Assuming each hole is independent from the next and that the probability of getting a hole in one remains constant from hole to hole, this question is simply a case of using the Binomial Distribution.

So the probability that exactly 4 of the 155 golfers got a hole-in-one on hole 6 is:

p(x)=(n/x) p^x (1-p)^n-x

where

(n/x)= n!/x!(n-x)!

so p(4)=0.0000001

It was basically a fluke!