Example 22. How many times do you need to roll two dice so that the probability of rolling at least one double six is greater than $1 / 2$? (This problem was first posed by the French mathematician and writer de Mere ( $1610-1684$ ), hence the problem is named after him).
Solution. Let the event $A_{i}$ be "rolling two sixes on the $i$-th throw". Since any of the six faces of the first die can match any of the six faces of the second die, there are $6 \cdot 6=36$ equally likely and mutually exclusive events. Only one of these - rolling a six on both the first and the second die - is favorable to the event $A_{i}$. Therefore,

$$
P\left(A_{i}\right)=\frac{1}{36}
$$

from which

$$
q=1-p=1-\frac{1}{36}=\frac{35}{36}
$$

The throws of the dice are independent trials, so we can use formula (1.8.17), which in this case takes the form

$$
1-\left(\frac{35}{36}\right)^{n}>\frac{1}{2}
$$

or

$$
\left(\frac{35}{36}\right)^{n}\frac{\ln 2}{\ln 36-\ln 35}=\frac{0.6931}{0.0284}=24.4
$$

Thus, to have the probability of rolling two sixes greater than $1 / 2$, the dice need to be thrown at least 25 times.