Answer to Question #344872 in Statistics and Probability for Akshay

Question

A gambler plays roulette at Monte Carlo and continues gambling, wagering the same amount each time on Red, until he wins for the first time. If the probability of Red is 18/38 and the gambler has only enough money for 5 trials, then what is the probability that he wins on the second trial?

Answer 1

$$P(second)=(1-\dfrac{18}{38})(\dfrac{18}{38})=\dfrac{90}{361}$$