Hm… well had to look up my solution and it didn’t run into any problem with that.
I think in this instance the expected_balls only represent part of all drawn balls - meaning it expects to draw those 3, while the fourth one can be whatever.
It’s like throwing a dice 10 times and looking for the chance to get (at least) three times a 6. It doesn’t matter what the other 7 results are.
Believe it or not, I was in my bathroom yesterday and thought about this solution (never mind the 4th ball).
But I smiled and discarded that idea.
I will try.
Thank you