No, I don't.
I will have to take on faith their statistic that for an amateur golfer, the probability of a hole-in-one are 1 in 12,500 per hole.
However, almost all hole-in-ones are achieved on par-3 holes. The typical course will have four such holes. So, the probability of a hole-in-one on a par 3 hole is (1/12500) × (4/18) = 1 in 2,778.
That said, the probability that exactly two golfers out of four get a hole-in-one on a par 3 hole is combin(4,2) × (1/12500)2 × (12449/12500)2 = 1 in 1,286,935.
Assuming four par-3 holes in a round, the probability of two hole-in-ones in the same hole are 4 × (1/1,286,935) = 1 in 321,734.
Where Golf Digest seems to have gone wrong is in assuming every hole has the same probability of a hole in one.
This question is raised and discussed in my forum at Wizard of Vegas.