Probability of 20 consecutive success in 100 runs. Suppose a chess player have a win rate equal 90%, what is the chance to have 20 consecutive wins (successes) playing 100 games? Consider that lose/draw = fail.

I've studied basic statistics in college and it seems like a binomial distribution problem (right?), but honestly I can't figure out a way to solve this problem considering "consecutive" successes.

Is there a statistical distribution for this kinda problem?

Thanks very much! I really appreciate any thoughts!
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.



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.
The probability of getting $20$ consecutive wins is:

$$0.9^{20}$$

The first win of these consecutive wins can be at any trial from $1$ to $80$, and the probability of it being any on any of these trials is evenly distributed, so the probability of getting $20$ consecutive wins out of $100$ is:

$$\frac{80}{100}*0.9^{20}$$

A generalization for this formula would be:

$$\frac{n-k}{n}p^k$$

Where $p$ is the probability, $n$ is the number of trials, and $k$ is the number of consecutive wins.