Probablity that the drawn number is greater than the previously drawn We've got $n$ different real numbers that are equally likely to be the maximum. We sample without replacement. Let $A_k$ be the event that $k^{th}$ drawn number is greater that all previously drawn. How to prove that $P(A_k)=\frac{1}{k}$?
I would appreciate any help. 
 A: Non-Rigorous/Intuitive Reasoning
$A_k$ is the event that the $k^{th}$ number is greater than all the numbers previously drawn. Assume that you are at time $n$.
That implies that you just drawed the $n^{th}$ number from the pool. Now, every number is equally likely to be the highest. Thus, there are n possible "highest numbers". The probability that the $n^{th}$ is highest is thus, $\dfrac{1}{n}$.
This can be generalized. The probability that the $k^{th}$ number drawn is the highest is thus, $\dfrac{1}{k}$
A: Let $H_i$ be the history of the first $i$ numbers. Define $x_k$ such that $x_k=1$ if the value of the $k^{th}$ number is greater than the other $k-1$ numbers prior to that and $x_k=0$ else. Use Bayes' theorem as follows.
$$P(x_k=1 \mid H_{k-1})=\frac {P(H_{k-1}\mid x_k=1)P(x_k=1) }{P( H_{k-1})}=P(x_k=1)=\frac{1}{k}$$
We have made use of the fact that the history after observing $k-1$ numbers is independent of the $k^{th}$ number (i.e., $P(H_{k-1}\mid x_k=1)=P(H_{k-1}))$. The last equality is because all $k$ observed numbers are equally likely to be the maximum.
