Suppose I have a deck of cards. This deck of cards has an infinite amount of cards that are labeled 1,2,3,4,5. You can assume there is an infinite amount of each card. The game works as follows: I pull two cards from this deck. You then must choose one of the two cards to add to your score (You can see the cards). Your score starts as 0, and we will play 3 rounds of this game. If your score ends up becoming ≥ 7, you have “busted” and your score will become and remain 0 for the rest of the game.
a. How would you play this game if your goal is to maximize your score?
b. What if we don’t know the distribution of the deck the cards are pulled from?
c. What if cards can be negative too?
d. What if there is one of each card, but they are replaced after each round?
e. What if you play an opponent and you get paid 10 if your score is strictly greater, otherwise nothing?
f. What if you can wager before each turn? When do you wager?
g. What if you can wager for the game against an opponent and you know their cards? What about if you don’t? What about if you know if they have wagered?
This was from an interview. For a strategy for the first pick, I thought that if the maximum of the two cards was less than 3, I'd pick the bigger one. If the maximum was greater than 3, I'd pick the smaller one. I calculated this by finding E(L), which would be the expected value of the lower card, and I found E(L) by summing up E(L | H = i)P(H = i) for all i, 1-5, where H is the value of the higher card. However, after calculating the expected value of the first pick using this strategy, I get that it's 3.8, and E(L) is 2.2, which means that you are always expected to lose, which doesn't sound right. I arrived at this strategy by reasoning that since your expected score over the last 2 picks will be 4.4, 1 or 2 will always be a safe choice for the first pick. However, if your max is greater than or equal to 3, then you would want to pick the smaller one to decrease your odds of busting as much as possible.
I got more lost on the followups, in part due to not understanding wording, i.e. for e), wouldn't your strategy just remain the exact same?
Currently the question is a tad ambiguous with its wording, so I will assume the game is identical to being given two $5$-sided dice with numbers $1$, $2$, $3$, $4$, $5$, which are assumed to be uniform for (a).
For (a), the key observation is that on the last round, if your score is $\leq n$ for some $n \in \{2, 3, 4, 5\}$, the probability you can choose a number that gets you to a score of $m$ for some $n < m \leq 6$ is always $\frac{2m - 3}{25}$ regardless of $n$. Hence, your expected score increases with a smaller score at the end of round 2. Thus, the optimal strategy is to always choose minimum number for round 1 and round 2, and choose the largest number keeping you under the threshold for round 3.
Calculating the probabilities and expected values explicitly:
Let $E_n$ denote the expected score after ending Round 2 on a score of $n$. Then \begin{align*} E_5 &= \frac{9}{25} \cdot 6 = \frac{54}{25}\\ E_4 &= \frac{9}{25} \cdot 6 + \frac{7}{25} \cdot 5 = \frac{89}{25}\\ E_3 &= \frac{9}{25} \cdot 6 + \frac{7}{25} \cdot 5 + \frac{5}{25} \cdot 4 = \frac{109}{25}\\ E_2 &= \frac{9}{25} \cdot 6 + \frac{7}{25} \cdot 5 + \frac{5}{25} \cdot 4 + \frac{3}{25} \cdot 3 = \frac{118}{25} \end{align*} Let $p_n$ denote the probability of ending round $2$ on a score of $n$, utilising the minimisation strategy: \begin{align*} p_2 &= \frac{9}{25}\cdot \frac{9}{25} = \frac{81}{625}\\ p_3 &= 2\cdot \frac{9}{25} \cdot \frac{7}{25} = \frac{126}{625}\\ p_4 &= 2\cdot \frac{9}{25} \cdot \frac{5}{25} + \frac{7}{25} \cdot \frac{7}{25} = \frac{139}{625}\\ p_5 &= 2\cdot \frac{9}{25} \cdot \frac{3}{25} + 2\cdot \frac{7}{25} \cdot \frac{5}{25} = \frac{124}{625} \end{align*} This gives overall expected score of \begin{align*} E &= \sum_{i = 2}^5 p_i E_i = \frac{42359}{15625} \approx 2.711 \end{align*}
For (b), if we don't know the distribution, I would expect this strategy to still be optimal since you maximise your chances of not disqualifying.
For (c), I would need more information regarding what the negative card values are but it means there are a lot more of the final round cases left to check, so its possible that "smaller score = better" idea doesn't hold anymore.
For (d), the key observation holds true again, except the probability is $\frac{m - 2}{10}$. Hence, the same strategy holds but the probabilities and expected values change: Let $E_n$ denote the expected score after ending Round 2 on a score of $n$. Then \begin{align*} E_5 &= \frac{4}{10} \cdot 6 = \frac{12}{5}\\ E_4 &= \frac{4}{10} \cdot 6 + \frac{3}{10} \cdot 5 = \frac{39}{10}\\ E_3 &= \frac{4}{10} \cdot 6 + \frac{3}{10} \cdot 5 + \frac{2}{10} \cdot 4 = \frac{47}{10}\\ E_2 &= \frac{4}{10} \cdot 6 + \frac{3}{10} \cdot 5 + \frac{2}{10} \cdot 4 + \frac{1}{10} \cdot 3 = 5\\ p_2 &= \frac{4}{10}\cdot \frac{4}{10} = \frac{4}{25}\\ p_3 &= 2\cdot \frac{4}{10} \cdot \frac{3}{10} = \frac{6}{25}\\ p_4 &= 2\cdot \frac{4}{10} \cdot \frac{2}{10} + \frac{3}{10} \cdot \frac{3}{10} = \frac{1}{4}\\ p_5 &= 2\cdot \frac{4}{10} \cdot \frac{1}{10} + 2\cdot \frac{3}{10} \cdot \frac{2}{10} = \frac{1}{5}\\ E &= \sum_{i = 2}^5 p_i E_i = \frac{3383}{1000} = 3.383 \end{align*}
For (e), we see that the (a) strategy should still be able to maximise your score, and you can't really do better than the strategy in (a), so I would keep with the strategy.
For (f), you can use Kelly Criterion to size your bets after round 2, so you should be making wagers if you end round 2 on $2, 3,$ or $4$ (assuming $1:1$ betting). You can do similar for the end of round 1 and before the start of the game.
For (g), knowing their cards changes the probabilities of victory but you can still use the Kelly Criterion to size the bets. If you don't know their cards, the best you can do is basically the same as (f), but now you account for probabilities of winning as well. If you know what they wager, then that again changes your betting strategy since the bet sizes gives you information on their cards (assuming they are a rational agent, and thus will also use the Kelly Criterion).