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 $\displaystyle n$ have the same number.

The probability that all the cars have different numbers is:

$\displaystyle p(0)=\frac{100!}{(100-n)!}\times\frac{1}{100^n}$

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

$\displaystyle p(1^*)=\frac{100!}{(100-n+1)!}\times\frac{1}{100^{n-1}}\times \frac{1}{100}$

This is the same as the probability that the $\displaystyle i$ and $\displaystyle j$-th cars have the same number and all the rest are different. There are $\displaystyle n(n-1)/2$ such pairs so:

$\displaystyle p(1)=\frac{100!}{(100-n+1)!}\times\frac{1}{100^{n-1}}\times \frac{n(n-1)}{200}$

If the above is correct (and I am issuing no guarantees at present) we need to find $\displaystyle n$ such that $\displaystyle p(0)+p(1)=0.5$

Now the nearest I can get to $\displaystyle 0.5$ is $\displaystyle 0.4967$ so presumably this is still not quite right (with $\displaystyle n=19 $) !?

CB