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:

For a single throw of a fair die, the probability of getting either 5 or 6 is $$\dfrac{2}{6}=\dfrac13$$ and the probability of getting any other number (1, 2, 3, 4) is $$\dfrac{4}{6}=\dfrac23.$$

The man follows this rule: he stops at the first appearance of 5 or 6, but in any case throws at most three times. We compute every possible stopping scenario, its probability, and the total money won or lost in that scenario.

Let $$p=\dfrac13 \;(\text{success}), \qquad q=\dfrac23 \;(\text{failure}).$$

Case 1: Success on the first throw.
Probability $$=p=\dfrac13,$$
Money outcome $$=+100.$$

Case 2: Failure on the first throw and success on the second.
Probability $$=q\,p=\dfrac23\cdot\dfrac13=\dfrac29,$$
Money outcome $$=-50 \;(\text{first throw}) + 100 \;(\text{second throw}) = +50.$$

Case 3: Failures on the first two throws and success on the third.
Probability $$=q^{2}p=\left(\dfrac23\right)^{2}\!\!\cdot\dfrac13=\dfrac49\cdot\dfrac13=\dfrac{4}{27},$$
Money outcome $$=-50-50+100=0.$$

Case 4: Failures on all three throws.
Probability $$=q^{3}=\left(\dfrac23\right)^{3}=\dfrac{8}{27},$$
Money outcome $$=-50-50-50=-150.$$

Now we apply the definition of expected value: $$E=\sum (\text{probability})\times(\text{money outcome}).$$

So,

$$\begin{aligned} E &= \left(\dfrac13\right)(100) \;+\; \left(\dfrac29\right)(50) \;+\; \left(\dfrac{4}{27}\right)(0) \;+\; \left(\dfrac{8}{27}\right)(-150) \\[6pt] &= \dfrac{100}{3} \;+\; \dfrac{100}{9} \;+\; 0 \;-\; \dfrac{1200}{27}. \end{aligned}$$

To add these, we convert each term to a common denominator of 27:

$$ \dfrac{100}{3}=\dfrac{900}{27},\qquad \dfrac{100}{9}=\dfrac{300}{27},\qquad -\dfrac{1200}{27} \text{ is already over } 27. $$

Adding, we obtain

$$E=\dfrac{900}{27}+\dfrac{300}{27}-\dfrac{1200}{27}=0.$$

The expected value is zero rupees; the game is fair in the long run.

Hence, the correct answer is Option D.