Very silly probability question. I found this question(which I think is quite easy) although it is not that easy at all. 

There are ten cards on a table, numbered from 1 to 10. If you pick the card numbered 10, you win 20\$ but if you pick any other card, you win 0\$. You pick a card. What is the expected amount you will win?

I thought that the answer must be 20\$ but it is not .

For a random variable $X$ that takes values $x_i$ with probability $\mathbb{P}(X=x_i)=p_i$ the expected value $\mathbb{E}[x]$ is defined as
$$
\mathbb{E}[X] = \sum_{i}{x_ip_i}=x_1p_1+x_2p_2+\ldots.
$$
In your case, $X$ takes the values $20$ with probability $1/10$ and $0$ with probability $9/10$. The expected value is thus
$$
\mathbb{E}[X]=20\times \frac{1}{10} + 0\times\frac{9}{10} = 2.
$$
