Generate multiple random numbers to equal a value

Question

So here is the deal: I want to (for example) generate 4 pseudo-random numbers, that when added together would equal 40. How could this be dome in python? I could generate a random number 1-40, then generate another number between 1 and the remainder,etc, but then the first number would have a greater chance of "grabbing" more.

Answer 1

Here's the standard solution. It's similar to Laurence Gonsalves' answer, but has two advantages over that answer.

1. It's uniform: each combination of 4 positive integers adding up to 40 is equally likely to come up with this scheme.

and

2. it's easy to adapt to other totals (7 numbers adding up to 100, etc.)

import random

def constrained_sum_sample_pos(n, total):
    """Return a randomly chosen list of n positive integers summing to total.
    Each such list is equally likely to occur."""

    dividers = sorted(random.sample(xrange(1, total), n - 1))
    return [a - b for a, b in zip(dividers + [total], [0] + dividers)]

Sample outputs:

>>> constrained_sum_sample_pos(4, 40)
[4, 4, 25, 7]
>>> constrained_sum_sample_pos(4, 40)
[9, 6, 5, 20]
>>> constrained_sum_sample_pos(4, 40)
[11, 2, 15, 12]
>>> constrained_sum_sample_pos(4, 40)
[24, 8, 3, 5]

Explanation: there's a one-to-one correspondence between (1) 4-tuples (a, b, c, d) of positive integers such that a + b + c + d == 40, and (2) triples of integers (e, f, g) with 0 < e < f < g < 40, and it's easy to produce the latter using random.sample. The correspondence is given by (e, f, g) = (a, a + b, a + b + c) in one direction, and (a, b, c, d) = (e, f - e, g - f, 40 - g) in the reverse direction.

If you want nonnegative integers (i.e., allowing 0) instead of positive ones, then there's an easy transformation: if (a, b, c, d) are nonnegative integers summing to 40 then (a+1, b+1, c+1, d+1) are positive integers summing to 44, and vice versa. Using this idea, we have:

def constrained_sum_sample_nonneg(n, total):
    """Return a randomly chosen list of n nonnegative integers summing to total.
    Each such list is equally likely to occur."""

    return [x - 1 for x in constrained_sum_sample_pos(n, total + n)]

Graphical illustration of constrained_sum_sample_pos(4, 10), thanks to @FM. (Edited slightly.)

0 1 2 3 4 5 6 7 8 9 10  # The universe.
|                    |  # Place fixed dividers at 0, 10.
|   |     |      |   |  # Add 4 - 1 randomly chosen dividers in [1, 9]
  a    b      c    d    # Compute the 4 differences: 2 3 4 1

Answer 2

There are only 37^4 = 1,874,161 arrangements of four integers in the range [1,37] (with repeats allowed). Enumerate them, saving and counting the permutations that add up to 40. (This will be a much smaller number, N).

Draw uniformly distributed random integers K in the interval [0, N-1] and return the K-th permutation. This can easily be seen to guarantee a uniform distribution over the space of possible outcomes, with each sequence position identically distributed. (Many of the answers I'm seeing will have the final choice biased lower than the first three!)

Answer 3

from numpy.random import multinomial
multinomial(40, [1/4.] * 4)

Answer 4

Assuming you want them to be uniformly distributed, and assuming you don't want repeats

addends = []
picks = range(1, 34)
while sum(addends) != 40:
    addends = random.sample(picks, 3)
    if sum(addends) > 39:
        continue
    addends.append(40 - sum(addends))

Answer 5

b = random.randint(2, 38)
a = random.randint(1, b - 1)
c = random.randint(b + 1, 39)
return [a, b - a, c - b, 40 - c]

(I assume you wanted integers since you said "1-40", but this could be easily generalized for floats.)

Here's how it works:

• cut the total range in two randomly, that's b. The odd range is because there are going to be at least 2 below the midpoint and at least 2 above. (This comes from your 1 minimum on each value).
• cut each of those ranges in two randomly. Again, the bounds are to account for the 1 minimum.
• return the size of each slice. They'll add up to 40.

Answer 6

Generate 4 random numbers, compute their sum, divide each one by the sum and multiply by 40.

If you want Integers, then this will require a little non-randomness.