I have this : "We roll a 7-sided dice n times. What is the lower value of n to have 60% of chance to have 2 same value"
And I don't know how I can resolve this ... I know that the result is 5 but I've try to use Binomial distribution but I don't find the right result.
On
You should try instead to count the number of possible rolls that have all values distinct. Note that there are 7 possible values for the first roll, 6 possible values for the second, et cetera. To compute the probability, divide by $7^n$ for the set of all possible rolls. Finally subtract the probability that all rolls are distinct from 1 to get the probability that at least one value is repeated.
Don't need the binomial theorem for this. Consider the chance of not having any two matching values. Each roll must then be one of the values not obtained by any of the previous rolls. In $n$ rolls this will be $$ Q(n) =\frac{7}{7} \cdot \frac{6}{7} \cdot \frac{5}{7} \cdots \frac{8-n}{7} $$ So you have to find the lowest $n$ such that $Q(n) < 0.4$. $$Q(1) = 1 \\ Q(2) = \frac{6}{7} \approx 0.85 \\ Q(3) = \frac{30}{49} \approx 0.61 \\ Q(4) = \frac{120}{343} \approx 0.35 < 0.4 $$ So the answer is that 4 rolls suffices to get the probability of a match above 60%.