Thousands of cars

Let’s say none of the cars had the same last four digits as you chose it, then the probability of that happening is

$\text{Probability it’s not the first car × Probability it’s not the second car × Probability it’s not the third car … Probability it’s not the 10000 car}$

All of these probabilities are equal to $\frac{9999}{10000}=0.9999$

So the probability none of the cars have the last four digits as you chose is $0.9999^{10000}$

And the probability $P$ atleast one of the cars has the same last four digits as you chose is $1-0.9999^{10000}$ (You can put this in a calculator and find the answer at this point itself)

According to the question, we are allowed to take $10000 ≈ ∞$

So let’s make $x=10000 ≈ ∞$

Rewriting $P$ in terms of $x$

We get $P=1-\left ( 1- \frac{x-1}{x} \right ) ^ {x} = 1- \left ( 1-\frac1x \right ) ^ {x}$

Now letting $x$ tend to $∞$

$P=1- \displaystyle\lim_{x ➝ ∞} \left ( 1-\frac1x \right ) ^ {x}$

Rearranging and taking the natural logarithm on both sides we get :

$\ln (1-P)= \displaystyle\lim_{x ➝ ∞} x\ln\left ( 1-\frac1x \right )$

$\ln (1-P)= \displaystyle\lim_{x ➝ ∞} \frac{\ln\left ( 1-\frac1x \right )}{\frac1x}$

Setting $y=\frac1x$

$\ln (1-P)= \displaystyle\lim_{y➝ 0} \frac{\ln\left ( 1-y\right )}{y}$

Using L'Hôpital's rule or the series expansion of the natural logarithm

$\ln (1-P)= -1$ $1-P=\frac1e$ $P=1-\frac1e=\boxed{0.632…}$