Susan and Brian are sitting down to play a friendly game. Each time they play Susan might win or Brian might win. They agree to repeatedly play the game until one of the following happens: Brian wins 2 games, Susan wins 3 games in a row, or 4 games have been played. How many different sequences of game results are possible?


To solve this problem, we need to count how many different sequences of game results are possible, given the conditions: 

1. **Brian wins 2 games.**
2. **Susan wins 3 games in a row.**
3. **A maximum of 4 games are played.**

Let's analyze this step by step.

### Step 1: Possible game outcomes
Each game can have two possible outcomes: either Brian (denoted as \(B\)) wins, or Susan (denoted as \(S\)) wins. We need to track the sequence of outcomes over multiple games, but there are constraints. 

### Step 2: Condition constraints
- **If Brian wins 2 games**, the sequence ends early.
- **If Susan wins 3 games in a row**, the sequence also ends early.
- **The maximum number of games is 4**, so after 4 games, the sequence stops if neither of the first two conditions is met.

### Step 3: List all valid sequences

#### Case 1: Brian wins 2 games
If Brian wins 2 games, the sequence must stop. Let's consider all possible sequences where Brian wins exactly 2 games:

- \( B B \) (in 2 games)
- \( S B B \) (in 3 games)
- \( B S B \) (in 3 games)
- \( S S B B \) (in 4 games)
- \( S B S B \) (in 4 games)
- \( B S S B \) (in 4 games)

These sequences total 6.

#### Case 2: Susan wins 3 games in a row
If Susan wins 3 games in a row, the sequence stops:

- \( S S S \) (in 3 games)

This is 1 sequence.

#### Case 3: 4 games are played without ending early
These are sequences where the game lasts all 4 rounds, without Brian winning twice and without Susan winning 3 in a row:

- \( S S S S \)
- \( S S B S \)
- \( S B S S \)
- \( S B B S \)
- \( B S S S \)
- \( B S B S \)
- \( B B S S \)

This gives 7 more sequences.

### Step 4: Total number of sequences
Now, we sum all the valid sequences:

- **Brian wins 2 games**: 6 sequences.
- **Susan wins 3 in a row**: 1 sequence.
- **4 games are played**: 7 sequences.

Thus, the total number of different sequences of game results is:
\[
6 + 1 + 7 = 14
\]

### Final Answer:
There are 14 different sequences of game results.

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Would you like any further details or clarifications?

Here are five related questions:
1. What would happen if Brian needed to win 3 games instead of 2?
2. How would the result change if Susan didn't need to win in a row?
3. What if the game allowed a maximum of 5 rounds instead of 4?
4. How can the sequence outcomes be visualized using a tree diagram?
5. Can this problem be solved using probability techniques?

**Tip:** When dealing with constrained sequences, always track early stopping conditions to avoid overcounting longer sequences.

Susan and Brian are sitting down to play a friendly game. Each time they play Susan might win or Brian might win. They agree to repeatedly play the game until one of the following happens: Brian wins 2 games, Susan wins 3 games in a row, or 4 games have been played. How many different sequences of game results are possible?

Explore the solution to a combinatorics problem where Susan and Brian play a game with constraints. We count how many different sequences of game outcomes are possible given conditions like Brian winning twice, Susan winning three times in a row, or the game lasting a maximum of four rounds. Suitable for Grades 9-12.

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