The problem states that team winning percentages are random, with an independent, uniform distribution between 0% and 100%. This problem statement is kind of nonsensical, because team winning percentages can't be independent since overall the league has to have an average winning percentage of 50%. Every time one team wins their opponent loses. You can't have a league where every team has a winning percentage over 50%, for example. Is independent really the correct problem statement?
The key words here are "For simplicity." If you want to solve a more realistic version of the puzzle in which teams play each other a large equal number of times and each team has a 50 percent chance of winning, then go for it! I'll try to include in the following week's write-up.
No. I simplified the problem by saying that teams have random winning percentages between 0 and 100 percent. Of course, it's impossible to have all teams above 50 percent. But if you assume they all follow the same distribution, and that no two teams have the same record, you'll get the same result.
That can't be right. Consider the extreme case where all games are intra-divisional. Then it's impossible for the worst team in one division to be better than the best team in the other. The same holds if every game except one is in division.
Your assumption seems to be that every total order on the set of ten teams is equally probable, but that can't be the case if the division system affects the schedule.
Hmmm…I might have mixed up my replies here. You get the same result as what’s in the puzzle if you assume all teams play all other teams with equal frequency, regardless of division.
Again, I simplified the problem here so that none of this matters.
In last week's solution, you write a linear combination as q·(p, 0) + (1−q)·(p, 1−p). Should this perhaps be q·(p, 0) + (1−q)·(0, 1−p) instead? (Or I might be misunderstanding.) Either way, thanks for the fun puzzles!
> The faces on one die can be whole numbers from one to six. But the faces on the other die can only be whole numbers from one to four.
I think you got this a bit wrong, the first die should be able to have any number (as stated its max value is 6, which means it's impossible to get a value over 10).
The problem states that team winning percentages are random, with an independent, uniform distribution between 0% and 100%. This problem statement is kind of nonsensical, because team winning percentages can't be independent since overall the league has to have an average winning percentage of 50%. Every time one team wins their opponent loses. You can't have a league where every team has a winning percentage over 50%, for example. Is independent really the correct problem statement?
The key words here are "For simplicity." If you want to solve a more realistic version of the puzzle in which teams play each other a large equal number of times and each team has a 50 percent chance of winning, then go for it! I'll try to include in the following week's write-up.
As long as there's symmetry in the puzzle, and as long as no two teams have the same record, you should get the same result anyway.
Do the teams all play each other an equal number of times no matter what division they are in?
No. I simplified the problem by saying that teams have random winning percentages between 0 and 100 percent. Of course, it's impossible to have all teams above 50 percent. But if you assume they all follow the same distribution, and that no two teams have the same record, you'll get the same result.
That can't be right. Consider the extreme case where all games are intra-divisional. Then it's impossible for the worst team in one division to be better than the best team in the other. The same holds if every game except one is in division.
Your assumption seems to be that every total order on the set of ten teams is equally probable, but that can't be the case if the division system affects the schedule.
Hmmm…I might have mixed up my replies here. You get the same result as what’s in the puzzle if you assume all teams play all other teams with equal frequency, regardless of division.
Again, I simplified the problem here so that none of this matters.
In last week's solution, you write a linear combination as q·(p, 0) + (1−q)·(p, 1−p). Should this perhaps be q·(p, 0) + (1−q)·(0, 1−p) instead? (Or I might be misunderstanding.) Either way, thanks for the fun puzzles!
Good catch -- fixed!
> The faces on one die can be whole numbers from one to six. But the faces on the other die can only be whole numbers from one to four.
I think you got this a bit wrong, the first die should be able to have any number (as stated its max value is 6, which means it's impossible to get a value over 10).
Thanks for that sanity check (oof). Updated!