This week's FiveThirtyEight Riddler asks the following question:
Assuming we don’t know anything about the strengths of the teams in the tournament, what are the chances that any pair of teams in a 32-team World Cup plays each other twice? (Given the way the World Cup bracket works, their first encounter would be in the round-robin, three-game group stage, with their second encounter in the final or the third-place game.)
Extra credit: What would those chances have been given FiveThirtyEight’s pre-tournament odds for this year’s field?Thanks to my pre-built World Cup simulator, it was pretty easy to adapt it to answer these questions. I set every team's strength equal and removed home field advantage (maybe this tournament is played on the moon) so that every single game is equal. After 100,000 simulations, the odds of having ANY two teams meet again in either the final and/or third place game is about 22% (this actually isn't too hard to calculate mathematically either; more on that later).
For the "extra credit" piece, I took FiveThirtyEight's pre-tournament rankings and ran them through my simulator as well. I used 0.5 goals for home field and an assumed standard deviation in scores of 1.62 (which appears to match FiveThirtyEight's assumptions after I fit my probability formulas against their output). Surprisingly, I actually got a different answer from them after 100,000 runs: 19.5%. I suspect my decreased likelihood is due to an assumption of more randomness in matches (thus decreasing the chance of the strongest teams meeting again).
The top frequencies of rematches (from before the start of the tournament) were as follows (with the top 5 outcomes still possible):
Now for the straight math answer. Since all opponents are exactly the same, you can assume any two teams make it out of the group, and all pairs of teams that advance have already played each other once in the round robin.
So the only math you need to do is on the bracket. From there, every team has a 50% chance of winning each game, which means any one team has a 12.5% of making the final or third place game. You can fix one team on one side of the bracket, and just need the probability that their counterpart makes it, which is 12.5% as well. With both the championship and third place match possible, you have 2 chances, or 12.5% + 12.5%. But these two events can also happen at the same time if a rematch happens in the final AND the third place game: 25%*25%*50% = 3.125%. So you have to subtract that occurrence: 12.5% + 12.5% - 3.125% = 21.875%.