Get a Random Boolean by percentage

Just return the test:

def reset(percent=50):
    return random.randrange(100) < percent

because the result of a < lower than operator is already a boolean. You do not need to give a starting value either.

Note that you need to use lower than if you want True to be returned for a given percentage; if percent = 100 then you want True all of the time, e.g. when all values that randrange() produces are below the percent value.

How about:

def reset(percent=50):
    return random.randrange(0, 100) > percent

@rjbez You asked for a function which returns true in X percentage of time. When X == 0% it should always return false. When X == 100% it should always return true. The current accepted answer is now fixed and has the proper relation

def reset(percent=50):
    return random.randrange(100) < percent

Just use PyProbs library. It is very easy to use.

>>> from PyProbs import Probability as pr
>>> 
>>> # You can pass float (i.e. 0.5, 0.157), int (i.e. 1, 0) or str (i.e. '60%', '3/11')
>>> pr.Prob('50%')
False
>>> pr.Prob('50%', num=5)
[False, False, False, True, False]