Card Game Strategy

Question

There is a regular deck of 52 playing cards and 3 cards are placed face-down on the table. The cards have values ranging from 1-13. The dealer secretly checks all face-down cards, i.e. you do not get to see them. The dealer then removes the lowest value card, without showing anyone the value, and replaces it with a new face-down card.

My friend was asked this question in an interview recently.

I know the initial expected value is simply 7 * 3 = 21. But now how de we find the new Expected Value conditional on the fact that the lowest value card has been removed and replaced

The interviewer after while gave a hint and mentioned using the Standard Deviation of a single card = $\sqrt 14 $ but I'm unsure what the logic of using std here is and if its correct to use std and how to proceed

Answer 1

Note that the game is equivalent to the following:

You randomly choose $4$ numbers between $1$ and $13$, add them up together, then look at the first $3$ and substract the lowest value.

Let $X$ be a random variable, which returns a value between $1$ and $13$ and let $X_1,X_2,X_3,X_4$ be independent variables of the same distribution as $X$. The result is then $$E\left(\sum_{i=1}^4 X_i -\min_{1\leqslant i\leqslant 3}X_i\right)=\sum_{i=1}^4E(X_i)-E\left(\min_{1\leqslant i\leqslant 3}X_i\right).$$

To calculate the expected value of this mininum, first observe that from independence we get $$\mathbb P\left(\min_{1\leqslant i\leqslant 3}X_i \geqslant t\right)=\mathbb P(X_1\geqslant t)\cdot\mathbb P(X_2\geqslant t)\cdot\mathbb P(X_3\geqslant t)=\left(\frac{14-t}{13}\right)^3,$$ where the last equality holds only for $t\in\{1,\ldots, 14\}$.

For any $Y$ with integer values $$\mathbb P(Y=t)=\mathbb P(Y\geqslant t)-\mathbb P(Y>t)=\mathbb P(Y\geqslant t)-\mathbb P(Y\geqslant t+1).$$

Hence $$E\left(\min_{1\leqslant i\leqslant 3}X_i\right)=\sum_{t=1}^{13}\left(\mathbb P(Y\geqslant t)-\mathbb P(Y\geqslant t+1)\right)\cdot t=$$$$=\sum_{t=1}^{13}\left(\left(\frac{14-t}{13}\right)^3- \left(\frac{13-t}{13}\right)^3\right)\cdot t.$$

I don't know of a connection to standard deviation here.

EDIT.

As noted in the comments the solution above comes from the interpretation that each card is drawn from another deck.

If the first 3 cards are drawn from the same deck, then $X_1$, $X_2$, $X_3$ are not independent and

$$\mathbb P\left(\min_{1\leqslant i\leqslant 3}X_i> t\right)=\frac{{{52-4t}\choose 3}}{{52\choose 3}}$$ for each $t\in\{0,\ldots, 12\}$.

Since it is not specified how the fourth card is chosen, I can't give the final formula.