In each of a set of games it is 2 to 1 in favour of the winner of the previous game. The chance that the player who wins the first game shall win three at least of the next four is

Options

A. `8/27`
B. `4/81`
C. `4/9`
D. None of these

Answer

To solve the problem, we need to find the probability that the player who wins the first game will win at least three out of the next four games, given that the odds are 2 to 1 in favor of the winner of the previous game.

### Step-by-Step Solution:

1. **Understanding the Odds**:
The odds of 2 to 1 in favor of the winner means that if the player wins, the probability of winning the next game is \( \frac{2}{3} \) and the probability of losing is \( \frac{1}{3} \).

2. **Winning All Four Games**:
- The probability of winning all four games is calculated as:
\[ P(\text{Win all 4}) = \left(\frac{2}{3}\right)^4 = \frac{16}{81} \]

3. **Winning Exactly Three Games**:
To find the probability of winning exactly three games, we need to consider the different scenarios in which the player can win three games and lose one. There are four different ways this can happen (the lost game can be in any of the four positions).

- **Case 1**: Lose the first game (LWWW)
- Probability:
\[ P(LWWW) = \frac{1}{3} \cdot \left(\frac{2}{3}\right)^3 = \frac{1}{3} \cdot \frac{8}{27} = \frac{8}{81} \]

- **Case 2**: Lose the second game (WLWW)
- Probability:
\[ P(WLWW) = \left(\frac{2}{3}\right) \cdot \frac{1}{3} \cdot \left(\frac{2}{3}\right)^2 = \frac{2}{3} \cdot \frac{1}{3} \cdot \frac{4}{9} = \frac{8}{81} \]

- **Case 3**: Lose the third game (WWLW)
- Probability:
\[ P(WWLW) = \left(\frac{2}{3}\right)^2 \cdot \frac{1}{3} \cdot \left(\frac{2}{3}\right) = \frac{4}{9} \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{8}{81} \]

- **Case 4**: Lose the fourth game (WWWL)
- Probability:
\[ P(WWWL) = \left(\frac{2}{3}\right)^3 \cdot \frac{1}{3} = \frac{8}{27} \cdot \frac{1}{3} = \frac{8}{81} \]

4. **Total Probability of Winning Exactly Three Games**:
- Adding the probabilities from all four cases:
\[ P(\text{Win exactly 3}) = \frac{8}{81} + \frac{8}{81} + \frac{8}{81} + \frac{8}{81} = \frac{32}{81} \]

5. **Total Probability of Winning at Least Three Games**:
- Now, we add the probability of winning all four games to the probability of winning exactly three games:
\[ P(\text{At least 3}) = P(\text{Win all 4}) + P(\text{Win exactly 3}) = \frac{16}{81} + \frac{32}{81} = \frac{48}{81} \]

6. **Simplifying the Probability**:
- Simplifying \( \frac{48}{81} \):
\[ \frac{48}{81} = \frac{16}{27} \]

### Final Answer:
The probability that the player who wins the first game will win at least three out of the next four games is \( \frac{16}{27} \).