Get a Random Boolean by percentage

Question

I'm trying to get a get a random boolean but with a weighted percentage. For instance, I want the user to pass in a percentage (i.e. 60) and the generator will randomly select true 60% of the time.

What I have is this:

def reset(percent=50):
    prob = random.randrange(0,100)
    if prob > percent:
        return True
    else:
        return False

Is there a better way to do this? This seems inefficient and cumbersome. I don't need it to be perfect since it is just used to simulate data, but I need this to be as fast as possible.

I've searched (Google/SO) and have not found any other questions for this.

Possible duplicate of [True or false output based on a probability](http://stackoverflow.com/questions/5886987/true-or-false-output-based-on-a-probability) — Martin Thoma, Jan 28 '17 at 08:49

Martijn Pieters · Accepted Answer · 2014-09-24T21:34:04.670

26

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.

edited Sep 24 '14 at 21:34

answered Jan 14 '13 at 18:40

Martijn Pieters

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I'd also note that in general, it's more normal to accept a float from `0` to `1` - percentages are a useful human concept, but in computing, this will probably mean multiplying stuff up for no reason. A float will give more range in chances allowed, and probably be quicker (just use `random.random()` instead of `randrange()`). – Gareth Latty Jan 14 '13 at 18:43
@Lattyware: Certainly, but if user input is already given in integer values, it doesn't matter much. – Martijn Pieters Jan 14 '13 at 18:44
And I'm just gonna go with 'it's monday' – rjbez Jan 14 '13 at 18:44
@rjbez: Is it really? Golly! :-P – Martijn Pieters Jan 14 '13 at 18:45
@MartijnPieters Of course, I was just saying for the more general case. – Gareth Latty Jan 14 '13 at 18:46
@MartijnPieters I think the relation is backwards, see my answer, and make the change. – Gregory Sep 24 '14 at 21:29
@Gregory: You are right; I never focused on that because I focused instead on simplifying the method, not wether or not it was correct. Updated now. – Martijn Pieters Sep 24 '14 at 21:36
@MartijnPieters no worries, that's what i figured. Up voted your answer. – Gregory Sep 24 '14 at 22:15

score 2 · Answer 2 · answered Jan 14 '13 at 18:40

2

How about:

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

answered Jan 14 '13 at 18:40

ecatmur

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Gregory · Answer 3 · 2014-09-24T22:16:40.603

1

@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

edited Sep 24 '14 at 22:16

answered Sep 24 '14 at 21:28

Gregory

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Thanks, I suppose I should have added a comment about that. I did figure that out long ago and updated the code but not the question. – rjbez Oct 29 '14 at 12:59

score 0 · Answer 4 · answered Mar 28 '22 at 20:20

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]

Get a Random Boolean by percentage

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