Jason throws $n$ darts at a dartboard, aiming for the center. The second dart lands farther from the center than the first, and so does the third, the fourth, ..., the n-1 th dart. If Jason throws the nth dart aiming for the center, what is the probability that the nth throw is further from the center than the first? Assume Jason's skillfulness is constant.
The answer in this post reasons as follows
$$\begin{align}&Prob(D_n > D_1 | D_2 > D_1, D_3 > D_1, \dots, D_{n-1} > D_1)\\&=Prob(D_n > D_1 | D_1 \text{ is the best among $(n-1)$ darts})\\&=1-Prob(D_n < D_1 | D_1 \text{ is the best among $(n-1)$ darts})\\&\underbrace{=}_{\text{HP: $D_i$ i.i.d.}}1-Prob(D_n \text{ is the best among $n$ darts})\end{align}$$
$$Prob(D_n \text{ is the best among $n$ darts}) \underbrace{\equiv}_{\text{HP: $D_i$ i.i.d.}} Prob(\text{Picking an object among $n$ available})=\frac{\text{# favorable outcomes}}{\text{# total outcomes}}= \frac{1}{n}$$
It is not clear to me why $$1-Prob(D_n < D_1 | D_1 \text{ is the best among $(n-1)$ darts})\underbrace{=}_{\text{HP: $D_i$ i.i.d.}}1-Prob(D_n \text{ is the best among $n$ darts})$$
1). Is it true that $Prob(D_n \text{ is the best among $n$ darts}) = Prob(D_n < D_1, D_n < D_2, \ldots, D_n < D_{n-1})$?
2). If one is true, why can't we jsut conclude that $Prob(D_n > D_1|D_2 > D_1, D_3>D_1, \ldots, D_{n-1} > D_1) = Prob(D_1 \text{is the best among the n darts})$?
3). Why is $Prob(D_n \text{is the best among n darts}) = \frac{1}{n}$
I may be interpreting the problem incorrectly.
Assuming that I am interpreting the problem correctly:
The problem can be re-stated as follows:
There are $n$ darts thrown at the dart board. What is the probability that the last dart thrown is the closest to the center?
It is to be assumed that the probability of any two darts being the exact same distance away from the center is $(0)$.
Presumably, the probability in the re-stated problem is $(1/n)$.
So, the real question is:
Why is the restated problem equivalent to the given problem?
In the first $(n-1)$ throws, before the $n$th dart was thrown, one of the previous throws had to be the closest. Whichever dart happened to be the closest should have no effect on whether the $n$th dart thrown is closer than all of the others.
Therefore, you can assume, without loss of generality, that the closest dart thrown, among the first $(n-1)$ darts is the first dart thrown.
Therefore,
the $n$th dart thrown will be closer than all of the other darts thrown
if and only if
the $n$th dart thrown is closer than the first dart thrown.
Therefore, the probability that the $n$th dart is farther from the center than the first dart thrown is $1 - (1/n).$
An alternative demonstration is:
Consider the $(n!)$ ways of permuting the (all distinct) elements of the set $\{x_1, x_2, \cdots, x_n\}$. Exactly $(n-1)!$ of these permutations will have the first element in the permutation being $x_n$.
Therefore, the probability that $x_n$ is the first element is $\dfrac{(n-1)!}{n!} = \dfrac{1}{n}.$
Now, consider the subset of permutations where $x_1$ proceeds all of the other elements, besides $x_n$.
This means that the pertinent subset of permutations may be represented visually as
$$\_ ~x_1 ~\_ ~x_{i_1} ~\_ ~x_{i_2} ~\_ \cdots ~\_ ~x_{i_{(n-2)}} ~\_ \tag1 $$
where $\{x_{i_1}, x_{i_2}, \cdots, x_{i_{(n-2)}}\}$ represent $\{x_2, x_3, \cdots, x_{n-1}\}$, in some order.
The elements $\{x_2, x_3, \cdots, x_{n-1}\}$ can be permuted in $(n-2)!$ ways.
Further, in (1) above, there are $n$ choices for which position will be taken by $x_n$.
Therefore, the total number of permutations where $x_1$ precedes $x_2, x_3, \cdots, x_{n-2}$ is
$n \times (n-2)!.$
Of these, the number of such permutations where $x_n$ specifically goes into the first spot is $(n-2)!$.
Therefore, if you are given that $x_1$ precedes $x_2, x_3, \cdots, x_{n-1}$, then the probability that $x_n$ comes first is
$\dfrac{(n-2)!}{n \times (n-2)!}.$