Given a $5$ out of $36$ lottery ($5$ unique numbers out of pool of $36$ numbers ranging $[1,2,…,36]$).
How to calculate probability that a draw has at least one pair of consecutive numbers (like $22, 23$ -a pair of numbers whose difference $23-22 = 1$)?
Every draw ($5$ numbers) has $10$ pairs. For dif1 we have total of $35$ pairs ($1$ and $2$, $2$ and $3$ ... $34$ and $35$). There are total of $630$ pairs (binomial(35, 2)).
Problem is I think cannot think like that:
$1$ concrete pair out of $630$ appears with $1/630$ chances. Probability to have any of $35$ dif1 pairs is $35/630$ (if I choose $2$ numbers randomly). But I choose $5$ numbers (which give $10$ pairs) - and it is not the same as just drawing $2$ pairs out of $630$ pairs.
I cannot figure out how to reason in this case.
The question is about not dif1 but also dif2, dif3... dif35 (there is only single such pair!).
How to mathematically calculate the probability?
Can I think of a single 5-out-of-36 draw as an equivalent of 10 independent "pick a pair out of all possible pairs"? It would give dif1 ($35$ dif1 pairs out of $630$ all pairs) as $((35/630)+(34/629)+(33/628)+(32/627)+(31/626)+(30/625)+(29/624)+(28/623)+(27/622)+(26/621))$. But it differs greatly from a real lottery, which makes me think that formula (and reasoning) above is not applicable!
P.S. Stars and bars method is described here (wiki), example how to use it is here