Drawing marbles from urns and expected values

Question

Suppose you have a variety of coloured marbles, $n_1$ of colour one, $n_2$ of colour two, up to $n_k$ of colour $k$, for a total of $n=n_1+\dotsb+n_k$ marbles. Suppose if you draw marble $i$ you get a payout of $v_i$ (positive or negative). You draw $j$ marbles. Find the expected payout. You can go through the definition and sum probability of a particular draw ($x$) value of that draw. Intuitively, the answer should be sum (vi)(ni/n)(j). Is this true and how do you prove it?

Answer 1

Assuming the payoff $v_i$ depends solely on the colour of the drawn marble and this is fixed we see that the expected payoff for the value of the first draw is $$\sum_{i=1}^k \frac{n_i}{n} v_i$$

Now if all the draws are with replacement it is trivial to prove that we can simply multiply the above answer by the amount of draws done, since they are independent.

If we are sampling without replacement we are essentially looking at the hypergeometric distribution for each colour. Noet that for a hypergeometric distribution the probability of drawing that colour does not change based over the amount of marbles drown. So the probability of drawing a marble of colour $i$ on the first draw is the same as for the fifth marble.

Therefore we once again achieve the same result by seeing that the probability stays at $\frac{n_i}{n}$ for each draw and by linearity the same result comes out as when we are replacing the marbles! (only assumption is that the amount of draws is less or equal than the total amount of marbles in the urn).

I hope this helps you to form a basis on which you can derive the proof yourself. I have intentionally not given you the full proof.