You have been chosen to play a game involving a 6-sided die. You get to roll the die once, see the results and may choose to stop or roll one more time. Your payoff is the sum of your rolls, unless the sum is greater than 7, in which case you “bust” and get nothing. What is your strategy for this game? This is, for each possible outcome at the first roll will you choose to stop or roll again. Why?

To develop a strategy for this game, we need to consider the possible outcomes of the first roll and decide whether it's better to stop or roll again based on the expected payoff.

### Rules Recap:
- You roll a 6-sided die.
- You can stop after the first roll or roll again.
- The payoff is the sum of the rolls unless the sum exceeds 7, in which case the payoff is 0 (bust).

### Strategy Analysis:
We'll consider each possible outcome of the first roll (1 to 6) and decide whether to stop or roll again.

#### If You Roll a 1:
- **Stop**: Payoff = 1.
- **Roll Again**: The second roll can result in:
  - 1 (Payoff = 2), 2 (Payoff = 3), 3 (Payoff = 4), 4 (Payoff = 5), 5 (Payoff = 6), 6 (Payoff = 7)
  - **Expected Value**: \(\frac{1}{6} \times (2 + 3 + 4 + 5 + 6 + 7) = \frac{27}{6} = 4.5\)
  - Since 4.5 > 1, **roll again**.

#### If You Roll a 2:
- **Stop**: Payoff = 2.
- **Roll Again**: The second roll can result in:
  - 1 (Payoff = 3), 2 (Payoff = 4), 3 (Payoff = 5), 4 (Payoff = 6), 5 (Payoff = 7), 6 (Bust)
  - **Expected Value**: \(\frac{1}{6} \times (3 + 4 + 5 + 6 + 7 + 0) = \frac{25}{6} \approx 4.17\)
  - Since 4.17 > 2, **roll again**.

#### If You Roll a 3:
- **Stop**: Payoff = 3.
- **Roll Again**: The second roll can result in:
  - 1 (Payoff = 4), 2 (Payoff = 5), 3 (Payoff = 6), 4 (Payoff = 7), 5 or 6 (Bust)
  - **Expected Value**: \(\frac{1}{6} \times (4 + 5 + 6 + 7 + 0 + 0) = \frac{22}{6} \approx 3.67\)
  - Since 3.67 > 3, **roll again**.

#### If You Roll a 4:
- **Stop**: Payoff = 4.
- **Roll Again**: The second roll can result in:
  - 1 (Payoff = 5), 2 (Payoff = 6), 3 (Payoff = 7), 4, 5, or 6 (Bust)
  - **Expected Value**: \(\frac{1}{6} \times (5 + 6 + 7 + 0 + 0 + 0) = \frac{18}{6} = 3\)
  - Since 4 > 3, **stop**.

#### If You Roll a 5:
- **Stop**: Payoff = 5.
- **Roll Again**: The second roll can result in:
  - 1 (Payoff = 6), 2 (Payoff = 7), 3, 4, 5, or 6 (Bust)
  - **Expected Value**: \(\frac{1}{6} \times (6 + 7 + 0 + 0 + 0 + 0) = \frac{13}{6} \approx 2.17\)
  - Since 5 > 2.17, **stop**.

#### If You Roll a 6:
- **Stop**: Payoff = 6.
- **Roll Again**: The second roll can result in:
  - 1 (Payoff = 7), 2, 3, 4, 5, or 6 (Bust)
  - **Expected Value**: \(\frac{1}{6} \times (7 + 0 + 0 + 0 + 0 + 0) = \frac{7}{6} \approx 1.17\)
  - Since 6 > 1.17, **stop**.

### Summary of Strategy:
- If you roll a **1, 2, or 3**, **roll again**.
- If you roll a **4, 5, or 6**, **stop**.

### Explanation:
The strategy is based on maximizing your expected payoff. For rolls of 1, 2, or 3, the potential gain from rolling again outweighs the risk of busting. For rolls of 4, 5, or 6, the risk of busting makes it more advantageous to stop and take the guaranteed higher payoff.

Do you want any more details, or have any questions?

Here are 5 related questions:
1. How would the strategy change if the die had more sides (e.g., an 8-sided die)?
2. What if the bust threshold was higher, say 9? How would that affect the strategy?
3. How do you calculate the expected value for different game strategies?
4. How would your strategy change if you were allowed three rolls instead of two?
5. What if the bust condition also applied if you rolled a total of 7?

**Tip:** Always compare the expected value of taking a risk with the guaranteed payoff to make the optimal decision in probabilistic scenarios.

Discover the optimal strategy for a game involving a 6-sided die where you decide whether to stop or roll again based on your first roll. Learn how to maximize your expected payoff and avoid busting with a detailed analysis of each possible outcome.

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