We call a dice roll successful if the rolled number is at least three. What is more likely: that at least one out of two rolls is successful, or that at least two out of four rolls are successful?
Solution. The number of possible rolls is 6, the successful rolls are: 3, 4, 5, 6. The probability that we roll one of these in a single roll is: $\frac{4}{6}=\frac{2}{3}$. The probability of an unsuccessful roll is $1-\frac{2}{3}=\frac{1}{3}$.

Let $A$ be the event in which at least one of two consecutive rolls is successful. Instead of calculating the probability of $A$, we calculate the probability of $\bar{A}$, which is the probability that both rolls are unsuccessful: $P(\bar{A})=\left(\frac{1}{3}\right)^{2}=\frac{1}{9}$. Since the outcomes of the rolls are independent and the occurrence of successful and unsuccessful rolls are mutually exclusive, their probabilities sum to 1. Therefore,

$$
P(A)=1-P(\bar{A})=\frac{8}{9}
$$

This is the probability that at least one of two rolls is successful.

Let $B$ be the event that at least two out of four rolls are successful. Then $\bar{B}$, the opposite of $B$, occurs when the number of successful rolls out of four is 0 or 1.

The probability that none of the four rolls are successful: $\left(\frac{1}{3}\right)^{4}=\frac{1}{81}$. There are 4 ways for one of the four rolls to be successful (it can be the first, second, and so on), so the probability of this is:

$$
4 \cdot\left(\frac{1}{3}\right)^{3} \cdot \frac{2}{3}=\frac{8}{81}
$$

Thus, the probability of the $\bar{B}$ event is:

$$
P(\bar{B})=\frac{1}{81}+\frac{8}{81}=\frac{9}{81}=\frac{1}{9}
$$

Therefore,

$$
P(B)=1-P(\bar{B})=1-\frac{1}{9}=\frac{8}{9}
$$

We can see that the probability of the two events is equal.