Probability this game ends in k darts thrown Marissa is playing a game where she throws darts at a dartboard until her most recent dart is further away from the bullseye than any dart she threw previously. What is the probability that this game ends in $k$ darts thrown?
My solution:
The probability is essentially equal to how many valid orderings of the darts there are over $k!$. Thus, label each dart from $1$ to $i$,corresponding to when it was thrown. Similarly, label the positions, according to how close they are to the bullseye(with $1$ being the position closest to the bullseye). We will call the first dart and first position $d_1$ and $p_1$ respectively. Now, we know that $d_k$ needs to be assigned to $p_k$, as the game ends at $k$ darts. Thus, we need to order the first $k-1$ darts. It's clear that $d_1$ must be assigned to $p_{k-1}$, as otherwise, there will be a time where the most recent dart thrown will be the furthest from the bullseye, ending the game before $k$. But after we assign $d_1$ to $p_{k-1}$, the rest of the ordering can be any which way, right? So it's $\frac{(k-2)!}{k!}$?
However, the answer is actually $\frac{k-1}{k!}$, and I fail to see how.
 A: Your reasoning is correct, and your answer, $\dfrac{(k-2)!}{k!}=\dfrac{1}{k(k-1)}$, is correct.
The supposed answer, $\dfrac{k-1}{k!}$, is wrong. The usual way to show an answer to a question like this is wrong, is to show that $\sum_{k=2}^\infty {\dfrac{k-1}{k!}}\neq1$, but in this case the sum equals $1$ by coincidence. So instead, let's consider the probability that the game ends in $3$ darts thrown. This means the darts, in order of increasing distance from the bullseye, were: second dart, first dart, third dart. The probability of this happening is $\dfrac{1}{3!}=\dfrac16$, which contradicts the supposed answer $\dfrac{k-1}{k!}=\dfrac13$.
By the way, another way to phrase the question is:

Marissa is playing a game where she throws darts at a dartboard until her most recent dart is further away from the first dart she threw. What is the probability that this game ends in $k$ darts thrown?

A: Assuming the game does not stop with the first dart:
You argue that for the game to end on toss $k$, that toss must be of greater measure (of length from the center) than for the first toss which must be greater than any arrangement of the $k-2$ sample measures between them.
That is soundly reasoned.
There are $(k-2)!$ such favoured arrangement of the $k!$ equally probable ways to theoretically reach that toss.   Therefore the probability of stopping on that toss is $$\mathsf P(N=k)=\dfrac{(k-2)!}{k!}\,\mathbf 1_{k\in[[2..\infty)]}$$
That is the correct answer for the question as given.
A: I think there are two different interpretations of the English language problem description at play here. Maybe you actually reformulated it?
You interpret “any” as “all” here. Note that as has been pointed out, this would mean that strictly speaking the game would always end after the first throw, so you need to make an exception for this.
The other interpretation is that the game ends if there is any previous dart that is nearer to the bullseye than the last dart. With that interpretation the last dart can be any except for the nearest ($k-1$ possibilities), but for the others there is only one possible order, namely descending distance from the center, so this yields the solution that you mention.
