Starting today, the big gambler Carlo Pietro decided to visit casinos daily. On the first day, he will bet on a game where the probability of winning is equal to $\frac{1}{2}$. On the second, third, and fourth days, he will bet on different games whose probabilities of winning are, respectively, equal to $\frac{1}{3}, \frac{1}{4}, \frac{1}{5}$, and so on for the following days.

a) What is the probability that he will not have won by the third day?

b) What is the probability that he will not have won by the fifth day?

c) What is the probability that he will not have won by the $2013^{\underline{0}}$ day?
Solution

a) For Pietro not to have won by the third day, it is necessary that he lost on the first, second, and third day. The probability that Pietro wins on the first day is $\frac{1}{2}$. Therefore, the probability that Pietro lost on the first day is

$$
1-\frac{1}{2}=\frac{1}{2}
$$

The probability that Pietro wins on the second day is $\frac{1}{3}$. Therefore, the probability that Pietro lost on the second day is

$$
1-\frac{1}{3}=\frac{2}{3}
$$

The probability that Pietro wins on the third day is $\frac{1}{4}$. Therefore, the probability that Pietro lost on the third day is

$$
1-\frac{1}{4}=\frac{3}{4}
$$

Thus, the probability that Pietro lost on the first, second, and third day is equal to the product of these probabilities, which gives us

$$
\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4}=\frac{1}{4}
$$

b) We repeat the same argument as before! Now up to the fifth day, which gives us the probability that Pietro has not won as:

$$
\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdot \frac{5}{6}=\frac{1}{6}
$$

c) Thinking inductively, the probability that Pietro has not won by the $2013^{\text{th}}$ day is equal to

$$
\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdots \frac{2011}{2012} \cdot \frac{2012}{2013} \cdot \frac{2013}{2014}=\frac{1}{2014}
$$