Probability - shoe shop stock out

To run out of stock, the nine customers must order seven or more of the same type of shoe. The chance they will order nine of one type is (1/9)^9. The chance they will order eight is (1/9)^8*(8/9). The chance they will order exactly seven is (1/9)^7*(8/9)^2. The sum of these three probabilities is 1.884E-07. There are nine sizes of shoe and it is not possible to run out of two types of shoe so multiply by nine to get a total probability of 1.696E-06.

I made an error in this post.

The chance they will order nine of one type is (1/9)^9.
The chance they will order eight is (1/9)^8*(8/9)*9.
The chance they will order seven is (1/9)^7*(8/9)^2*(9*8/2)

After multiplying by nine you get 5.522E-05

Thanks Engineer.

Is there an easy way to scale this, for example to:

A shoe shop has a shoe in 80 sizes and they keep 3 of each size in stock.
They expect 26 customers per week of this shoe - these can be any size.
What is the probability they will run out of stock of at least one size?

I can work out that p(at least 2 of same size) = 1-p(all different)
p(all different) = 80/80 * 79/80...55/80 = (80!/54!)/80^26 = 0.0103
So P(at least 2 of same size) = 1-0.0103 = 0.9897

But not sure how to scale up to at least 4 of the same size (which will be a stock out)

This is a "balls and bins" type of problem. You have a different take on it since you want no more than X balls in a bin, but the general problem is the same. This site has a lot of these types of questions, but I don't see your exactly. It should give you some insights though.

https://math.stackexchange.com/questions/tagged/balls-in-bins?sort=frequent&pageSize=50

Thanks, I've posted the question there and will report back if I get an answer I understand.

Thanks, I'm interested in the answer.