Suppose we use a hash function $H$ to hash $N$ distinct balls into $M$ distinct bins. Assuming simple uniform hashing, what is the expected number of collisions?
Note that a collision is defined by adding a ball to an already occupied bin. If the already occupied bin has $k$ balls in it, then the number of collisions upon adding a new ball is $k.$
By using expectation, I tried as :
=> 1 × Probability of collision in first insertion +
2 × Probability of collision in second insertion + .......... +
n × Probability of collision in nth insertion
=> $(1 ∗ 0) + (2 ∗ 1/m) + (3 ∗ 2/m) + (4 ∗ 3/m) +…+ (n ∗ n−1/m)$
Actually, The answer is $(n^2 - n)/2m$
But, I am not getting the answer. Where am I wrong here ?
