Expectation value of sum of 3 poker cards (variation)

Question

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!

Answer 1

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.

Answer 2

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 $13.5$