The Expected Number of Bernoulli Trials Needed for $n$ successes in a row for any $p$ was answered early by a member and also found in this writeup on page two: https://courses.cit.cornell.edu/info2950_2012sp/mh.pdf
I did not want to crowd the previous post which had two questions and many comments, so this focuses on the standard deviation.
Let $A_n$ be the expected/average/mean number of coin flips/bernoulli trials with probability of success $p$ to get $n$ number of heads in a row.
$A_n=\frac{p^{-n}-1}{1-p}$. $A_n$ take from the pdf:
Do the Binomial Distribution formulas for the mean, variance and standard deviation work for this $A_n$ formula that was found?
Can I assume the variance is just $\sigma^2 = \mu(1-p)$ with $\mu=A_n$?
If yes, then the standard deviation is $\sigma = \sqrt{\mu(1-p)}$ with $\mu=A_n$
I am not sure it is this straightforward because our expression for the mean is more complex than just $np$.