Binomial Approximation of Gaussian Distribution
Let $X \sim B(n, p)$. At the third layer $\begin{bmatrix}1 &amp; 3 &amp; 3 &amp; 1\end{bmatrix}$ of the Pascal's triangle, we have $X \sim B(3, \frac{1}{2})$. Which means $P(X = k) = {3 \choose k} (\frac{1}{2})^k (\frac{1}{2})^{3-k} = \frac{{3 \choose k}}{2^3}$. The variance is given by $\sigma^2 = n p q = 3 \cdot \frac{1}{2} \frac{1}{2} = \frac{3}{4}$ and the std by $\sigma = \sqrt{\frac{3}{4}}$. You can derive the variance by looking at $n$ independent Bernoulli trials.

In general, a 1d binomial filter/kernel of size $n$ with $p = \frac{1}{2}$ is given by $\begin{bmatrix}P(X = 0) &amp; P(X = 1) &amp; \cdots &amp; P(X = n-1)\end{bmatrix}$. As $n \to \infty$, this kernel will approximate the normal distribution with mean $\frac{n}{2}$ and variance $\sigma^2 = \frac{n}{4}$. If you want a different variance/std, you should change $p \neq 0.5$.

As somebody already wrote in the comments, you should look at the De Moivre-Laplace theorem to prove this. However, this is only a special case of the more general central limit theorem. By showing that the binomial distribution is asymptotically equivalent to the normal distribution as $n\to\infty$, I would argue that we have shown already both directions.