Probabilty that none of n i.i.d uniform samples from R^2 are larger in both coordinates than first. I am randomly sampling n elements from a square within R^2. They are independently uniformly distributed. What is the probability that for any arbitrarily but fixed previously selected sample index, it is true that no other sample is greater in both coordinates? I do not know anything else about the preselected sample.
This is part of a larger problem which I have reduced to this question, I can explain it should it be neccessary.
 A: When has Wolfram Alpha been wrong?  :) Yes the answer is $H_n / n$ where $H_n$ is the $n$th Harmonic number.
(I am curious how you asked WA.  Did you type in the integral?  Do you mind sharing the link?)
Consider all $n$ different $x$-coordinates of the $n$ points.  Since there is perfect symmetry between the points, your specific point $P = (x_P, y_P)$ is equally likely to have the smallest $x$, or second smallest, or third smallest, etc., up to being the largest.  I.e., let $M$ be the number of points whose $x$-coordinate exceeds $x_P$, then $M$ is uniform among $\{0, 1, 2, \dots, n-1\}$, each choice with probability $1/n$.
Now, when will point $P$ meet your criterion?  Consider the set containing $P$ and all $M$ points whose $x$-coordinate exceeds $x_P$ (the other points don't matter any more).  Among this set of $M+1$ points, point $P$ must have the highest $y$-coordinate for $P$ to meet your criterion.  Due to symmetry between these $M+1$ points, the probability is $1/(M+1)$.  (Here we used the fact $x$ and $y$ coordinates are independent.)  Therefore:
$$Prob(P \text{ meets your criterion}) = \sum_{m=0}^{n-1} \frac1{m+1} Prob(M=m) = \frac1n \sum_{m=0}^{n-1} \frac1{m+1}  = {H_n \over n}$$ 
And you are right: linearity of expectation implies that the expected number of points meeting your criterion (i.e. which you need to store) is $H_n$.
