Why we don't need to cover all possibilities in calculating expected value

Question

We choose 10 cards at random from a standard deck of 52 cards. Find EV of the number of aces.
If we pick a card and then replace it back EV is 10*1/13.
Why the EV is also 10*1/13 when we get an ace and don't put it back , why don't we need to cover all possibilities here?
EV : Expected Value

Answer 1

Expectations are additive , that is for random variables $\ X,Y\ $ we have $$E(X+Y)=E(X)+E(Y)$$

If we define $\ X_j=1\ $ if the $\ j\ $-th card is an ace and $\ 0\ $ otherwise, the sum $$X_1+\cdots +X_{10}$$ is the number of aces. Since every $\ X_i\ $ has expectation $\ \frac{1}{13}\ $, the total expectation is $\ \frac{10}{13}\ $