https://en.wikipedia.org/wiki/Birthday_problem
Give it a brief read and draw your own conclusions regarding hash collision.
Edit: Complementing with some pre-digested math (very informal, btw, I'm no mathematician):
Given a pseudorandom function Hash X P(N) -> H(M) (N length of the password, M length of the hash, M > N), there are 2^N possible passwords and therefore at most 2^N possible hashes (that implies that X(P) is not dense, that is, it doesn't spawn to all 2^M possible values of (M)).
Given two randomly generated hashes x1 and x2 for X unknown, each one has roughly (2^N)/(2^M) of chance of being the hash of an actual password, and then 1/(2^N) that is is the hash of a certain specific password.
Given x1 the hash of your desired password, and x2 a 'randomly generated' valid hash, x2 has about 1/(2^N) chance of being from the same password. If x1 and x2 are equal up to the Zth bit, then there is a 1/(2^(N - Z*N/M)) or (2^(N/M)^Z)/(2^N) chance of them being the same password, which goes up pretty fast with Z.
Include the fact that your acceptable password space is actually a small subset of P(N) (not everyone used randomly generated binary passwords) and then the N on the denominator of that chance becomes smaller than the N on top, increasing your odds.
All this is very inaccurate and mostly 'intuitive' but hopefully you get the idea.
Give it a brief read and draw your own conclusions regarding hash collision.
Edit: Complementing with some pre-digested math (very informal, btw, I'm no mathematician):
Given a pseudorandom function Hash X P(N) -> H(M) (N length of the password, M length of the hash, M > N), there are 2^N possible passwords and therefore at most 2^N possible hashes (that implies that X(P) is not dense, that is, it doesn't spawn to all 2^M possible values of (M)).
Given two randomly generated hashes x1 and x2 for X unknown, each one has roughly (2^N)/(2^M) of chance of being the hash of an actual password, and then 1/(2^N) that is is the hash of a certain specific password.
Given x1 the hash of your desired password, and x2 a 'randomly generated' valid hash, x2 has about 1/(2^N) chance of being from the same password. If x1 and x2 are equal up to the Zth bit, then there is a 1/(2^(N - Z*N/M)) or (2^(N/M)^Z)/(2^N) chance of them being the same password, which goes up pretty fast with Z.
Include the fact that your acceptable password space is actually a small subset of P(N) (not everyone used randomly generated binary passwords) and then the N on the denominator of that chance becomes smaller than the N on top, increasing your odds.
All this is very inaccurate and mostly 'intuitive' but hopefully you get the idea.