If all teams were equal, the chance of a perfect 64-team March Madness bracket would be 1 in 2

^{63}and the chance of a perfect Premier League would be 1 in 20! - and, happily, 2

^{63}and 20! are actually very similar to each other (one's 10

^{18.3}and the other's 10

^{18.9}).

However, all teams are not equal. The chance of all 63 favourites winning their March Madness matches has been estimated by 538 at somewhere around the 1 in 10

^{9}to 1 in 10

^{10}region, which would seem to give an upper bound for the chance of a perfect bracket in practice. In a similar vein, how would you go about working out the chance a perfect Premier League prediction?

My approach is: chance of favourite winning * chance of (favourite of the remaining 19 winning among the remaining 19) * chance of (favourite of the remaining 18 winning among the remaining 18) * ... etc., and looking at bookmakers' odds for naming the top four in the correct order, for naming the bottom three in the correct order, naming the best-place finisher outside six favourites and so on, I'd estimate an upper bound of something like 1 in 10

^{11}to 1 in 10

^{12}.

Is this the sort of question that people here would enjoy thinking about? Is there a better way of doing it?

Quote:chris.m.dicksonA UK bookmaker is offering a promotion where people can bet at (up to) 25 million to 1 against being able to rank the twenty English Premier League soccer teams' finishing positions exactly correctly. (Each team plays each of the other 19 teams in the league at home and away over the course of the season.) There's a guaranteed prize that goes to the player who is closest at Christmas.

If all teams were equal, the chance of a perfect 64-team March Madness bracket would be 1 in 2^{63}and the chance of a perfect Premier League would be 1 in 20! - and, happily, 2^{63}and 20! are actually very similar to each other (one's 10^{18.3}and the other's 10^{18.9}).

However, all teams are not equal. The chance of all 63 favourites winning their March Madness matches has been estimated by 538 at somewhere around the 1 in 10^{9}to 1 in 10^{10}region, which would seem to give an upper bound for the chance of a perfect bracket in practice. In a similar vein, how would you go about working out the chance a perfect Premier League prediction?

My approach is: chance of favourite winning * chance of (favourite of the remaining 19 winning among the remaining 19) * chance of (favourite of the remaining 18 winning among the remaining 18) * ... etc., and looking at bookmakers' odds for naming the top four in the correct order, for naming the bottom three in the correct order, naming the best-place finisher outside six favourites and so on, I'd estimate an upper bound of something like 1 in 10^{11}to 1 in 10^{12}.

Is this the sort of question that people here would enjoy thinking about? Is there a better way of doing it?

Welcome to the forum, Chris. I think that's a very interesting question, and I hope you generate good discussion from the math guys on here.

Sometimes bookies do offer over the odds betting opportunities to get players to sign up - once Coral were offering 5/1 (max £5) that Man United would have a corner against QPR ( https://www.freebets.com/enhanced-odds/coral-5-1-man-united-to-win-a-corner ) - essentially free money!