I want to know the probability that in a week's lottery, the pairwise differences between the five drawn numbers are all at least five.
Let $k$ denote the number of quintuples in which the pairwise differences are all at least 5. Then the desired probability is $p=\frac{k}{\binom{90}{5}}$.

Let $1 \leq a_{1}<a_{2}<a_{3}<a_{4}<a_{5} \leq 90$, and $a_{i+1}-a_{i} \geq 5(i=1,2,3,4)$. For the numbers $b_{i}=a_{i}-4(i-1)$ (where $i=1, 2, \ldots, 5$), it holds that $1 \leq b_{1}<b_{2}<b_{3}<b_{4}<b_{5} \leq 90-16=74$, so to each favorable quintuple, we can assign a combination of five numbers from the first 74 numbers using the mapping $\left\{a_{i}\right\} \rightarrow\left\{b_{i}\right\}$; it is evident that different $a_{i}$ quintuples correspond to different $b_{i}$ quintuples.

Conversely, every $1 \leq b_{1}<\ldots<b_{5} \leq 74$ quintuple can be represented as the image of the mapping $\left(a_{i}=b_{i}+4(i-1)\right)$, so the number of corresponding $a_{i}$ quintuples is $\binom{74}{5}$, and therefore $p=\binom{74}{5} /\binom{90}{5}=0.367$.

Seress Âkos (Budapest, Fazekas M. Gyak. Gimn., IV o. t.)

Note. The probability in question for 1, 2, 3, etc., instead of 5, is as follows in thousandths: 1000, 792, 629, 480, 367, 275, 203, 147, 104, 72, 48, 31, 19, 11, 6, 3, 2, 1. The last value corresponds to 18, and 19, 20, 21, 22 are theoretically possible but very rare, with only 8568, 2002, 252, 6 cases out of 43949 268, respectively.