We roll a die $n$ times. What is the probability that there are two equal numbers among the rolled numbers?
Solution. If $n \geq 7$, then $P=1$ with certainty that we roll the same number.

If $n=2$, then the total number of possibilities is $6^{2}$, and the favorable outcomes are $6^{2}-6 \cdot 5=6$.

$$
P_{2}=\frac{6}{6^{2}}=\frac{1}{6}
$$

If $n=3$, then the total number of possibilities is $6^{3}=216$, and the favorable outcomes are $6^{3}-6 \cdot 5 \cdot 4=96$.

$$
P_{3}=\frac{96}{216} \approx 0.444
$$

If $n=4$, then the total number of possibilities is $6^{4}=1296$, and the favorable outcomes are $6^{4}-6 \cdot 5 \cdot 4 \cdot 3=936$.

$$
P_{4}=\frac{936}{1296} \approx 0.722
$$

If $n=5$, then the total number of possibilities is $6^{5}=7776$, and the favorable outcomes are $6^{5}-6 \cdot 5 \cdot 4 \cdot 3 \cdot 2=7056$.

$$
P_{5}=\frac{7056}{7776} \approx 0.907
$$

If $n=6$, then the total number of possibilities is $6^{6}=46656$, and the favorable outcomes are $6^{6}-6!=45936$.

$$
P_{6}=\frac{45936}{46656} \approx 0.985
$$

Generally, for $(2 \leq n \leq 6)$:

$$
P_{n}=\frac{6^{n}-\frac{6!}{(6-n)!}}{6^{n}}
$$