Expected value of drawing three cards with replacement

Question

You have a standard deck of cards. Each card is worth its face value (A=$1$, K=$13$).

What is the expected value (EV) of drawing three cards with replacement (cards are placed back into the deck after each being drawn)?

Here is my attempt:

The expected value of an event is the average of the values of each outcome.

In this case, the outcome is drawing a card and the value of the outcome is the value of the card. Since the cards are replaced, each outcome has the same probability. The probability of drawing a card with value $x$ is 1/13​.

Therefore, the expected value of drawing one card with replacement is:

$E = \frac{1}{13}(1 + 2 + 3 + ... + 13) = 7$

To find the expected value (EV) of drawing three cards with replacement, can I just multiply by $3$?

so $EV(3 cards) = 7 * 3 = 21$

Answer 1

Yes , you can. One way to see this is let $Χ_1,Χ_2,Χ_3$ be the random variable that is the outcome of the $i$ draw.

Clearly in your case if $X$ is the total outcome:

$$ X = X_1+X_2+X_3 $$

and thus:

$$ \mathbb{E}[X] = \mathbb{E}[X_1]+\mathbb{E}[X_2]+\mathbb{E}[X_3] = 3 \cdot 7 = 21 $$

in a more intuition based manner, you do a random experiment 3 times, with the same conditions, so it is only natural that the total EV is just the sum of the EV's of each event and as you have the same conditions these must all be the same (This is where the replacement comes in, if you did not do the replacement you cannot do this) .