In the book, "A Practical Guide to Quantitative Finance Interviews", the following question is posed on page 75:
Jason throws two darts at a dartboard, aiming for the center. The second dart lands farther from the center than the first. If Jason throws a third dart aiming for the center, what is the probability that the third throw is further from the center than the first? Assume Jason's skillfulness is constant.
At first glance, it seems the second throw is irrelevant – since his skillfulness is constant, the second throw in no way influences his third throw. Furthermore, it seems like this question can only be answered in terms of the distance from the center of the first toss. Indeed, denote this distance by $D_1$. Then assuming the area of the dart board is $1$, the probability that the third throw is further than the first is just the area of the annulus, which is given by $1 – \pi D_1^2$.
However, the solution given in this text is $\frac{2}{3}$, which is arrived at by enumerating the possible outcomes. Let $D_i$ be the distance of the $i$th throw. Then the possible outcomes assuming the second throw is further than the first are:
$$
D_3 < D_1 < D_2 \\
\boxed{D_1 < D_3 < D_2} \\
\boxed{D_1 < D_2 < D_3}
$$
The two boxed outcomes satisfy our event in question, so the probability must be $\frac{2}{3}$. But, this seems to be answering a different question, namely,
What's the probability the third throw is not the best out of three, given the second throw is worse than the first?
Are the two block-quoted questions really asking the same thing, and I've misunderstood the first one?
Best Answer
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.
$\mathsf P(D_1<D_2) = \dint_0^R f(x) \dint_x^R f(y)\operatorname d y\operatorname d x \\ \mathsf P(D_1<D_3\mid D_1< D_2) = \frac{\int_0^R f(x) \left(\int_x^R f(y)\operatorname d y\right)^2\operatorname d x}{\int_0^R f(x) \int_x^R f(y)\operatorname d y\operatorname d x}$