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Assume you have a deck of 52 cards with the values K = 13, Q = 12, J = 11, 10 = 10,..., 2 = 2 and A = 1. Suppose you draw 3 cards $X_1, X_2$ and $X_3$ at random from the deck of 52 cards.

Then you find out the minimum value from the 3 chosen cards and then remove it from the choice and then uniformly at random choose another card from the remaining deck and put it with your other 2 chosen cards.

What is the Expected Value of the Final 3 chosen cards:

a) The 3rd card is drawn such that the minimum valued card is put back into the deck

b) The 3rd card is drawn such that the minimum valued card is NOT put back into the deck

Source : This was asked to me in a quant trading firm interview, so I am guessing that instead of the brute force method of calculating the PMFs for each scenario, there must be a better method which leverages the linearity of the expectation operator.

My Views

For scenario (a), what I feel is that atleast for the first scenario - the answer should be 7$\times$3 = 21 since even if the minimum card is resampled, it would not affect the other two chosen cards (thus both of them would have an individual expectation of 7) and then since the new card would be chosen from the remaining deck, it would have the same distribution as the minimum valued card But any of the 3 positions in the tuple could have been the minimum value, so essentially all the 3 card choices would have the same distribution, thus the same expectation (= 7) and the answer should be 7$\times$3 = 21.

I'm not at all sure how I would proceed with scenario (b) But what I feel is, if we can get the expected value of the minimum of 3 draws, we can subtract it from 7 and that would most probably be the expected value of the last card drawn.

For this question, I would really appreciate some different viewpoints or entire solutions themselves, thank you!

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  • $\begingroup$ What have you tried? $\endgroup$
    – Math Lover
    Aug 17, 2021 at 6:34
  • $\begingroup$ This was asked to me in a quant trading firm interview, so I am guessing that instead of the brute force method of calculating the PMFs for each scenario, there must be a better method which leverages the linearity of the expectation operator. $\endgroup$ Aug 17, 2021 at 6:40
  • $\begingroup$ Please edit the question to add the source as you mentioned, your thoughts on it, how you would solve it and what you are seeking. As it stands, your question misses context and has a high chance of getting downvoted / closed. Please read how to ask a good question - math.meta.stackexchange.com/questions/9959/… $\endgroup$
    – Math Lover
    Aug 17, 2021 at 6:43
  • $\begingroup$ What I feel is that atleast for the first scenario - the answer should be 7*3 = 21 since even if the minimum card is resampled, it would not affect the other two chosen cards (thus both of them would have an individual expectation of 7) and then since the new card would be chosen from the remaining deck, it would have the same distribution as the minimum valued card But any of the 3 positions in the tuple could have been the minimum value, so essentially all the 3 card choices would have the same distribution, thus the same expectation (= 7) and the answer should be 7*3 = 21. $\endgroup$ Aug 17, 2021 at 6:43
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    $\begingroup$ Just to clarify: are the first three cards drawn with replacement, i.e. you can draw a card more than once? It's not clear from your wording, but your answer tells me they are. $\endgroup$ Aug 17, 2021 at 10:52

2 Answers 2

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Let's say the values of the three cards drawn are $X_1, X_2$ and $X_3$, with order statistics $X_{(1)} \le X_{(2)} \le X_{(3)}$. The expected value of the minimum card drawn is $$E(X_{(1)}) = \sum_{n=0}^{12} P(X_{(1)} > n) = \sum_{n=0}^{12} \frac{\binom{52-4n}{3}}{\binom{52}{3}} = 3.70588$$ After the minimum card is discarded, the remaining two cards have an expected total $$E(X_{(2)} + X_{(3)}) = 3 \cdot 7 - E(X_{(1)}) = 17.2941$$ If the minimum card is put back in the deck, then the expected value of a card in the deck is $$\frac{49 \cdot 7 + E(X_{(1)})}{50} = 6.93412$$

If the minimum card is not put back, then the expected value of a card in the deck is still $7$.

Add the expected value of a card in the deck, with or without replacement of the discarded card, to $E(X_{(2)} + X_{(3)})$ in order to find the expected value of the final three cards in a) or b) as appropriate.

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Hint
Pick three random numbers from the interval $[0,1]$, and sort them. Their expected values are $1/4, 2/4$ and $3/4$.
These cards are uniformly distributed between $0.5$ and $13.5$

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  • $\begingroup$ Are you sure about your first claim? Can you provide a reference? As for your second claim, it is simply false. $\endgroup$
    – TonyK
    Aug 17, 2021 at 21:00
  • $\begingroup$ The pdf of the $m^{th}$ of $n$ numbers is $p(x)={n\choose m-1,1,n-m}x^{m-1}(1-x)^{n-m}$. The mean value is $\int_0^1 xp(x)dx$, and Wolfie says $\int_0^1 x^a(1-x)^bdx=a!b!/(a+b+1)!$ $\endgroup$
    – Empy2
    Aug 18, 2021 at 8:13

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