A, B, C, and D play dice games where the loser doubles the amount of money of all other players. Four games are played in all, and A, B, C, and D lose these games in this sequence only. Each player has Rs. 96 at the end of the fourth game. What is the amount with C at the end of the third game?

Let's solve the problem step by step.

### Understanding the Problem
- **Initial Condition:** All players have an equal amount of money initially. Let's denote this amount by \( x \).
- **Game Rules:** 
  - The loser doubles the amount of money of all other players.
  - Four games are played, and the losers are A, B, C, and D, respectively.
- **Final Condition:** Each player has Rs. 96 at the end of the fourth game.

### Step-by-Step Calculation

#### After Game 1:
- **Loser:** A
- **Other Players' Money Doubles:** 
  - If A loses, B, C, and D's money doubles.
- **Equations:**
  \[
  2(B + C + D) = 3x \quad \text{(since B, C, and D's money doubles)}
  \]
  A's money is \( x - 3x = -2x \) after doubling others' money, but A had to match the others' gain, so A would end up with the same amount as each other player.

So after the first game, the amount with each player will be:

\[
A = x - 3x + x = x \quad \text{(due to matching others' gain)}, \quad B = 2x, \quad C = 2x, \quad D = 2x.
\]

#### After Game 2:
- **Loser:** B
- **Other Players' Money Doubles:** 
  - A, C, and D’s money doubles.
- **Equations:**
  \[
  A = 2x, \quad C = 4x, \quad D = 4x, \quad B = \frac{x}{2}
  \]

#### After Game 3:
- **Loser:** C
- **Other Players' Money Doubles:** 
  - A, B, and D’s money doubles.
- **Equations:**
  \[
  A = 4x, \quad B = 2x, \quad D = 8x, \quad C = \frac{x}{4}
  \]

#### After Game 4:
- **Loser:** D
- **Other Players' Money Doubles:** 
  - A, B, and C’s money doubles.
- **Equations:**
  \[
  A = 8x, \quad B = 4x, \quad C = \frac{x}{2}, \quad D = \frac{x}{8}
  \]

Given that after the fourth game, each player has Rs. 96:

\[
8x = 96 \implies x = 12
\]

### Calculation of Amount with C After the Third Game
After the third game, the amount with C was:

\[
\frac{x}{4} = \frac{12}{4} = 3
\]

Thus, after the third game, C had Rs. 3.

### Answer:
The amount with C at the end of the third game was **Rs. 3**.

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**Related Questions:**
1. What was the amount with B at the end of the second game?
2. How much money did D have after losing the second game?
3. What was the initial amount each player had before the games started?
4. How much money did each player have after the first game?
5. What would have been the result if B lost the first game instead of A?

**Tip:** To solve problems involving sequential changes like this, always track the changes step by step for each participant after every round or event.

Explore the solution to a sequential dice game problem involving players A, B, C, and D. Understand how each player's money changes after each game, leading to Rs. 96 for each player by the end of the fourth game. Focus on calculating the amount with player C after the third game.

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