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…

Probability of success i.e. getting a $5$ or $6=P\left(s\right)=\frac{2}{6}=\frac{1}{3}$.

So, probability of failure i.e. getting any number other than $5$ or $6=P\left(f\right)=1-P\left(s\right)=1-\frac{1}{3}=\frac{2}{3}$.

Let $X$ be the amount that a man wins in a maximum of three throws.

So, $X$ can take values :

Case I : First two throws are $5$ or $6$, game stops

$\Rightarrow X=100\Rightarrow P\left(X=100\right)=\frac{1}{3}$

Case II : First throw is other than $5$ or $6$ and second and third throws are $5$ or $6$, game stops

$\Rightarrow X=50\Rightarrow P\left(X=50\right)=\frac{2}{3}\times \frac{1}{3}=\frac{2}{9}$

Case III : First two throws are other than $5$ or $6$ and last throw is $5$ or $6$

$\Rightarrow X=0\Rightarrow P\left(X=0\right)=\frac{2}{3}\times \frac{2}{3}\times \frac{1}{3}=\frac{4}{27}$

Case IV : All three throws are other than $5$ or $6$

$\Rightarrow X=-150\Rightarrow P\left(X=-150\right)=\frac{2}{3}\times \frac{2}{3}\times \frac{2}{3}=\frac{8}{27}$

Therefore, expected gain or loss

$=E\left(X\right)=100\times \frac{1}{3}+50\times \frac{2}{9}+0\times \frac{4}{27}-150\times \frac{8}{27}=0$