If by "the exact same spot" you mean a point, the chances are zero even with a countable infinity of throws, as there are uncountably many points. Any countable set of points has no area. If you mean within a small error, take the ratio of the area to the area of the dartboard (presumably you are assuming a uniform distribution over the dartboard and nowhere else) and that is the chance on the next throw. Of course, those who can hit the dartboard every time don't have a uniform distribution of throws over it, so this is unrealistic.
Yes.
The key to the question is that "skillfulness is constant" means that the three distances are independent and identically distributed. It is also almost impossible for them to be tied, since they are continuous RV.
This means that even without knowing what the distribution of distances may be, we do know that the $6$ order arrangements are equally probable.
$$\boxed{\boxed{D_1{<}D_2{<}D_3 \;,\; D_1{<}D_3{<}D_2} \;,\; D_3{<}D_1{<}D_2}\\ D_2{<}D_1{<}D_3 \;,\; D_2{<}D_3{<}D_1 \;,\; D_3{<}D_2{<}D_1$$
Now we were asked to evaluate the probability that the third throw is placed further than the first throw, when given that the first throw is placed closer than the second. That is: $\mathsf P(D_1<D_3\mid D_1<D_2)$. Clearly this is: $2/3$.
If doing it the easy way isn't convincing, we shall do it the hard way:
Let $f_1(r), f_2(r), f_3(r)$ be the probability density functions of the distance from the center of the three darts, respectively, and $F_1(r), F_2(r), F_3(r)$ be their cumulative distribution functions. (These of course will be co-equal because the darts are i.i.d., but the indices are a readability guide.)
Without knowing the skill of the dart thrower we may as well assume the darts are uniformly distributed over the board (why? why not?). Then ...,
$$\newcommand{\dint}{{\displaystyle \int}}
\begin{align}
f_1(r) & = \dfrac{2r}{R^2} & =f_2(r)=f_3(r)
\\[2ex]
F_1 (r) & = \dfrac{r^2}{R^2} & =F_2(r)=F_3(r)
\\[2ex]
\mathsf P(D_1<D_2)
& = \int_0^R f_1(r)\,\big(1-F_2(r)\big) \operatorname d r
\\[1ex] & = \frac 2{R^4}\int_0^R rR^2-r^3\operatorname d r
\\[1ex] & = \dfrac 1 2
\\[2ex]\mathsf P(D_1<D_3\mid D_1<D_2)
& = \int_0^R f_1(r \mid D_1<D_2)\; \big(1-F_3(r)\big) \operatorname d r
\\[1ex] ~ & = \dfrac{
\dint_0^R f_1(r) \, \big(1-F_2(r)\big) \, \big(1-F_3(r)\big) \operatorname d r
}{
\dint_0^R f_1(r) \, \big(1-F_2(r)\big)\operatorname d r
}
\\[1ex] ~ & = \dfrac{
\frac{2}{R^6} \dint_0^R r \, (R^2-r^2)^2 \operatorname d r
}{
\frac 1 2
}
\\[1ex] ~ & = \dfrac 2 3
\end{align}$$
As an exercise for the student, try this with some other assumption of the dart thrower's skill, such as $f_1(r)=\frac 1 R$.
Finally examine the situation when you make no assumption, and convince yourself of the symmetry argument.
$\mathsf P(D_1<D_2) = \dint_0^R f(x) \dint_x^R f(y)\operatorname d y\operatorname d x \\ \mathsf P(D_1<D_3\mid D_1< D_2) = \frac{\int_0^R f(x) \left(\int_x^R f(y)\operatorname d y\right)^2\operatorname d x}{\int_0^R f(x) \int_x^R f(y)\operatorname d y\operatorname d x}$
Best Answer
Consider a dartboard with $n$ sectors and respective sector probabilities $p_0, \dots, p_{n-1}$.
The probability of hitting sector $i$ twice in a row is $p_i^2$, and the combined probability of hitting the same sector twice in a row is $q := \sum_{i=0}^{n-1} p_i^2$.
Let $[i+1]_n$ denote $(i+1)$ modulo $n$. The probability of hitting sector $i$ and then sector $[i+1]_n$ is $p_i p_{[i+1]_n}$. The total probability of such an event occurring is $$r := \sum_{i=1}^n p_i p_{[i+1]_n} \le \left( \sum_{i=1}^n p_i^2 \sum_{i=1}^n p_{[i+1]_n}^2 \right)^{1/2} = \sum_{i=1}^n p_i^2 = q,$$ where we used the Cauchy-Schwarz inequality. The inequality holds with equality if and only if $p_i p_{[i+1]_n} = p_i^2$ for all $i$ which is equivalent to requiring that $p_i$ is the same for all $i$.
So it is more likely to hit the same sector twice than to hit the two different sectors as specified in the question.