How many experiments do we need to perform at least so that the probability of rolling with 3 dice

a) once a 15,

b) once at least a 15,

is greater than $1 / 2$
The possible cases for the 3rd class repetitive variations of 6 numbers ranging from 1 to 6, which number is $V_{6}^{2,3} = 6^{3} = 216$.

a) Since $15 = 6 + 6 + 3 = 6 + 5 + 4 = 5 + 5 + 5$, the 15 can be achieved in $P_{3}^{2} + P_{3} + 1 = \frac{3!}{2!} = 3 + 3! + 1 = 3 + 6 + 1 = 10$ ways, which means the number of non-15 rolls is $216 - 10 = 206$.

Therefore, the probability that we never roll a 15 in $x$ rolls is

$$
\left(\frac{206}{216}\right)^{x} = \left(\frac{103}{108}\right)^{x}
$$

The desired probability will be greater than $\frac{1}{2}$ if the opposite is less than $\frac{1}{2}$, that is,

$$
\left(\frac{103}{108}\right)^{x} < \frac{1}{2}
$$

Solving for $x$:

$$
x > \frac{\lg 2}{\lg 108 - \lg 103} \approx 14.6
$$

Thus, we need to roll at least 15 times.

b) Favorable cases here include the favorable cases from part a) plus the cases $6 + 6 + 4 (=16)$, $6 + 5 + 5 (=16)$, $6 + 6 + 5 (=17)$, and $6 + 6 + 6 (=18)$ and their permutations. Therefore, the total number of favorable cases is

$$
10 + 3 \cdot P_{3}^{2} + 1 = 10 + 3 \cdot \frac{3!}{2!} = 10 + 9 = 19
$$

The number of unfavorable cases is $216 - 19 = 197$. Thus, the probability that we roll at most 14 in $x$ rolls is $\left(\frac{197}{216}\right)^{x} = \left(\frac{197}{216}\right)^{x}$. The desired probability will be greater than $\frac{1}{2}$ if the opposite is less than $\frac{1}{2}$, that is,

$$
\left(\frac{197}{216}\right)^{x} < \frac{1}{2}
$$

Solving for $x$:

$$
x > \frac{\lg 2}{\lg 216 - \lg 197} \approx 7.1
$$

Thus, we need to perform at least 8 trials.

Géza Biczó (Bp., II., Rákóczi g. II. o. t.)