# Power of 2 choices balls and bins - Expected number of empty bins

I am looking at the standard balls and bins example and more precisely to the expected number of empty bins. Given $$n$$ balls and $$n$$ bins, we throw balls into bins sequentially uniformly at random. So the probability for a ball $$i$$ to fall into a given bin is exactly, $$P(\text{ball } i \text{ fall into bin } i ) = \frac{1}{n}.$$

If $$X$$ denotes the number of empty bins then, we know that $$E[X] = n \left( 1 - \frac{1}{n} \right)^n \approx \frac{n}{e}.$$

However, I was wondering if there were known results concerning the 2 choices method? Instead of choosing one bin uniformly at random, we select 2 potential bins and place the ball in the least loaded one, breaking ties arbitrarily.

I would be tempted to say that $$E[X] = n \left( 1- \frac{2}{n} \right)^n \approx \frac{n}{e^{2}}$$

but is it true?

Moreover, are there known results when the probability is not uniform? For example, what would happen if the probability to choose a bin depends on the number of ball inside it?