Ytukyg

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?

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: $mathsf P(D_1<D_3mid D_1<D_2)$.   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^3operatorname d r
\[1ex] & = dfrac 1 2
\[2ex]mathsf P(D_1<D_3mid 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.

$mathsf P(D_1<D_2) = dint_0^R f(x) dint_r^R f(y)operatorname d yoperatorname d x \ mathsf P(D_1<D_3mid D_1< D_2) = frac{int_0^R f(x) left(int_r^R f(y)operatorname d yright)^2operatorname d x}{int_0^R f(x) int_r^R f(y)operatorname d yoperatorname d x}$

...it seems like this question can only be answered in terms of the distance from the center of the first toss.

Why? I think you have almost all the information to answer it. I say almost because you have to interpret the "Jason's skillfulness is constant" phrase.

I try to answer it by using simple probabilistic arguments and notation.

Translate Question in Mathematical Terms

First we need to translate the question in probabilistic terms by writing (I use your notation for $D_i$):

$Prob(D_3 > D_1 | D_2 > D_1) = ?$

Interpretation of "Jason's skillfulness is constant"

The fact that "Jason's skillfulness is constant" can be interpreted in different ways. Does it mean he's a "normal"/average player (e.g., his probability distribution is a 2D normal pdf centered in the dartboard center) or a "skillful" professional player (e.g., his probability distribution has an higher kurtosis than the normal pdf) or is he a monkey throwing darts on the wall without much aim in doing so?

We don't know it and we don't care, we simply interpret this as the fact that the distances $D_i$ of Jason's darts from the center constitute a sequence of random variables i.i.d. (independent and identically distributed), whatever that distribution is (normal pdf, leptokurtic, "monkey" pdf, etc.).

Answer to the question

Now that we have clarified the question and made the relevant assumptions we can answer it. As I said, I'll try to make it is simple as possible by using trivial logic/probability.

1. $$begin{align}Prob(D_3 > D_1 | D_2 > D_1)&=Prob(D_3 > D_1 | D_1 text{ is the best among 2 darts})\&=1-Prob(D_3 < D_1 | D_1 text{ is the best among 2 darts})\&underbrace{=}_{text{HP: $D_i$ i.i.d.}}1-Prob(D_3 text{ is the best among 3 darts})end{align}$$


2. As you can see above, I've used the i.i.d. assumption (in the last step). Now I translate the probability over $D_3$ to a more intuitive one and I use again the i.i.d. assumption:
   $$Prob(D_3 text{ is the best among 3 darts}) underbrace{equiv}_{text{HP: $D_i$ i.i.d.}} Prob(text{Picking an object among 3 available})=frac{text{# favorable outcomes}}{text{# total outcomes}}=frac{1}{3}$$


3. Hence the result is:
   $$begin{align}Prob(D_3 > D_1 | D_2 > D_1)&=1-Prob(D_3 text{ is the best among 3 darts})\&=1-frac{1}{3}\&=frac{2}{3}end{align}$$

Extension to Generic $n$

You can extend this result to a generic number $n$ of darts, with the further (similar) assumption that $D_1$ is the best dart among $(n-1)$ darts:

1. $$begin{align}&Prob(D_n > D_1 | D_2 > D_1, D_3 > D_1, dots, D_{n-1} > D_1)\&=Prob(D_n > D_1 | D_1 text{ is the best among $(n-1)$ darts})\&=1-Prob(D_n < D_1 | D_1 text{ is the best among $(n-1)$ darts})\&underbrace{=}_{text{HP: $D_i$ i.i.d.}}1-Prob(D_n text{ is the best among $n$ darts})end{align}$$


2. $$Prob(D_n text{ is the best among $n$ darts}) underbrace{equiv}_{text{HP: $D_i$ i.i.d.}} Prob(text{Picking an object among $n$ available})=frac{text{# favorable outcomes}}{text{# total outcomes}}= frac{1}{n}$$


3. Hence the result is:
   $$begin{align}Prob(D_n > D_1 | D_2 > D_1, D_3 > D_1, dots, D_{n-1}>D_1)&=1-Prob(D_n text{ is the best among $n$ darts})\&=1-frac{1}{n}\&=frac{n-1}{n}end{align}$$