Well, this is similar to asking how many times in a day does the hour hand point to 12.
If one meant "in a day" to mean a single instance of a day, then the end points of the day are included and the answer is 3 because it starts at 12, passes 12 at noon, and ends at 12.
But, if "in a day" was meant to mean over some series of days, then you have to adjust the idea of a day length so as to not count the 12's twice as the ending of one day and the beginning of the next. Each day length interval has one end point included and the other open... with an additional interval end point either at the beginning or end of the series.
So you set a convention that either says 12 is the beginning of a day, or 12 is the end of a day when two days are contiguous. That gives each day 2 12's but leaves on extra 12 either at the beginning or end of the series of days, depending on which convention you choose.
So for one day you would have 3 12's (1x2)+1... (stealing the formula from below, but this is not how the calculation would be for the one day instance; but its consistent)
For two days, 5 12's (2x2)+1
three days, 7 12's (3x2)+1
So number of days=N, then number of 12's is (Nx2)+1
The series trends to 2/day...
1 day->3/1=3
2 days->5/2=2.5
3 day->7/3=2.3333
100 days-> (201 12's)/100=2.01
1000 days-> (2001)/1000=2.001
10,000 days->20,001/10,000=2.0001
...where the indefinite length of the series of days allows the number of 12's "per/day" to be general ("2"), for any finite segment of the indefinite series where either convention gives each day one inclusive and one exclusive end point for its length interval.
But that must be seen similar to a "rate". The number of actual 12's counted in a finite series of days standing apart from the indefinite series must include the additional 12 that is the beginning or end of the finite series, depending on the convention chosen.