In the last 3 months I have seen 10 different cars with the license plates ending in the same four digits (all different cars, with different front 3 letters). I live in Texas, where it is 3 letters followed by 4 numbers (excluding specialty, government plates, etc). In Texas plates are issued sequentially and given in batches to various local DMV offices. In other words they start at AAA0000 and go to AAA9999 before moving on to AAB0000. There are rules on certain numbers that can't be used but let's assume that every number is fair game, so there should theoretically be 10000 possible plates for each unique three letter combo. Given that they are released in batches you will often see cars with the same preceding letters in the same area, but the numbers are random.
Let's assume that the number in question is 2911 (although it could be any 4 digit number, but only that specific number and not just the odds of seeing 10 duplicates of any number). In about a 3 month time frame I've seen 10 cars with this number. Let's call it 100 days for easier math. I do notice a lot of plates while driving but not every plate. Let's assume I notice 100 plates in a day, also for easier math. What would be the odds of seeing one specific number on 10 different cars in that time frame?
Best Answer
You can solve this using Binomial Random Variable. Actually, you can continue my answer even if you are not familiar with the Binomial Random Variable.
Let $X$ be a random variable that takes value one if you see your desired number(2911) on the number plate.
$$\ \text{Total Number of Plates you see} = 100\text{days}\times 100\text{ plates/day}=10000\\ Pr(X=1) = {10000 \choose 10} \times ({ 1 \over 10000})^{10} \times ({ 9999 \over 10000})^{9990} \approx 1.0101\times 10^{-7} $$
We see that odds of seeing this number on ten different number plates is indeed very less.