A geometrically distributed random variable where the first success probability is different Your first success probability $p_a$ is higher on the first trial, than on the remaining trials (where it is $p_b$, constantly). How does this impact the mean of the general experiment ($1/p$ in the general, fixed geometric case) and the standard deviation ($\sqrt{(1-p)/p^2}$ in the general, fixed geometric case)?
EDIT: I think I got the mean:
$p_a + (1-p_a)(1/p_b+1)$
But I'm still unsure about the stdev
 A: It seems you've already figured out how to adjust the expected value: Consider the experiment as one trial with probability $p_a$, and with probability $1-p_a$ a standard series of Bernoulli trials with parameter $p_b$ plus one extra trial.
Since the variance can be written as the difference of two expectation values, you can do the same thing for the variance. From $\langle n\rangle=1/p$ and $\sigma^2=(1-p)/p^2=\langle n^2\rangle-\langle n\rangle^2$ in the unmodified case, we get $\langle n^2\rangle=(1-p)/p^2+1/p^2=(2-p)/p^2$. Thus in the modified case, we have
$$
\begin{align}
\langle n'\rangle&=p_a+(1-p_a)\langle n+1\rangle\\
&=p_a+(1-p_a)(1/p_b+1)\\
&=1+(1-p_a)/p_b\;,\\
\langle n'-1\rangle^2&=\left(\langle n'\rangle-1\right)^2\\
&=(1-p_a)^2/p_b^2\;,\\
\langle(n'-1)^2\rangle&=p_a\cdot0+(1-p_a)\langle n^2\rangle\\
&=(1-p_a)(2-p_b)/p_b^2\;,
\end{align}
$$
and thus
$$
\begin{align}
\sigma'^2
&=\langle(n'-1)^2\rangle-\langle n'-1\rangle^2
\\
&=
(1-p_a)(2-p_b)/p_b^2-(1-p_a)^2/p_b^2\\
&=
\frac{1-p_a}{p_b^2}(1+p_a-p_b)\;.
\end{align}
$$
To check the result, note that it yields the correct values $0$ for $p_a=1$, $(1-p_b)/p_b^2$ for $p_a=0$ and $(1-p_a)p_a$ for $p_b=1$.
