The question is as follows:

Draw 4 cards from a deck of 54 cards (with 2 jokers), what is the expected value of the 4 cards? Assume the value of a J is 11, Q is 12 and K is 13, a joker is 100.

The answer provided is:

We need to compute:

$ \quad \mathbb{E}(X_1+X_2+X_3+X_4) = \mathbb{E}(X_1)+ \mathbb{E}(X_2)+ \mathbb{E}(X_3)+ \mathbb{E}(X_4)$ = $4\mathbb{E}(X_1)$

My question is why do we have $\quad \mathbb{E}(X_1)=\mathbb{E}(X_2)=\mathbb{E}(X_3)=\mathbb{E}(X_4)$?

If we draw one card each time and with replacement, we draw the next one, I can understand the equation. But if we draw four times consecutively without replacement, do we still have the equal expectations for four times? Why? Thanks!


2 Answers 2


First I would like to give you the intuitive answer the your specific question than a very formal one. This is a great question!

  • Intuition: If you draw a card from the deck and don't look at it and then draw another card and do look at it, it should be clear that each of the 54 cards were equally likely to be that second card, in the same way they were equally likely to be the first.

  • Formal: You are totally right that $X_1$ and $X_2$ are not independent. I think this is what is bugging you. The reason that doesn't matter here is actually all of the summing that is going on. Here I will do the computation for 2 cards to let you see why this summing matters. The summations are over cards $i, j$ and $v(i)$ equals the value of card $i$.

\begin{eqnarray} \mathbb{E}[X_1 + X_2] &=& \sum_i \sum_j P(X_1=i, X_2=j)(v(i) + v(j)) \\ &=& \left[ \sum_i\sum_j P(X_1=i, X_2=j)v(i) \right] + \left[ \sum_i\sum_j P(X_1=i, X_2=j)v(j) \right] \\ &=& \left[ \sum_i P(X_1=i)v(i) \right] + \left[ \sum_j P(X_2=j)v(j) \right]\\ &=& \mathbb{E}[X_1] + \mathbb{E}[X_2]. \end{eqnarray}

So because we're summing these RVs the sum can be broken up and only the marginal distributions of the random variables matter. So the fact that $P(X_2|X_1) \neq P(X_2)$ does not matter. Does this make sense?


Say you draw one card and look at it and it's the Queen of Hearts. Now you know that the odds of drawing a Queen for the second card are a little bit lower, but only for that sample. When you calculate the expected value for subsequent cards, you have to take into account that any of the 54 cards could have been chosen first.

If you draw a million samples of 4 cards from the deck, the odds of the Queen of Hearts appearing in the first spot are the same as appearing in the second, third, and fourth. The odds of any card appearing in any spot are the same.


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