This is an extension of Dart Throw Probability from "A Practice Guide To Quantitative Finance Interviews." Now we extend the problem to a generic $n+1$ darts, where we are given (1) Dart #1 is thrown closest to the center (2) Dart #2 to $n$ is thrown further from the center than the 1st dart.
What is the probability that the Dart #(n+1) is farther from the center than the first dart?
This question is equivalent to asking what is the probability that the $n+1-th$ dart is not the best among $n+1$ darts or the complement of "what is the probability that the $n+1$-th dart is the best among the $n+1$ darts.
When $n$ is large, doing the sequence enumeration or the integration in Graham Kemp's answer isn't practical. The author of this book provided an explanation that I'm struggling to understand:
This is what I understand:
(1) If we let $D_i$ denote the radial distance from the center of the i-th dart, then $D_i$ and $D_j$ are independent for $i \neq j$ and $i,j \in \{1,\cdots, n+1\}$, i.e., each dart's location on the board is independent.
(2) The event $D_i > D_1$ and $D_j > D_1$ are NOT independent any $i,j$ such that $i \neq j$ and $i \neq 1 \neq j$, i.e., knowing that a dart is further from the center than the first dart affects the knowledge of whether another dart is further from the center than the first dart.
(3) $D_k$ is independent of all events $D_j > D_i$ where $i \neq j$, $j \neq k$, $i \neq k$, i.e., where the k-th dart lands does not depend on the ordering of all darts (excluding k).
(1)-(3) are basically what the solution states, if I understand correctly. What I don't understand is why the answer $\frac{n}{n+1}$ is valid and how it relates to (1)-(3). Could someone explain?
In addition, I feel like the solution $\frac{n}{n+1}$ doesn't use the information that Darts 2-n are further away from the center than the first dart. If we neglect this information and instead ask the question "We will throw $n+1$ darts, what is the probability that the last dart is not the best given that each dart's radial location is IID, but we aren't given any information about the ordering of the first $n$ darts when we throw the n+1 dart?" I feel like the answer to this question is also $\frac{n}{n+1}$?
another edit (basically summarizing my comments below)
I seem to have come to the conclusion that conditioning on the fact that $D_{2,\cdots,n} > D_1$ does not change the answer. There are $(n+1)!$ permutations. $n!$ of these permutations will have the $(n+1)-th$ dart as the closest to the center. $\frac{n!}{(n+1)!} = \frac{1}{n+1}$, and $1-\frac{1}{n+1} = \frac{n}{n+1}$. So we arrive at the same conclusion without conditioning on $D_{2,\cdots,n} > D_1$. Now if we were to condition on this, we reduce the number of total permutations by $\frac{2}{3}$, but we also reduce the number of permutations where the $(n+1)-th$ dart is the best by $\frac{2}{3}$. So the probability is unchanged, i.e., $\frac{\frac{2}{3}n!}{\frac{2}{3}(n+1)!} = \frac{1}{n+1}$. Actually, the reduction by $\frac{2}{3}$ isn't correct, it's only correct for the case of 4 darts. The reduction factor should be a function of $n$, but both the numerator and denominator are reduced by this reduction factor. I'll come back and edit when I figured out the general reduction factor.
I believe the reduction factor is $\frac{(n+1)! - (n! + (n-1)!)}{(n+1)!}$. I found this by considering permutations of the set $\{D_1, \cdots, D_{n+1}\}$. In order to satisfy the condition that $D_{2,\cdots,n} > D_1$, $D_1$ must be the first in the sequence or it must be the second in the sequence given that $D_{n+1}$ is the first in the sequence. We see that there are $n! + (n-1)!$ instances of this. Hence the total number of permutations is reduced by the factor $\frac{(n+1)! - (n! + (n-1)!)}{(n+1)!}$. Now if we were to consider the total number of permutations where the $(n+1)-th$ dart is the best and condition on $D_{2,\cdots,n} > D_1$, this would require that all sequences look like $(D_{n+1}, D_1, \text{any permutation of } \{D_2, \cdots, D_n\})$. So there are $(n-1)!$ permutations of this. So the total number of permutations where the $(n+1)-th$ dart is the best and conditioned on $D_{2,\cdots,n} > D_1$ is reduced by $\frac{n! - (n-1)!}{n!}$, which is $=\frac{(n+1)! - (n! + (n-1)!)}{(n+1)!}$
Here's a thorough answer with fleshed out probability-theoretic arguments.
