Problem 1236

1. The Setup:
- We are given a sequence of random variables $X_1, X_2, \ldots, X_n$ representing the outcomes of rolling a dice $n$ times.
- We define another sequence $Y_1, Y_2, \ldots, Y_n$ by:
$$Y_1 = X_1, \quad Y_k = X_k + \frac{1}{Y_{k-1}} \quad \text{for} \quad k = 2, \ldots, n$$
- We need to find the probability $p_n$ such that:
$$\frac{1 + \sqrt{3}}{2} \leq Y_n \leq 1 + \sqrt{3}$$

2. Initial Observations:
- Fact 1: $Y_k \geq 1$ for all $k$. This is because the minimum value of $X_k$ is 1 and $\frac{1}{Y_{k-1}} \geq 0$.
- Fact 2: $Y_n$ is never irrational. This follows from the fact that $X_k$ are integers and $\frac{1}{Y_{k-1}}$ is a rational number.

3. Probability Analysis:
- For $\frac{1 + \sqrt{3}}{2} \leq Y_n \leq 1 + \sqrt{3}$ to hold, $X_n$ must be either 1 or 2.
- Denote $q_n$ as the probability that $1 \leq Y_n \leq \frac{1 + \sqrt{3}}{2}$.

4. Case Analysis:
- **Case 1: $X_n = 1$**
- We need:
$$\frac{\sqrt{3} - 1}{2} \leq \frac{1}{Y_{n-1}} \leq \sqrt{3}$$
- Flipping the inequalities, we get:
$$\frac{1}{\sqrt{3}} \leq Y_{n-1} \leq 1 + \sqrt{3}$$
- The probability that $Y_{n-1}$ lies in this range is $p_{n-1} + q_{n-1}$.
- Therefore, the probability for this case is:
$$\frac{1}{6} (p_{n-1} + q_{n-1})$$

- **Case 2: $X_n = 2$**
- We need:
$$\frac{1}{Y_{n-1}} \leq \sqrt{3} - 1$$
- Flipping the inequality, we get:
$$Y_{n-1} \leq \frac{1 + \sqrt{3}}{2}$$
- The probability for this case is:
$$\frac{1}{6} (1 - q_{n-1})$$

5. Combining Cases:
- The total probability $p_n$ is given by:
$$p_n = \frac{1}{6} (p_{n-1} + q_{n-1}) + \frac{1}{6} (1 - q_{n-1})$$
- Simplifying, we get:
$$p_n = \frac{1}{6} (1 + p_{n-1})$$

6. Initial Condition:
- For $n = 1$, $Y_1 = X_1$. The probability that $X_1$ lies in the range $\frac{1 + \sqrt{3}}{2} \leq X_1 \leq 1 + \sqrt{3}$ is:
$$p_1 = \frac{1}{6}$$

7. Solving the Recurrence:
- The recurrence relation is:
$$p_n = \frac{1}{6} (1 + p_{n-1})$$
- This can be solved iteratively or using standard methods for solving linear recurrence relations.

The final answer is $\boxed{ p_n = \frac{1}{6} (1 + p_{n-1}) }$ with $p_1 = \frac{1}{6}$.