Why does mean() of Numpy.randint(0,100) converge to 49.49?

Question

I wanted to test how evenly distributed the numpy randint function is, so I did this.

>>> a = np.random.randint(0, 100, 10000)
>>> a.mean()
`49.1685`
>>> a = np.random.randint(0, 100, 1000000)
>>> a.mean()
`49.494202`
>>> a = np.random.randint(0, 100, 1000000000)
>>> a.mean() <br>`49.49944384`

I was confused about why it was reaching 49.49 as an average. I figured someone else out there would have the same question.

Answer 1

Took me a minute to realize.
The range(0, 100) only includes numbers ranging from 0-99.
The code speaks for itself:

>>> a = np.random.randint(0, 100, 1000000000)
>>> a.mean()
`49.49944384` 
>>> a = np.random.randint(0, 101, 1000000000)
>>> a.mean()
`50.000402272`

Answer 2

Because np.random.randint(0, 100) generates a number between 0 (included) and 100 excluded , ie the max is 99.

So the mean is at (0 + 99) / 2 = 49.5 .

Illustration that upper bound is not included:

np.randint(0, 2, 100)   # only 0s and 1s

Side note: this behavior is consistent with various functions and syntax in Python. For instance, with the range function:

[i for i in range(0, 5)]    # [0, 1, 2, 3, 4]