计算三个骰子的频率

Question

I need to create a 6 x 6 x 6 table that has all the 216 possible three dice results. Considering that the same dice values in whatever order they come is the same result (eg (1,4,6) is the same as all the 1, 4, 6 transformations, ie (6,4,1) or (4,6,1), and so on).

Then I want to calculate the frequency of different results and print them sorted by increasing incidence.(maybe with dictionary?)

I made this list:

mylist = [[[(x,y,z) for z in range (1,7)]for y in range(1,7)]for x in range (1,7)]

and I have all the 216 possible results.

But I can't make it work for counting them and find the same results...

Can you help me?

Answer 1

I think you do not have to use 3-dimensional table in this case, I would do it following way

mylist = []
for x in range(1,7):
    for y in range(1,7):
        for z in range(1,7):
            mylist.append((x,y,z))
mylist = [tuple(sorted(i)) for i in mylist]
for i in list(set(mylist)):
    print(i,'occurs',mylist.count(i),'times')

Answer 2

You can use itertools.product to produce all the possible results from rolling 3 dices, pass the results to collections.Counter and then to frozenset to count the composition of the results while ignoring orders, and then use the Counter.elements() method to reconstruct the counts back to readable tuples as keys in a dict comprehension:

from itertools import product
from collections import Counter
{tuple(sorted(Counter(dict(t)).elements())): c for t, c in Counter(frozenset(Counter(p).items()) for p in product(range(1, 7), repeat=3)).items()}

This returns:

{(1, 1, 1): 1,
 (1, 1, 2): 3,
 (1, 1, 3): 3,
 (1, 1, 4): 3,
 (1, 1, 5): 3,
 (1, 1, 6): 3,
 (1, 2, 2): 3,
 (1, 2, 3): 6,
 (1, 2, 4): 6,
 (1, 2, 5): 6,
 (1, 2, 6): 6,
 (1, 3, 3): 3,
 (1, 3, 4): 6,
 (1, 3, 5): 6,
 (1, 3, 6): 6,
 (1, 4, 4): 3,
 (1, 4, 5): 6,
 (1, 4, 6): 6,
 (1, 5, 5): 3,
 (1, 5, 6): 6,
 (1, 6, 6): 3,
 (2, 2, 2): 1,
 (2, 2, 3): 3,
 (2, 2, 4): 3,
 (2, 2, 5): 3,
 (2, 2, 6): 3,
 (2, 3, 3): 3,
 (2, 3, 4): 6,
 (2, 3, 5): 6,
 (2, 3, 6): 6,
 (2, 4, 4): 3,
 (2, 4, 5): 6,
 (2, 4, 6): 6,
 (2, 5, 5): 3,
 (2, 5, 6): 6,
 (2, 6, 6): 3,
 (3, 3, 3): 1,
 (3, 3, 4): 3,
 (3, 3, 5): 3,
 (3, 3, 6): 3,
 (3, 4, 4): 3,
 (3, 4, 5): 6,
 (3, 4, 6): 6,
 (3, 5, 5): 3,
 (3, 5, 6): 6,
 (3, 6, 6): 3,
 (4, 4, 4): 1,
 (4, 4, 5): 3,
 (4, 4, 6): 3,
 (4, 5, 5): 3,
 (4, 5, 6): 6,
 (4, 6, 6): 3,
 (5, 5, 5): 1,
 (5, 5, 6): 3,
 (5, 6, 6): 3,
 (6, 6, 6): 1}

Answer 3

One way to do it using the stuff I've mentioned:

from collections import Counter

a = [(i,j,k) for i in range(1, 7) for j in range(1, 7) for k in range(1, 7)]
c = Counter(frozenset(x) for x in a if len(set(x)) == 3)

counter_most_common = c.most_common(1)[0][1]
for combination, counter in c.items():
    if counter == counter_most_common:
        print(f"{combination} is one of the most common combinations! It exists {counter} times")

As you can see, they're all equally likely, which makes perfect sense if you think about it