In a gambling, a man will win $\$10$ if he is managed to get 5 or 6 facing up in rolling a fair die, he will need to pay $\$5$ otherwise. If the man continues to play the game until he wins 3 times, what is his expected gain?
Here's my working:
{5 or 6} =2/6=1/3
(1 or 2 or 3 or 4} =4/6=2/3
Expected gain = (1/3) * $$10 + (2/3) * -$5 = $0 for each round
Best Answer
Using the negative binomial distribution (the law that counts how many trials are needed to get $r$ successes) you get that the average numbers of draws is
$$\mathbb{E}[X]=\frac{r}{\theta}=\frac{3}{\frac{1}{3}}=9$$
Thus the expected gain is
$$3\times \$10+6\times(-\$5)=\$0$$