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!

I'm guessing if you only calculate based on the digits, the probability is going to be slightly different than the real one, because you only have a finite number of plates you can choose from.

I'm glad you enjoyed! It was a real game I played when driving around.

Is your real name also Seth? This is wholesome and hilarious

Yes, the other name in the problem is my sister's name :)

Sounds like the birthday paradox problem. Is it?

Nearly, but not 8 digits of precision worth.

Can you reuse a plate with 500?

The wording seems to strongly imply no; you need two separate plates with 500 on them.

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).