Recompute expected value of the sum of value of five face down cards when the lowest value card is replaced by a new face down card Recently I heard of this puzzle/riddle type question that was asked in an application interview and I am unable to solve it. The problem is as follows

Stage 1: The interviewer holds a regular deck of 52 playing cards, excluding jokers, and places 5 cards face-down on the table. The cards have values ranging from $1$ through $13$ where an ace has value $1$ and a king has value $13$, etc. The interviewee is asked to give the expected value of the sum of the values of the face-down cards on the table.
Stage 2: Same setting as in stage $1$ but this time the interviewer secretly checks all face-down cards, i.e. the interviewee does not get to see them. The interviewer then removes the lowest value card, without showing anyone the value, and replaces it with a new face-down card. The interviewee is then asked to give the new expected value of the sum of the values of the remaining cards.

So, stage $1$ is simple. The expected value of a face-down playing card is $7$, therefore the expected value of the sum of the values of $5$ face-down cards, is $7*5=35$.
However, stage $2$ seems too difficult to solve due to lack of information. I understand that the expected value of the sum of the values of the $5$ face-down cards goes up, but how can one know by how much? I would say that the expected value is still approximately $35$, but this turned out to be incorrect according to the interviewer. The hint the interviewer gave was that one can use the standard deviation of one face-down playing card, which to my knowledge is $\sqrt{14}$. Note that you are supposed to solve this problem without difficult computation because you do not have paper or a calculator at hand.
Any help is appreciated!
 A: I do not understand the hint that was given; I feel that any argument involving standard deviation will not give an exact answer since this is a discrete problem. But I might be wrong. The solution I give below requires a calculator, but I may be overlooking some more elegant argument.

Let $X_1, \ldots, X_5$ be the values of the five cards. You have shown $E[X_1 + \cdots + X_5] = 35$.
You are asked to compute $E[X_1 + \cdots + X_5 - \min_i X_i + X_6]$ where $X_6$ is drawn from the remaining $52-5=47$ cards. It suffices to compute $E[\min_i X_i]$ and subtract this expectation from $E[X_1 + \cdots + X_5 + X_6] = 6 \cdot 7 = 42$.
By the tail sum formula for expectation,
$$E[\min_i X_i] = \sum_{j=1}^{13} P(\min_i X_i \ge j)
\color{red}{\approx} \sum_{j=1}^{13} \left(\frac{14-j}{13}\right)^5 \approx 2.7$$
where I have used a calculator to compute the last sum.
Edit: As mentioned in the comments, $P(\min_i X_i \ge j)$ does not equal $\left(\frac{14-j}{13}\right)^5$ since the first five cards are drawn without replacement, but it is a close approximation. The exact probability is $\binom{4(14-j)}{5} / \binom{52}{5}$, and plugging that into the above sum yields $\approx 2.6$.
