Lets just call the last two digits the cars number

So we have that the bet is that two or more cars in the next

have the same number.

The probability that all the cars have different numbers is:

The probability that the first and second cars have the same number and all the others are different is:

This is the same as the probability that the

and

-th cars have the same number and all the rest are different. There are

such pairs so:

If the above is correct (and I am issuing no guarantees at present) we need to find

such that

Now the nearest I can get to

is

so presumably this is still not quite right (with

) !?

CB