Picking two cards from a box randomly

Question

There are N cards (numbered from 1 to N) in a box. N is a positive integer. We choose two cards randomly, and we divide the the smaller number on the cards with the other number (so the smaller one is in the numerator...). $Z$ denotes the ratio calculated as we discussed before. What is $\mathbb{E}(Z)$?

Here is what I've done so far:

$\mathbb{E}(X_{1})=\mathbb{E}(X_{2})=N/2$ where $X_{1},X_{2}$ are the numbers on the cards.

$\xi=min(X_{1},X_{2})$ and $\eta=max(X_{1},X_{2})$

The question is: $\mathbb{E}(Z)=\mathbb{E}(\frac{\xi}{\eta})$

I think the following is right:

$\mathbb{E}(\xi)+\mathbb{E}(\eta)=\mathbb{E}(X_{1})+\mathbb{E}(X_{2})=N$

What should I do next? I dont even know how to calculate $\mathbb{E}(Z)=\mathbb{E}(\frac{\xi}{\eta})$ easily, because $\xi$ and $\eta$ aren't independent, are they?

Answer 1

What is

$$ \sum_{i =2}^N \frac{1}{i} + 2 \cdot\sum_{i = 3}^N \frac{1}{i} + \ldots + (N - 1) \cdot \sum_{i = N - 1}^N \frac{1}{i}?$$

Because you're going to get one of those $\frac{N \cdot (N - 1)}{2}$ numbers at random with equal probability.

(Hint: how often does each principal fraction occur?)

Answer 2

There are $\frac{N(N-1)}{2}$ equally likely combinations. These add up to

$\frac{1}{2}+\frac{1+2}{3}+...+\frac{1+2+...+(N-1)}{N}=\frac{1}{2}(1+2+...+(N-1))=\frac{N(N-1)}{4}.$

So the expected score is $\frac{1}{2}.$