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World Cup combinatorics again

June 16, 2026 at 10:12 PM by Dr. Drang

At the end of yesterday’s post about the combinatorics of the World Cup group stage, I said “this is good enough for me.” Turns out that was a lie. Today I rewrote some of the code to come up with a slightly more detailed result.

The last bit of work in that post was to generate the 40 unique point totals that can come from a group. These are the possible totals, with no regard for which team gets which total:

9 6 3 0     7 6 3 1     7 3 2 2     5 5 3 2
9 6 1 1     7 6 2 1     6 6 6 0     5 5 3 1
9 4 4 0     7 5 4 0     6 6 4 1     5 5 2 2
9 4 3 1     7 5 3 1     6 6 3 3     5 4 4 3
9 4 2 1     7 5 2 1     6 5 4 1     5 4 4 2
9 3 3 3     7 4 4 1     6 5 2 2     5 4 3 2
9 2 2 2     7 4 3 3     6 4 4 3     5 3 3 2
7 7 3 0     7 4 3 2     6 4 4 2     4 4 4 4
7 7 1 1     7 4 3 1     5 5 5 0     4 4 4 3
7 6 4 0     7 4 2 2     5 5 4 1     3 3 3 3

As discussed early in yesterday’s post, there are 36=729 ways the six games of a group can turn out, so today’s script associates each of the above totals with the game results that add up to that total. I built a dictionary in which the keys are the totals shown above—expressed as tuples, like (7, 7, 1, 1)—and the values are lists of lists of game result lists, like

[[[3, 0, 0, 0],
  [3, 0, 0, 0],
  [1, 0, 0, 1],
  [0, 1, 1, 0],
  [0, 0, 0, 3],
  [0, 0, 0, 3]],
 [[3, 0, 0, 0],
  [1, 0, 1, 0],
  [3, 0, 0, 0],
  [0, 0, 3, 0],
  [0, 1, 0, 1],
  [0, 0, 3, 0]],
 [[0, 3, 0, 0],
  [0, 0, 3, 0],
  [1, 0, 0, 1],
  [0, 1, 1, 0],
  [0, 3, 0, 0],
  [0, 0, 3, 0]],
 [[0, 3, 0, 0],
  [1, 0, 1, 0],
  [0, 0, 0, 3],
  [0, 3, 0, 0],
  [0, 1, 0, 1],
  [0, 0, 0, 3]],
 [[1, 1, 0, 0],
  [3, 0, 0, 0],
  [3, 0, 0, 0],
  [0, 3, 0, 0],
  [0, 3, 0, 0],
  [0, 0, 1, 1]],
 [[1, 1, 0, 0],
  [0, 0, 3, 0],
  [0, 0, 0, 3],
  [0, 0, 3, 0],
  [0, 0, 0, 3],
  [0, 0, 1, 1]]]

The innermost lists in this output are the points accumulated in a single game. For example, [1, 1, 0, 0] represents a game in which the first two teams in the group played to a draw. So the above result tells us that there are six ways the six games in a group can lead to a (7, 7, 1, 1) table at the end of the stage.

Here’s the Python code that built the dictionary and printed out a summary of the results:

python:
 1:  #!/usr/bin/env python3
 2:  
 3:  from itertools import product
 4:  from collections import defaultdict
 5:  import numpy as np
 6:  
 7:  # Possible point distributions from each game.
 8:  game1 = [[3, 0, 0, 0], [0, 3, 0, 0], [1, 1, 0, 0]]
 9:  game2 = [[3, 0, 0, 0], [0, 0, 3, 0], [1, 0, 1, 0]]
10:  game3 = [[3, 0, 0, 0], [0, 0, 0, 3], [1, 0, 0, 1]]
11:  game4 = [[0, 3, 0, 0], [0, 0, 3, 0], [0, 1, 1, 0]]
12:  game5 = [[0, 3, 0, 0], [0, 0, 0, 3], [0, 1, 0, 1]]
13:  game6 = [[0, 0, 3, 0], [0, 0, 0, 3], [0, 0, 1, 1]]
14:  
15:  # Enumerate all possible six-game results, and put them in a dictionary.
16:  # The keys are the sorted point totals and the values are the list of
17:  # game results.
18:  games = defaultdict(list)
19:  results = np.array(list(product(game1, game2, game3, game4, game5, game6)))
20:  for r in results:
21:    points = tuple(sorted(r.sum(axis=0).tolist(), reverse=True))
22:    games[points].append(r.tolist())
23:  
24:  # Show how many game results lead to each tuple of sorted point totals.
25:  count = 0
26:  p = sorted(games.keys(), reverse=True)
27:  for k in p:
28:    count += len(games[k])
29:    print(f'({" ".join(str(p) for p in k)}): {len(games[k]):3d}')
30:  print(f'    Total: {count:3d}')

