如何获得多项式情况下每个可能结果的概率？

Question

Using stats.binom.pmf(np.arange(0,11),10,0.5) , I can get an array of numbers representing the probability of an event happening n times(0 to 10) in 10 times given the probability of a single occurrence 0.5, like below:

array([0.00097656, 0.00976563, 0.04394531, 0.1171875 , 0.20507813,
       0.24609375, 0.20507813, 0.1171875 , 0.04394531, 0.00976563,
       0.00097656])

Now I want to get a similar output for a multinomial case.

Saying I'm rolling a dice three times, how can I get such a probability array representing the probability of the total points(3 to 18) using an existing package?

Answer 1

The trick is to consider the appropriate Laplace space, where each event has the same probability. In the case of three dice, this means considering the tuples (1, 1, 1), (1, 1, 2), ... (6, 6, 6) as elementary events. So you can count up how many of these tuples result in each of the sums from 3 to 18, and then divide those numbers by the total number of tuples to get the probabilities:

from itertools import product
from collections import defaultdict

counts = defaultdict(int)
for dice in product(range(1, 7), repeat=3):
    counts[sum(dice)] += 1 
    
probabilities = {n: counts[n] / 6 ** 3 for n in range(3, 19)}
print(probabilities)

{3: 0.004629629629629629,
 4: 0.013888888888888888,
 5: 0.027777777777777776,
 6: 0.046296296296296294,
 7: 0.06944444444444445,
 8: 0.09722222222222222,
 9: 0.11574074074074074,
 10: 0.125,
 11: 0.125,
 12: 0.11574074074074074,
 13: 0.09722222222222222,
 14: 0.06944444444444445,
 15: 0.046296296296296294,
 16: 0.027777777777777776,
 17: 0.013888888888888888,
 18: 0.004629629629629629}

Answer 2

To illustrate my comment, here is the function:

def discrete_uniform_sum_pmf(a, b, n):
    du_pmf = {i: 1/(b-a+1) for i in range(a, b+1)}
    du_sum_pmf = {0: 1}

    for i in range(n):
        new_sum_pmf = defaultdict(float)
        for prev_sum, dice in product(du_sum_pmf, du_pmf):
            new_sum_pmf[prev_sum + dice] += du_sum_pmf[prev_sum] * du_pmf[dice]
        du_sum_pmf = new_sum_pmf

    return du_sum_pmf

It would be a lot simpler with numpy and has a lot of space for improvement (using CDF to avoid nested loops, exploiting symmetry, etc.) - but should be sufficient to illustrate the idea.

The output of discrete_uniform_sum_pmf(1, 6, 3) is the same shown by Arne