Expected number of buckets with at least one ball

Question

This question is related to this question: Given K balls and N buckets what is the expected number of occupied buckets

The question is:

Given $K$ balls and $N$ buckets how do you calculate the expected number of buckets with at least $1$ ball. Each ball is put in a bucket chosen at random with a uniform probability distribution. Assume also $K ≤ N$.

The first part of the answer goes like this:

I will assume that we are throwing balls sequentially towards the buckets, with at any stage each bucket equally likely to receive a ball, and independence between throws. Then the probability that bucket $i$ has no balls in it after $K$ balls have been thrown is equal to
$$\left(\frac{N-1}{N}\right)^K.$$

I had a different way of calculating this probability: counting the number of ways the $K$ balls can be put into $N-1$ buckets (all except bucket $i$), and dividing that by the number of ways $K$ balls can be put into $N$ buckets. I can use stars and bars to calculate each. Why does this method not work in this case? I'm a bit confused.

Answer 1

First off, you've mixed up $N$ and $K$ between the given answer and your method. For the rest of this answer, I'll assume you meant

I had a different way of calculating this probability: counting the number of ways the $K$ balls can be put into $N−1$ buckets (all except bucket $i$), and dividing that by the number of ways $K$ balls can be put into $N$ buckets. I can use stars and bars to calculate each. Why does this method not work in this case? I'm a bit confused.

For this to make any sense to compute the probability, all the outcomes you're counting need to be equally likely. Is this true? Well, let's just take $N=2$ and $K=2$ for example. Using your logic, the outcomes are

1. Two balls in bucket 1
2. One ball in bucket 1, one ball in bucket 2
3. Two balls in bucket 2

Are these equally likely to occur? No. Outcomes 1 and 3 each occur with probability $1/4,$ but outcome 2 occurs with probability $1/2.$