Players A and B play a sequence of independent games. Player A throws a die first and wins on a 6. If he fails, B throws and wins on a 5 or 6. And so on. Find the probability of each player winning the sequence. 

To solve the problem, we need to determine the probability that Player A wins the sequence and the probability that Player B wins the sequence, given their respective winning conditions:

- **Player A wins if he rolls a 6 on his turn.**
- **Player B wins if he rolls a 5 or 6 on his turn, but only if Player A does not win in the preceding round.**
  
Let's define the following probabilities:
- \( P(A) \): Probability that Player A wins.
- \( P(B) \): Probability that Player B wins.

### Step-by-Step Solution:

1. **Probability of Player A winning on his first roll**:
   - Player A rolls a 6: \( P(\text{A wins on first roll}) = \frac{1}{6} \).

2. **Probability that Player A does not win on his first roll**:
   - Player A does not roll a 6: \( P(\text{A does not win}) = \frac{5}{6} \).

3. **Probability of Player B winning on his first roll**:
   - Player B rolls a 5 or 6: \( P(\text{B wins on first roll}) = \frac{2}{6} = \frac{1}{3} \).

4. **Probability that Player B does not win on his first roll**:
   - Player B rolls something other than 5 or 6: \( P(\text{B does not win}) = \frac{4}{6} = \frac{2}{3} \).

### Recursive Probability Calculations:

Let \( P(A) \) and \( P(B) \) denote the overall probabilities of Player A and Player B winning the sequence, respectively.

**Player A's Probability**:
- Player A can win immediately on his first roll with probability \( \frac{1}{6} \).
- If Player A doesn't win on the first roll, Player B has a chance to win. If Player B doesn't win, the sequence resets to the original state (with Player A rolling first again).

Thus, the probability that Player A eventually wins is:
\[
P(A) = \frac{1}{6} + \left(\frac{5}{6}\right)\left(\frac{2}{3}\right)P(A)
\]

Simplifying this equation:
\[
P(A) = \frac{1}{6} + \frac{10}{18}P(A)
\]
\[
P(A) - \frac{10}{18}P(A) = \frac{1}{6}
\]
\[
\frac{8}{18}P(A) = \frac{1}{6}
\]
\[
P(A) = \frac{1}{6} \times \frac{18}{8} = \frac{3}{8}
\]

**Player B's Probability**:
- Since the probabilities must sum to 1 (either Player A or Player B must eventually win), we have:
\[
P(B) = 1 - P(A) = 1 - \frac{3}{8} = \frac{5}{8}
\]

### Final Probabilities:
- **Probability that Player A wins the sequence**: \( \frac{3}{8} \).
- **Probability that Player B wins the sequence**: \( \frac{5}{8} \).

Would you like further clarification on any part of this solution, or do you have any other questions?

Here are 5 related questions to consider:
1. What would be the probabilities if Player A had to roll a 5 or 6 to win?
2. How would the game change if Player B could only win by rolling a 6?
3. What is the expected number of rounds until a player wins?
4. How does the probability change if Player B rolls first instead of Player A?
5. What is the probability of each player winning if Player A needs a 6, but Player B needs a 4, 5, or 6 to win?

**Tip:** Recursive probabilities often arise in games where outcomes depend on the results of repeated trials. Solving such problems typically involves recognizing the recursive structure and setting up the appropriate equation.

Explore the probability of each player winning a sequence of games where Player A wins on a 6 and Player B wins on a 5 or 6 if Player A fails. Understand the recursive probability calculations involved in determining the outcomes. Suitable for high school students studying probability theory.

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