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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!

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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 $\approx 2.6$.

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  • $\begingroup$ Do you mean $X_1+\cdots +X_6$ where $X_6$ denotes the value of the card which replaces the minimum of the first $5$? $\endgroup$ – lulu Feb 13 at 20:14
  • $\begingroup$ @lulu I misread the problem, I did not realize there was a replacement card. I'll edit my answer. $\endgroup$ – angryavian Feb 13 at 20:17
  • $\begingroup$ @angryavian That looks good, thanks alot. However, as you said it requires a calculator. I upvoted you and unless someone posts a solution that can be computed without use of calculator of paper, I'll mark your answer as accepted! $\endgroup$ – S. Crim Feb 13 at 20:28
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    $\begingroup$ @S.Crim The "min" appears because you state that we are removing the smallest card among the 5. The event "min $\ge j$" is the same as "all 5 cards are $\ge j$," which equals $(\frac{14-j}{13})^5$ if the cards are drawn with replacement (which isn't the case here), and $\binom{4(14-j)}{5}/\binom{52}{5}$ if the cards are drawn without replacement. $\endgroup$ – angryavian Feb 15 at 18:07
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    $\begingroup$ @S.Crim Your problem says nothing about repeating the procedure, but if you want to think about that you have to tell us whether you will replace any removed cards in the deck, or throw them away permanently. In either case I don't see why the expectation increases to infinity, since the sum of 5 cards is bounded by 64. In the situation where you replace removed cards back into the deck, I think $13 \cdot 4 + 7 = 59$ is the limiting expectation. $\endgroup$ – angryavian Feb 15 at 18:10

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