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Let's say you have a deck of 4 cards: Ace, King, Queen, Jack. (Worth 40, 30, 20, 10 points respectively)

The deck is shuffled and you are given two choices for each card: to view or to discard. If you view the card, you gain the points the card is worth. If you discard a card without looking, you gain 25 points. What is expected number of points given optimal strategy?

I am trying to do calculate this conditioned upon the first card. If the first card is 40, we can say the average of the remaining cards is 20. Since this is less than discard value we discard the remaining cards for 75 points and get a total of 115. However maybe we would think of this differently for the 30 case. If the first card is worth 30 we could discard the remaining 3 and pocket 105 points or we can say there is a $\frac13$ chance of drawing Ace next and getting 120 points and $\frac23$ of getting 100 points which results in $106.67$ points. This makes it seem like the only way to find the value of this game is to brute force the calculations. What is the best way to go about finding the fair value?

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  • $\begingroup$ Have you tried the 2-card and 3-card case? How would you generalize that approach? $\endgroup$
    – Calvin Lin
    Commented Dec 28, 2023 at 18:22
  • $\begingroup$ Yes, you need brute force. Probably it is more simple if you start by the end. Example if you know that the 2 remaining cards are 10 and 40, (because you started with 20+30 or 30+20), what is the good strategy, and what is the expected number of points. $\endgroup$
    – Lourrran
    Commented Dec 28, 2023 at 18:31

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I will first simplify this problem. We see that this problem is equivalent to $A, K, Q, J$ being worth $3, 1, -1, 3$ points, and discarding giving no points. Observe for the optimal strategy, we will never keep a card once we have discarded a card at some point since the probability distribution for the cards in the deck is the exact same.

We now set up the situation as keeping cards until we decide to stop. Using recursion, we see that the value of a state if we choose to keep the card is the sum of the values of the subsequent states, plus the sum of the cards, divided by the number of cards in the deck currently. If we chose to discard, then the value of the state is $0$.

Hence, we can determine the value of each state, and thus determine the exact optimal strategy. Let $0$ denote the card not being present in the deck, and $1$ denote the card being present. We have \begin{align*} \begin{array}{c|c|c|c|c} A & K & Q & J & \text{Value}\\ \hline 1 & 0 & 0 & 0 & 3\\ 0 & 1 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 1 & 1 & 0 & 0 & 4\\ 1 & 0 & 1 & 0 & \frac{5}{2}\\ 1 & 0 & 0 & 1 & \frac{3}{2}\\ 0 & 1 & 1 & 0 & \frac{1}{2}\\ 0 & 1 & 0 & 1 & 0\\ 0 & 0 & 1 & 1 & 0\\ 1 & 1 & 1 & 0 & \frac{10}{3}\\ 1 & 1 & 0 & 1 & \frac{13}{6}\\ 1 & 0 & 1 & 1 & 1\\ 0 & 1 & 1 & 1 & 0\\ 1 & 1 & 1 & 1 & \frac{13}{8}\\ \end{array} \end{align*} Hence, the value of this simplified game is $\frac{13}{8}$, so the expected number of points in the optimal strategy is \begin{align*} \boxed{\frac{865}{8}} = 108.125 \end{align*} Using the table to calculate the exact optimal strategy, we get the following branches to follow before stopping: \begin{align*} &A,\\ &KA, KQA, KQJA, KJA, KJQA,\\ &QA, QKA, QKJA, QJAK, QJKA,\\ &JAK, JAQK, JKA, JKQA, JQAK, JQKA \end{align*}

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