I read on Wikipedia that the number of combinations leading to a tie between the second and third place in an initial group (of four teams) in the FIFA World Cup is $207$. How is this calculated?

I understand that if we consider all possible outcomes (win, draw, loss) for all six matches in a group of four teams, there are $3^{6}$ combinations possible. Only one team out of the four in the group can win all three games, and so score $9$ points ($3$ points are allocated for a win, $1$ for a draw and $0$ for a loss). Let's say one team in this group loses all three its games, and the remaining two teams each has one win and one draw, then we have a tie between second and third place, and there are twelve ways this can happen.

Similarly, the winning team in the group (say $t_0$) could win two games (beating $t_2$ and $t_3$ say) and draw one (with $t_1$). If there was a draw between $t_2$ and $t_3$, $t_1$ beat $t_2$ and $t_3$ beat $t_1$, $t_1$ and $t_3$ end up in a tie for second and third place with four points each. There are again twelve ways this can happen.

I can't quite get to 207 though?

  • $\begingroup$ What number did you get to, and how? $\endgroup$
    – Henry
    Commented Apr 5, 2023 at 13:01
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    $\begingroup$ So it's second and third place in the initial groups, not in the world cup as a whole, maybe you should make that a little clearer. $\endgroup$ Commented Apr 5, 2023 at 13:35
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    $\begingroup$ I think there are $36$ ways you can have 1 team with two wins one draw, 2 teams with one win one draw one loss, and 1 team with one draw two losses (a slight expansion of your second example). Overall I see $14$ points distributions leading to your second/third place issue and indeed $207$ overall possibilities out of the $40$ points distributions with $729$ in total $\endgroup$
    – Henry
    Commented Apr 5, 2023 at 14:55

1 Answer 1


I doubt that you can avoid a rather tedious case-by-case analysis, which is best checked with a computer.

Suppose the teams are called A, B, C, D, and that we list the games in the order A–B, A–C, A–D, B–C, B–D, C–D. Then, for example, the results 11XX11, 1X11X1 and X1121X (and no others) lead to the teams A, B, C, D obtaining 7, 4, 4, 1 points, respectively. This then tells you all the ways of getting a table where the best team has 7 points, the second and third are tied at 4 points, and the last team has 1 point, namely you can choose which teams that should “play the roles of A and D” above, and this can be done in 4·3=12 ways. For each of these 12 choices, you have the three scenarios above, so you get 3·12=36 ways of having a tied situation with the distribution of points (7,4,4,1).

Below is the complete list of possible outcomes with a tie for second and third place. The listed game results are the ones that will result in A, B, C, D (in that order) having the points listed, and then you have to multiply that number of cases by the suitable combinatorial factor to get the stated total number of ways of getting that distribution of points with the teams in any order.

  • XXXXXX (3,3,3,3) 1 way
  • 12X1XX / 21X2XX (4,4,4,3) 8 ways
  • 1X21X1 / 21XX21 / X1221X / X2112X / 12XX12 / 2X12X2 / 2X12X2 (4,4,4,4) 6 ways
  • XX1XXX (5,3,3,2) 12 ways
  • 1XX1X1 / X1X21X (5,4,4,2) 24 ways
  • XX1121 / XX1212 (5,4,4,3) 24 ways
  • XX1X11 (5,5,5,0) 4 ways
  • 1211XX / 2112XX (6,4,4,2) 24 ways
  • 112X11 / 211X21 / 121X12 (6,4,4,3) 36 ways
  • 121111 / 211211 (6,6,6,0) 8 ways
  • 11XX11 / 1X11X1 / X1121X (7,4,4,1) 36 ways
  • 111XXX (9,2,2,2) 4 ways
  • 111121 / 111212 (9,3,3,3) 8 ways
  • 111X11 (9,4,4,0) 12 ways

In all, 1+8+6+12+24+24+4+24+36+8+36+4+8+12 = 207 ways.


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