Let $X$ be the number of hits. Then $X$ can take on values $0$, $1$, $2$, or $3$.
We want to find the probability of $k$ hits, that is, $\Pr(X=k)$, for $k=0,1,2,3$.
To save space, write for example $hmm$ for hit then miss then miss.
The probability that $X=0$ is the probability of $mmm$. This is $\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{3}{4}$, that is, $\frac{9}{24}$.
There are three ways we can have $1$ hit. They are $hmm$, $mhm$, and $mmh$.
The probabilities are $\frac{1}{3}\cdot\frac{1}{2}\cdot\frac{3}{4}$, $\frac{2}{3}\cdot \frac{1}{4}\cdot\frac{1}{2}$ and $\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{1}{4}$. These add up to $\frac{8}{24}$.
The probability that $X=3$ is $\frac{1}{3}\cdot\frac{1}{2}\cdot\frac{1}{2}$, which is $\frac{2}{24}$.
We skipped $X=2$, because the probabilities must add up to $1$. So the probability that $X=2$ is $\frac{5}{24}$. It might be a good idea to find $\Pr(X=2)$ the long way, by listing cases, as a check.
In summary, $\Pr(X=0)=\frac{9}{24}$ and $\Pr(X=1)=\frac{8}{24}$ and $\Pr(X=2)=\frac{5}{24}$ and $\Pr(X=3)=\frac{2}{24}$.
Now finding the mean is simple:
$$E(X)=\frac{9}{24}\cdot 0+\frac{8}{24}\cdot 1+\frac{5}{24}\cdot 2+\frac{2}{24}\cdot 3.$$
For the variance, if is easiest to use the fact that $\text{Var}(X)=E(X^2)-(E(X))^2$. And we have
$$E(X^2)=\frac{9}{24}\cdot 0^2+\frac{8}{24}\cdot 1^2+\frac{5}{24}\cdot 2^2+\frac{2}{24}\cdot 3^2.$$
Let $q_i = P(X=i), i=0,1,2,\ldots$. Assume $i\gt 0$ and consider the outcome of the first dart. To obtain $i$ bullseyes it must either be a bullseye, in which case we need a further $i-1$ bullseyes, or a non-bullseye hit, in which case we still need $i$ bullseyes. Therefore, using the law of total probability,
\begin{eqnarray*}
q_i &=& \dfrac{p}{2}q_{i-1} + \dfrac{1-p}{2}q_i \\
\therefore (p+1)q_i &=& pq_{i-1} \\
q_i &=& \dfrac{p}{p+1}q_{i-1} \\
q_i &=& \left(\dfrac{p}{p+1}\right)^iq_{0}\qquad\text{solving the simple recurrence relation.} \\
\end{eqnarray*}
We now require $q_0$. To obtain $0$ bullseyes we need $0$ or more non-bullseye hits followed by a miss. Therefore,
\begin{eqnarray*}
q_0 &=& \dfrac{1}{2} \sum_{n=0}^\infty{\left(\dfrac{1-p}{2}\right)^n} \\
&=& \dfrac{1}{2}\cdot \dfrac{1}{1-\frac{1-p}{2}} \qquad\text{using geometric series formula} \\
&=& \dfrac{1}{p+1}.
\end{eqnarray*}
Therefore, $$q_i = \dfrac{p^i}{\left(p+1\right)^{i+1}}, \quad i=0,1,2,\ldots.$$
Best Answer
For the intuition, take an extreme case, where the better player is almost sure to hit, and the other player is quite a bit less accurate. If the bad player goes first, there are two chances for an end by the time $3$ shots have been taken. If the good player goes first, an end by the time $3$ shots have been taken is less likely.