[SethTro]

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.

[am17an]

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!

[vhcr]

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.

[SethTro]

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

[dpacmittal]

Sounds like the birthday paradox problem. Is it?

[knollimar]

Can you reuse a plate with 500?

[stevefan1999]

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.

[dekhn]

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).