How to generate a list of random numbers so their sum would be equal to a randomly chosen number

Question

I want to generate a list of random distribution of numbers so their sum would be equal to a randomly chosen number. For example, if randomly chosen number is 5, the distrib

Answer 1

Consider doing it continuously first. And for a moment we do not care about final number, so let's sample uniformly X_i in the interval [0...1] so that their sum is equal to 1

X_1 + X_2 + ... X_n = 1

This is well-known distribution called Dirichlet Distribution, or gamma variate, or simplex sampling. See details and discussion at Generating N uniform random numbers that sum to M. One can use random.gammavariate(a,1) or for correct handling of corners gamma variate with parameter 1 is equivalent exponential distribution, with direct sampling code below

def simplex_sampling(n):
    r = []
    sum = 0.0
    for k in range(0,n):
        x = random.random()
        if x == 0.0:
            return (1.0, make_corner_sample(n, k))

        t = -math.log(x)
        r.append(t)
        sum += t

    return (sum, r)

def make_corner_sample(n, k):
    r = []
    for i in range(0, n):
       if i == k:
           r.append(1.0)
       else:
           r.append(0.0)

    return r

So from simplex_sampling you have vector and the sum to be used as normalization.

Thus, to use it for, say, N=5

N = 5

sum, r = simplex_sampling(N)

norm = float(N)/sum

# normalization together with matching back to integers
result = []
for k in range(N):
    # t is now float uniformly distributed in [0.0...N], with sum equal to N
    t = r[k] * norm 
    # not sure if you could have zeros,
    # and check for boundaries might be useful, but
    # conversion to integers is trivial anyway:
    # values in [0...1) shall be converted to 0,
    # values in [1...2) shall be converted to 1, etc
    result.append( int(t) )