How can I create three random integers which sum to a specific value? (Python)

Question

Let's say bob = 6

I want to create 3 random integers that have a sum of 106 (100 + whatever bob's original integer is. It could be 10, but in this case, it's 6).

I've got:

from random import *
bob = 6
bob1 = (randint(0,100))
bob2 = (randint(0,100))
bob3 = (randint(0,100))
print bob1
print bob2
print bob3

I can generate integers, but how can I make sure the sum of them = 100 + original integer? (106 total). If the sum doesn't = 106, then I want the script to keep going until it makes 106.

What if `bob=201`? how long would you like your script keep going until it makes 301? — gboffi, Apr 24 '16 at 14:11
Hi gboffi, i'm keeping it simple for now, but yes, in theory I would like the script to handle any number. But for now, i'm just keeping it basic. — BubbleMonster, Apr 24 '16 at 14:13

ayhan · Answer 1 · 2016-04-24T15:07:52.740

The general approach to generate numbers that add up to a certain number is like this:

import random
bob = 6
numbers = sorted(random.sample(range(100+bob), 2))
bob1 = numbers[0]
bob2 = numbers[1] - numbers[0]
bob3 = 100 + bob - numbers[1]

It selects two cut points between 0 and 100 + bob and assigns the numbers as illustrated:

This will also ensure all three numbers have the same distribution (simulated with 1m trials):

mean    34.700746   35.639730   35.659524
std     24.886456   24.862377   24.861724

As opposed to the numbers generated dependently:

mean    50.050665   27.863753   28.085582
std     29.141171   23.336316   23.552992

And their histograms:

score 3 · Accepted Answer · edited Apr 24 '16 at 14:20

3

Just calculate the third value:

from random import randint
bob = 6
bob1 = randint(0, 100)
bob2 = randint(0, min(100, 100 + bob - bob1))
bob3 = 100 + bob - bob1 - bob2
print bob1
print bob2
print bob3

edited Apr 24 '16 at 14:20

Delgan

18,571
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answered Apr 24 '16 at 14:07

Daniel

42,087
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Thanks this is what I wanted. I just had to add an extra ) closing bracket on line 4. But otherwise this seems perfect. Thanks. – BubbleMonster Apr 24 '16 at 14:12

score 0 · Answer 3 · answered Apr 24 '16 at 14:08

0

bob1 = (randint(0,100))
bob2 = (randint(0,(100-bob1)))
bob3 = 100 - (bob1 + bob2)

answered Apr 24 '16 at 14:08

Keiwan

8,031
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Reblochon Masque · Answer 4 · 2016-04-24T14:40:58.020

Here is a general function that will always randomly generate 3 numbers in [0, n] that add to n; as the intermediate values are determined by "bob"'s initial value, we pass it this value and a target total number. The function returns a tuple of 3 numbers that together with bob-initial-value add up to the target:

import random

def three_numbers_to_n(bob, target):
    n = target - bob
    a = random.randrange(0, n+1)
    b = random.randrange(a, n+1) - a
    c = n - a - b
    return a, b, c

for _ in range(5):
    bob = 6
    result = three_numbers_to_n(bob, 106)
    print(bob, result, sum(result) + bob)

sample output:

6 (13, 3, 84) 106
6 (45, 49, 6) 106
6 (27, 2, 71) 106
6 (44, 18, 38) 106
6 (100, 0, 0) 106

If you wish, you could random.shuffle(a, b, c) before returning to remove the predictability that the first number is likely larger than the second likely larger than the 3rd.

How can I create three random integers which sum to a specific value? (Python)

4 Answers4

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