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.
$\mathsf P(D_1<D_2) = \dint_0^R f(x) \dint_x^R f(y)\operatorname d y\operatorname d x \\ \mathsf P(D_1<D_3\mid D_1< D_2) = \frac{\int_0^R f(x) \left(\int_x^R f(y)\operatorname d y\right)^2\operatorname d x}{\int_0^R f(x) \int_x^R f(y)\operatorname d y\operatorname d x}$
Assuming that the results of each dart throw are independent*, then the probability of hitting double-double-treble 20, in that order, is indeed simply the product of the individual probabilities that you’ve derived from your simulation. I would argue that the resulting very small value does in fact match your experience: you’ve computed the probability of hitting that particular combination while aiming for the bullseye. The relevant real-life experience to which you should compare this isn’t the overall frequency with which you can land this particular combination, but instead the number of times that you’ve gotten it accidentally while trying to shoot bulls. I’ll go out on a limb here and say that’s never happened. The first two darts both have to land in a fairly small area that’s half a board width away from your aiming point.
Note, too, that your simulated probability of a treble 20 is almost three times that of a double 20. This certainly makes sense for Gaussian scatter from the center of the board, but is backwards from what one might expect when actually trying to hit those regions. Cetera paribus the probability of a treble 20 should be smaller since it covers a significantly smaller area than the double 20.
* Assuming independence for such small regions of the dart board doesn’t seem like a good approximation to me. Each dart that lands significantly reduces in interesting ways the available area for the next one. There’s not a lot of leeway for three darts in any of the treble zones, for instance.
Best Answer
This shouldn’t be too surprising given this very simple model. The underlying assumption is that the probability is uniformly distributed across the total area so that the probabilities are ratios of areas. In particular, any factor of $\pi$ in the expression for the area of one of the rings will cancel with the $\pi$ in the total area. Moreover, the shape and location of the subregions are irrelevant in this model, as is the point at which the player is aiming.
This isn’t a particularly realistic model, even assuming a novice player, but the point is to illustrate how a uniform distribution works, not to model a real darts player.