In California, non-commercial vehicle license plates have a single-digit series number, which for my purposes we will disregard; followed by three letters, then three digits. The letters are always together in a cluster, same with the digits.

Like this:

(A) BBB CCC.

What I’d like to know is, how many potential unique license plates does that make, assuming 26 possibilities for each B and 10 possibilities for each C?

And especially, what is the statistical probability of a given plate having both a double letter and a double digit?

Part 2:
The odds of a double letter is *roughly *2/26. Whatever the first letter is, the second letter has a 1/26 chance of matching it. Then the third letter has a 1/26 chance of matching the second number. I assume you’re only talking about successive letters matching, that the first and third letter being the same doesn’t count.

The odds of a double number is *roughly *2/10, for the same reason.

The odds of both of those occurring is (2/26)*(2/10) = 4/260 = 2/130 = 1.5%. There’s some minor modifications to that because of some small probability cases I’m not bother to include here, they’re negligible.

The first question is easy: 2626261010*10 = 17,576,000.

(Most states with letters on their plates leave out letters that look like numbers. So, no I’s S’s, or O’s at the very least. But let’s ignore that for now.)
There are 262626 = 17,576 possible letter combinations, of which some have double letters. There are two forms that double letters can take: AAB or BAA where AA is the double letter and B is the other letter.

Both A and B have 26 possible values, so the total number of double letter combinations of the form AAB is 26*26 = 676. Double that for the BAA combos and you get 1352.

The same logic can be applied to the numbers: 10102 = 200.

So, the probability of getting a double letter and a double number should be ( 1352 / 17,576 ) * ( 200 / 1000 ) = 0.015384… – call it about 1.5%.
I think.

Well, BBB would have 26[sup]3[/sup] possible combinations, and CCC would have 10[sup]3[/sup] combinations, so you’d have 26[sup]3[/sup] x 10[sup]3[/sup] possible unique plates which is 17,576,000 possibilities.

The second question is a bit harder, and I don’t remember the exact forumulae off the top of my head, but they should be easy enough to derive. The problem is, since your wording is a little vague, I’ll have to make some assumptions.

You can have double anywhere in there (so order doesn’t matter), they don’t have to be sequential, and you’re not excluding triple cases. So in that case, you’d have n[sup]2[/sup]/n[sup]3[/sup] possibilities. So that should be (26[sup]2[/sup]/26[sup]3[/sup]) x (10[sup]2[/sup]/10[sup]3[/sup]), or 0.385%.

I chose that, quite frankly, because it’s the easiest to calculate, but if you want to give tighter restrictions about what you mean, I can adjust accordingly.

Not quite as you double counted the license plate with a triple letter. There are 2625 plates of the form AAB where B is different from A. There are another 2625 plates of the form ABB with A and B distinct and 26 plates of the form AAA so 1301 all together.

Okay, I failed miserably, and I see it’s because I’m an idiot. It’s (3 * n[sup]2[/sup] - 2 * n) / n[sup]3[/sup], so that’s 3.15%. And, for the same assumptions everyone else is making, which I suppose is the intuitive way, it’s (2 * n[sup]2[/sup] - n) / n[sup]3[/sup], that’s 1.43%. Sorry for the screw up.

The probability question assumes that letters and numbers are randomly assigned. That’s unlikely; I would think that at least they’re assigned consecutively, and perhaps with some other non-random scheme (parts of the letters signify the county, or something).

Most schemes would probably result in a higher probability of doubled letters/digits. For instance, in a scheme that just assigns them consectutively (from AAA-001 to ZZZ-999), until more than 26,000 plates have been assigned, the probability of a double letter is one.

The occurrence of double characters is not probabilistic, true, but the probability of a double letter or number being assigned to a particular person is as described.

This is getting into tiny nitpicking of course, but why else is the Dope here?

So I nitpickingly disagree: the scheme for assigning plate IDs affects the probability of a double letter being assigned to a person. Two examples: As I mentioned above, if it’s a sequential assignment and less than 26000 plates have been assigned, then it’s 100% chance of having a double letter.
For another example, at least until some time recently (and maybe still), the chance of having a doubled digit in your US area code was zero, though simple probability would say it was (about) 20%. (All area codes had a zero or one as the middle digit, none had zero or one as the first, and all x-1-1 codes were reserved and I think x-0-0 was also not given out).
For all I know, CA explicitly does not give out double letter plates for some reason, just like AT&T didn’t give out doubled digit area codes. (maybe they’re easier to make mistakes typing them into computers).

What I’m saying is, if we don’t know CA’s assignment scheme, then the only way to assign a probability is to assume completely random assignment. But if we do know the assignment scheme, we can make a better (possibly very different) estimate of the chances.

To add to Quercus, I can’t speak for California, but I know at least Virginia does not assign plates in a strickly AAA-0000 to ZZZ-9999 fashion because soon after they switched to the 7 character plate, I saw both Axx-#### and Zxx-####. As I understand, they assigned different starting points for sequential distribution in different areas and that the ones that started at AAA counted forward, the ones from ZZZ counted backwards, and there were others in the middle and I’m not quite sure which ways they went.

It wouldn’t surprise me if California had a similar distribution method. It also wouldn’t surprise me if they specifically avoided double characters for the reason Quercus outlined. However, we have no reason to assume they’re specifically dodging certain letter and number combinations, making any assumptions would be purely arbitrary, and thus add unnecessary complication. We can also safely say that the number of plates with the pattern outlined in the OP is probably much much higher than the aforementioned 26,000 plates and that as the number of plates grows, the ratios will approach those calculated using random distribution assumption. So, without knowing California’s plate distributions methods or where they are along in that methodology, I think the random assumption is as accurate as we can get.

Again, I don’t know about California, but I do know that at least in Virginia they don’t exclude any numbers or letters and, as an anecdote, I’ve seen some interesting vanity plates as a result along the lines of II11I1I or VVWVWVW or O00O0O0. I guess they don’t bother since, at least with the generic plate, the letters and numbers aren’t mixed so it’s obvious which one it is.

Either way, if California does exclude certain letters or numbers, you can just take the formula I did, or one of the other methodologies and change the 26 to 24, or 10 to 8, or whatever.