Translation: A and B play rock-paper-scissors, with the winner determined by each round, playing continuously until one person wins twice in total. What is the probability that the game can be divided, and each has won one round?



Let's break down the problem and solve it.

### Problem Analysis
In this problem, A and B play rounds of rock-paper-scissors, where:
1. They continue playing rounds until one of them wins **two rounds in total**.
2. We want to find the probability that, **before either player wins the two rounds needed to end the game, each player has won exactly one round**.

### Step-by-Step Solution
1. **Define Outcomes for Each Round:** Each round of rock-paper-scissors can result in:
   - A wins the round.
   - B wins the round.
   - The round is a tie (not relevant here because ties do not contribute to the win count).

2. **Define Our Event of Interest:**
   - We want each player, A and B, to have won one round before any player has won two rounds.

3. **Path to Reach the Event:**
   - The only sequence of rounds that allows both players to win once before either reaches two wins is:
     1. A wins the first round.
     2. B wins the second round (or vice versa).

   - After this point (each player has one win), the game would continue until one player wins their second round, which would end the game.

4. **Calculate Probability of Each Scenario:**
   - For any single round, the probability of either A or B winning is 1/2, assuming a fair game.
   - The probability of A winning the first round and B winning the second round (or vice versa) is:
     \[
     P(\text{A wins, then B wins}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}
     \]
     Similarly, the probability of B winning the first round and A winning the second round is also:
     \[
     P(\text{B wins, then A wins}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}
     \]

5. **Combine Probabilities:**
   - Since the two sequences (A wins first, then B; or B wins first, then A) are mutually exclusive and cover all ways for each player to win one round each, we add these probabilities:
     \[
     P(\text{each wins one round}) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}
     \]

### Conclusion
The probability that each player has won exactly one round before either wins a second round (which would end the game) is:
\[
\frac{1}{2}
\]

Would you like more details, or have any questions?

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### Follow-up Questions
1. How would the probability change if ties were allowed and counted as rounds?
2. What if the players needed to win three rounds to end the game? How would that affect the probability?
3. How does the probability change if A has a slight advantage (e.g., A wins each round with a probability of 0.6)?
4. Can we generalize this probability calculation for other games with different win requirements?
5. How would the probability differ if the game allowed for multiple rounds to result in a tie?

### Tip
In probability problems involving sequences of outcomes, it can be helpful to outline all possible paths and then sum their probabilities. This approach works especially well for short sequences.

A and B play rock-paper-scissors, with the winner determined by each round, playing continuously until one person wins twice in total. What is the probability that the game can be divided, and each has won one round?

Explore the probability calculation for a rock-paper-scissors game where A and B play until one wins two rounds. Understand key probability concepts, including the multiplication rule, in determining the likelihood that each player wins one round before the game ends. Suitable for Grades 9-11.

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