin is unfair.