# How do the following Hash Functions compare?

Consider the two hash functions used to map IP addresses. $x_i$ represents a octave (or "bit field") of the address.

Hash Function 1: $$h_a(x_1, x_2, x_3, x_4) = \sum^{4}_{i=1} a_ix_i \bmod n$$

Where $n$ is a prime number closest to the total number of IP addresses being mapped. The probability of two IP addresses being mapped to the same bucket is $\frac{1}{n}$ using this hash function.

Hash Function 2:

$$h_a(x_1, x_2, x_3, x_4) = \sum^{4}_{i=1} x_i \bmod n$$

What would the probability to two IP addresses being mapped to the same bucket be using the second hashing function?

The reason is that your second function is a specific hash function, while the first formulation is a family of hash functions (indexed by $a$).
How can you compare the two? Choose $a$ in random and compare the two instances? Compare for the best/worst $a$, maybe? Compare on average (for a uniform $a$)?
As for your other question, the answer is no, in the second hash, the probability is not equal $1/n$ to be in each bucket. Assume that each $x_i$ is uniform in [0..255], then $x_1+x_2$ is not uniform in [0..510]. For instance $P(0)=1/256^2$ (both are 0) while $P(4) = P(0,4) + P(1,3) + P(2,2) + P(3,1) + P(4,0) = 5/256^2$. There's no reason that the sum will be uniform $\mod n$.
Last comment: As explained above, for the first hash, a probability of $1/n$ is over the choice of $a$.