tag:blogger.com,1999:blog-264226589944705290.post5846210055879086885..comments2021-12-14T05:53:12.175-08:00Comments on God Plays Dice: Reconstructing World Cup resultsMichael Lugohttp://www.blogger.com/profile/15671307315028242949noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-264226589944705290.post-74567420198132831532010-06-27T02:34:45.102-07:002010-06-27T02:34:45.102-07:00It's no surprise that when teams tie in points...It's no surprise that when teams tie in points, it will often be impossible to reconstruct the individual match results from the points -- the usual difficulty of nontrivial automorphism groups that mathematicians are well-accustomed to struggling with!<br /><br />So it's probably worth mentioning that in four-team groups with 3 points for a win and 1 point for a draw, if the four teams end up with distinct point totals then it's always possible to reconstruct who won (or drew) each match. <br /><br />This fails with five-team groups and any point assignments. For instance even if you can tell that the five individual records were:<br /><br />A: 3W 1T <br />B: 2W 2L<br />C: 1W 2T 1L<br />D: 1W 1T 2L<br />E: 1W 3L<br /><br />then E could have beaten any of B, C, D, and there's exactly one set of outcomes consistent with each of those possibilities.ztbbhttps://www.blogger.com/profile/12186937753366214272noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-46810062679505564262010-06-24T13:35:04.319-07:002010-06-24T13:35:04.319-07:00Sune, you cannot determine wins, losses, and draws...Sune, you cannot determine wins, losses, and draws from just the points if there are 5 or more teams per group. There are tons of different point distributions that demonstrate this, but I like 10-4-4-4-4. In that case, one team drew all the rest, but you can't tell which.<br /><br />In some cases, like 6-6-6-3-3, you can't even determine the win-loss digraph up to isomorphism.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-46164518624215652182010-06-23T23:50:27.924-07:002010-06-23T23:50:27.924-07:00"In fact you can always reconstruct the numbe..."In fact you can always reconstruct the number of wins, draws, and losses from the number of points, except in the case of three points, which can be a win and two losses, or three draws."<br />Even in this case, it is possible to reconstruct the number of wins, draws, and losses for each team, if you know the other teams points: The total number of points given for a win is 3 and for a draw it is 2, so you can calculate the number of draws in the group. This way, you can calculate how many of the teams with 3 point who got them by playing 3 draws. And this must be either all the teams with 3 point or none of them, since if a team plays 3 draws, all the other teams played at least one draw.<br />Is the same thing possible for 5 teams?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-88690314760869813892010-06-23T14:19:55.616-07:002010-06-23T14:19:55.616-07:00According to my (computerized) efforts so far, non...According to my (computerized) efforts so far, none of Groups A through D in this World Cup is uniquely determined by the score table. Groups A, C, and D have two possibilities (for the scores of *all* of the matches), while Group B has three possibilities.<br /><br />According to the same program, the number of different possibilities for the match scores for Groups A through H in the previous World Cup are 11, 1, 5, 3, 2, 6, 1, and 8. That is, I claim only Groups B and G are uniquely recoverable.<br /><br />This is modulo potential errors and the fact that I'm not using advanced tie-breaking reasoning. I hope it's right ...Anonymousnoreply@blogger.com