I wrote Problem 371, https://projecteuler.net/problem=371 , as a high school student in 2012!

I'm so happy to have spent twenty years of my life learning math and solving problems on Project Euler and elsewhere.

This is one of my favourite problems, I still remember that it has a very real edge case even though I solved it more than 10 years ago. Thank you for the problem!

That sounds like a combinatorial problem...alphabets from AAA to ZZZ, numbers from 000 to 999.

That means one of the total sum of possible car plates is 26^3.

Since we want to find pairs (x, y) that x + y = 1000. That means the total sum would also add up sum([1 for x in range(1000) for y in range(1000) if x + y == 1000])/2 since there is a symmetry.

But wait, find the expected number of plates he needs to see for a win. So maybe we need to borrow something from statistics (Possion/chi-squared distribution) or queueing theory...?

Edit: ah I saw the solution, it is a Markov chain.

Interesting- I ask a license plate question (when will california run out of plates in its current serialization format, based on a couple of plates observed in two different years). It's a much simpler question, though (just linear extrapolation).