In a game, a man wins Rs 100 if he gets 5 or 6 on a throw of a fair die and loses Rs 50 for getting any other number on the die. If he decides to throw the die either till he gets a five or a six or to a maximum of three throws, then his expected gain/loss (in rupees) is: (a) `400/3` gain (b) `400/9` loss (c) 0 (d) `400/3` loss

Options

A. `(400)/(3)` gain
B. `(400)/(3)` loss
C. 0
D. `(400)/(9)` loss

Answer

To solve the problem, we need to calculate the expected gain/loss for the man based on the outcomes of throwing a fair die.

### Step-by-Step Solution:

1. **Define the Outcomes**:
- The man wins Rs 100 if he rolls a 5 or a 6.
- He loses Rs 50 if he rolls a 1, 2, 3, or 4.

2. **Calculate the Probabilities**:
- The probability of rolling a 5 or a 6 (winning) is:
\[ P(\text{win}) = P(5) + P(6) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \]
- The probability of rolling a 1, 2, 3, or 4 (losing) is:
\[ P(\text{lose}) = 1 - P(\text{win}) = 1 - \frac{1}{3} = \frac{2}{3} \]

3. **Calculate Expected Gain/Loss for Each Throw**:
- If he wins (with probability \(\frac{1}{3}\)), he gains Rs 100.
- If he loses (with probability \(\frac{2}{3}\)), he loses Rs 50.

The expected value (E) can be calculated as:
\[ E = P(\text{win}) \times \text{Gain} + P(\text{lose}) \times \text{Loss} \]
Substituting the values:
\[ E = \left(\frac{1}{3} \times 100\right) + \left(\frac{2}{3} \times (-50)\right) \]
\[ E = \frac{100}{3} - \frac{100}{3} = 0 \]

4. **Consider Maximum Throws**:
- The man can throw the die a maximum of 3 times. However, since the expected value for one throw is already 0, throwing the die multiple times does not change the expected outcome.
- If he rolls a 5 or 6 in any of the three throws, he stops. If he does not, he will have lost Rs 50 for each unsuccessful throw.

5. **Final Calculation**:
- The expected gain/loss remains 0, as calculated above.

### Conclusion:
The expected gain/loss for the man after throwing the die up to three times is **0**.