$D_1,\ldots,D_{n+1}$ are iid random variables which measure the distance of each throw to the center. We naturally assume that $P(\{D_1=D_2\})=0$ (this is the case e.g., when the underlying distribution of $D_1$ is continuous). As a consequence $P\big(\bigcap_{i\neq j} \{D_i\neq D_j\}\big)=1.$
The question is to compute $P\big(\{D_{n+1}>D_1\} \big| \{D_{n}>D_1\}\cap \ldots\cap\{D_{2}>D_1\} \big)$, i.e., the ratio $$ \frac{P\big(\{D_{n+1}>D_1\} \cap \{D_{n}>D_1\}\cap \ldots\cap\{D_{2}>D_1\} \big)}{P\big( \{D_{n}>D_1\}\cap \ldots\cap\{D_{2}>D_1\} \big)}.$$
It suffices to compute the numerator. The denominator will then follow by replacing $n+1$ with $n$. Note that $$\begin{align} P\big(\{D_{n+1}>D_1\} \cap \ldots\cap\{D_{2}>D_1\} \big) &= P\big(\{D_1 < \min(D_2,\ldots,D_{n+1})\} \big) \\&= P\big(\{D_1 < \min(D_2,\ldots,D_{n+1})\} \cap \bigcap_{i\neq j} \{D_i\neq D_j\} \big) \\&= P\big(\bigsqcup_{\sigma \in Sym(\{2,\ldots,n+1\}) } \{D_1 < D_{\sigma(2)}<\ldots < D_{\sigma(n+1)}\} \big) \\&= \sum_{\sigma \in Sym(\{2,\ldots,n+1\}) } P\big( \{D_1 < D_{\sigma(2)}<\ldots < D_{\sigma(n+1)}\} \big) , \end{align} $$ where $Sym(\{2,\ldots,n+1\}$ is the set of permutations on $\{2,\ldots,n+1\}$.
For each fixed $\sigma$, by the iid assumption, the vector $(D_1,D_{\sigma(2)},\ldots, D_{\sigma(n+1)})$ has the same distribution as $(D_1,D_{2},\ldots, D_{n+1})$. Consequently, $$P\big( \{D_1 < D_{\sigma(2)}<\ldots < D_{\sigma(n+1)}\} \big) = P\big( \{D_1 < D_{2}<\ldots < D_{n+1}\} \big),$$ and $P\big(\{D_{n+1}>D_1\} \cap \ldots\cap\{D_{2}>D_1\} \big) = n! \, P\big( \{D_1 < D_{2}<\ldots < D_{n+1}\} \big).$
Similarly, $$\begin{align} 1&= P\big(\bigsqcup_{\sigma \in Sym(\{1,\ldots,n+1\}) } \{D_{\sigma(1)} < D_{\sigma(2)}<\ldots < D_{\sigma(n+1)}\} \big) \\&= (n+1)! \, P\big( \{D_1 < D_{2}<\ldots < D_{n+1}\} \big) , \end{align} $$ thus $P\big( \{D_1 < D_{2}<\ldots < D_{n+1}\} \big) = \frac 1{(n+1)!}$ and finally $$P\big(\{D_{n+1}>D_1\} \cap \ldots\cap\{D_{2}>D_1\} \big) = \frac {n!}{(n+1)!} = \frac 1{n+1}.$$
The ratio we are interested in is therefore $$\frac{1/(n+1)}{1/n} = \frac{n}{n+1}.$$