## Oregon licence plates consist of three letters followed by a three digit number (each digit can be from [0..9]).

While driving to work Seth plays the following game:

Whenever the numbers of two licence plates seen on his trip add to 1000 that’s a win.

E.g. MIC-012 and HAN-988 is a win and RYU-500 and SET-500 too (as long as he sees them in the same trip).

Find the expected number of plates he needs to see for a win.

Give your answer rounded to 8 decimal places behind the decimal point.

Note: We assume that each licence plate seen is equally likely to have any three digit number on it.

### The probability of having any specific three-digit number is 1/1000 for each license plate Seth sees (because the digits can range from 000 to 999 inclusive, so there are 10 x 10 x 10 = 1,000 different possible numbers).

Since he is looking for two numbers that add up to 1000, once he has seen the first number (let’s call it n), he needs to find a second number such that the second number + n = 1000. There is only one such number (1000 – n).

Since each license plate is assumed to be equally likely to have any three-digit number on it, the chance that any subsequent license plate has the number 1000 – n on it is 1/1000.

Thus, after seeing the first license plate (which can be any number), the expected number of plates he needs to see in order to find a “winning” pair is 1,000.

However, we have only considered the scenario starting from the second license plate seen. We still need to account for the first license plate seen. Since there’s no restriction on the first license plate’s number, it’s certain to have some number on it. Hence, Seth needs to see 1 + 1000 = 1001 license plates in total to get a win.

So, the expected number of plates he needs to see for a win is 1001.00000000 (rounded to 8 decimal places).

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