In independent Bernoulli trials with probability $p$ of success on each trial, let $X$ be the number of failures before the $n$th success. Then
$$
\Pr(X=x) = \binom{-n}{\phantom{+}x} (-q)^x p^n \text{ for } x=0,1,2,3,\ldots,
$$
where $q=1-p={}$probability of failure on each trial and
$$
\binom m x = \frac{\overbrace{m(m-1)(m-2)\cdots(m-x+1)}^\text{$x$ factors}}{x!} \text{ for any } m\in \mathbb R,
$$
i.e. $X$ has a <b>negative binomial</b> distribution. Then
\begin{align}
\operatorname E(X) & = nq/p, \\[5pt]
\operatorname{var}(X) & = nq/p^2. 
\end{align}
Thus $$\Pr(X>nq/p + 2\sqrt{nq} / p) \tag1$$ is the probability that $X$ is more than two standard deviations above the mean.

With $p=0.8= 1-q$ I graphed the probability in line $(1)$ as a function of $n$ for $n=1,\ldots,300$, using these R commands:

    x <- seq(300)
    y <- y <- 1 - pnbinom(x * (1-0.8)/0.8 + 2*sqrt(x*(1-0.8)/0.8^2), x, 0.8)
    plot(x,y)

[![enter image description here][1]][1]

Here is the part from $230$ to $280$:

[![enter image description here][2]][2]

This looks neat and orderly. Now enter this R command:

    lines(xx,yy)

(where $\texttt{xx}$ is the sequence from $230$ to $280$ and $\texttt{yy}$ is the corresponding set of $\texttt{y}$ values).

[![enter image description here][3]][3]

This looks like approximate periodicity with period $7,$ with the lowest points corresponding to indices congruent to $1 \bmod 7.$ So I looked at

    xxx <- seq(43)*7 + 1

etc., and got this:

[![enter image description here][4]][4]

So my question is: Why should we expect all this?

  [1]: https://i.sstatic.net/b7F1r.png
  [2]: https://i.sstatic.net/ZTwaB.png
  [3]: https://i.sstatic.net/mkPBU.png
  [4]: https://i.sstatic.net/ThMhl.png