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
Best Answer
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.