The output is

(9 6 3 0):  24
(9 6 1 1):  12
(9 4 4 0):  12
(9 4 3 1):  24
(9 4 2 1):  24
(9 3 3 3):   8
(9 2 2 2):   4
(7 7 3 0):  12
(7 7 1 1):   6
(7 6 4 0):  24
(7 6 3 1):  24
(7 6 2 1):  24
(7 5 4 0):  24
(7 5 3 1):  24
(7 5 2 1):  24
(7 4 4 1):  36
(7 4 3 3):  24
(7 4 3 2):  24
(7 4 3 1):  24
(7 4 2 2):  24
(7 3 2 2):  12
(6 6 6 0):   8
(6 6 4 1):  24
(6 6 3 3):  24
(6 5 4 1):  24
(6 5 2 2):  12
(6 4 4 3):  36
(6 4 4 2):  24
(5 5 5 0):   4
(5 5 4 1):  24
(5 5 3 2):  12
(5 5 3 1):  12
(5 5 2 2):  12
(5 4 4 3):  24
(5 4 4 2):  24
(5 4 3 2):  24
(5 3 3 2):  12
(4 4 4 4):   6
(4 4 4 3):   8
(3 3 3 3):   1
    Total: 729

where the numbers after the colons are the number of ways the games can go to result in the given totals. There is, for example, only one way to get (3, 3, 3, 3), which is for every game to end in a draw.

The code is not clever in any way. I build the games dictionary (actually a defaultdict) one step at a time via the loop in Lines 20–22. As with yesterday’s script, I generate all the 729 game results using the product function from the itertools library and use NumPy’s sum function to add up the points from each of those results. The output comes from Lines 25–30.

If I run this script in an interactive environment, like IPython or Jupyter, I can pull out the games that lead to any set of points. I think the (5, 4, 3, 2) total is fun because it’s the only straight among the 40 possibilities. After some reformatting, here’s what games[(5, 4, 3, 2)] returns:

(3 0 0 0) (0 0 3 0) (1 0 0 1) (0 1 1 0) (0 1 0 1) (0 0 1 1) 
(3 0 0 0) (1 0 1 0) (0 0 0 3) (0 1 1 0) (0 1 0 1) (0 0 1 1) 
(3 0 0 0) (1 0 1 0) (1 0 0 1) (0 3 0 0) (0 1 0 1) (0 0 1 1) 
(3 0 0 0) (1 0 1 0) (1 0 0 1) (0 1 1 0) (0 3 0 0) (0 0 1 1) 
(0 3 0 0) (3 0 0 0) (1 0 0 1) (0 1 1 0) (0 1 0 1) (0 0 1 1) 
(0 3 0 0) (1 0 1 0) (3 0 0 0) (0 1 1 0) (0 1 0 1) (0 0 1 1) 
(0 3 0 0) (1 0 1 0) (1 0 0 1) (0 0 3 0) (0 1 0 1) (0 0 1 1) 
(0 3 0 0) (1 0 1 0) (1 0 0 1) (0 1 1 0) (0 0 0 3) (0 0 1 1) 
(1 1 0 0) (3 0 0 0) (0 0 0 3) (0 1 1 0) (0 1 0 1) (0 0 1 1) 
(1 1 0 0) (3 0 0 0) (1 0 0 1) (0 0 3 0) (0 1 0 1) (0 0 1 1) 
(1 1 0 0) (3 0 0 0) (1 0 0 1) (0 1 1 0) (0 1 0 1) (0 0 3 0) 
(1 1 0 0) (0 0 3 0) (3 0 0 0) (0 1 1 0) (0 1 0 1) (0 0 1 1) 
(1 1 0 0) (0 0 3 0) (1 0 0 1) (0 3 0 0) (0 1 0 1) (0 0 1 1) 
(1 1 0 0) (0 0 3 0) (1 0 0 1) (0 1 1 0) (0 1 0 1) (0 0 0 3) 
(1 1 0 0) (1 0 1 0) (3 0 0 0) (0 1 1 0) (0 0 0 3) (0 0 1 1) 
(1 1 0 0) (1 0 1 0) (3 0 0 0) (0 1 1 0) (0 1 0 1) (0 0 0 3) 
(1 1 0 0) (1 0 1 0) (0 0 0 3) (0 1 1 0) (0 3 0 0) (0 0 1 1) 
(1 1 0 0) (1 0 1 0) (0 0 0 3) (0 1 1 0) (0 1 0 1) (0 0 3 0) 
(1 1 0 0) (1 0 1 0) (1 0 0 1) (0 3 0 0) (0 0 0 3) (0 0 1 1) 
(1 1 0 0) (1 0 1 0) (1 0 0 1) (0 3 0 0) (0 1 0 1) (0 0 3 0) 
(1 1 0 0) (1 0 1 0) (1 0 0 1) (0 0 3 0) (0 3 0 0) (0 0 1 1) 
(1 1 0 0) (1 0 1 0) (1 0 0 1) (0 0 3 0) (0 1 0 1) (0 0 0 3) 
(1 1 0 0) (1 0 1 0) (1 0 0 1) (0 1 1 0) (0 3 0 0) (0 0 0 3) 
(1 1 0 0) (1 0 1 0) (1 0 0 1) (0 1 1 0) (0 0 0 3) (0 0 3 0)

Notice that in every set of six games (i.e., each row), there are two games that end in a win and four that end in a draw. The team with five points has a win and two draws; the team with four points has a win, a draw, and a loss; the team with three points has three draws; and the team with two points has two draws and a loss.

OK, now I’m done.

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