Probability of 20 consecutive success in 100 runs.

Feller has this all worked out on p. 325 of An Introduction to Probability Theory and Its Applications, 3rd Edition, equation 7.11: $$q_n \sim \frac{1-px}{(r+1-rx)q} \cdot \frac{1}{x^{n+1}}$$ where $q_n$ is the probability of no success run of length $r$ in $n$ trials, $p$ is the probability of success, $q=1-p$, and $x$ is the root near 1 of

$$ 1-x + q p^r x^{r+1} = 0 $$

With your data, we find $x \approx 1.017502$ and $q_{100} \approx 0.2247$.

So the probability that the chess player will have at least one run of 20 successes is $0.7753$, approximately.

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Edit: Actually "$x$ is the root near 1 ..." is slightly misleading. The equation has two positive roots, and we must peak the one that is not $1/p$. Details here.