That was a thread in the poll board. Whether or not all 12 cupless teams will win by 2050. I decided to do a little probability. There are 2 assumptions that I know are false but were made so that this exercise would be plausible and simple:
1) I assume a 1/30 probability for all teams to win a cup in a given season, even though I know that's silly to assume that, it's probably worse odds that it'll be that simple given how long it takes bottom feeders (e.g. Buffalo and Florida) to become cup contenders. Really I don't think you can possibly get this probability.
2) I used a sample of 37 years even though the playoffs already started this year. Just much easier.
So without further ado I'd be interested if someone that understand probability well can tell me if my methodology is correct:
1) There's a possibility of all 12 playoff-less teams winning once to satisfy this query. However there's also a chance that there will be all 12 playoff-less teams winning once and then another winning a second time, with 24 teams that currently have cups occupying the other years. This can go on to 12 current cupless teams winning once and 25 current cupless years winning in the rest of the 25 years. So 12 current cupless teams and 25 current cup winners to 12 current cupless teams and the rest being repeats of any of those cupless teams. So these are the possibilities for me:
12 first time cup winners and 25 current cup teams:
(12/30)*(11/30)*)(10/30)*(9/30)*(8/30)*(7/30)*(6/30)*(5/30)*(4/30)*(3/30)*(2/30)*(1/30)*(18/30)^25
12 first time cup winners and 1 repeat of those 12 first time cup winners and 24 current cup teams:
(12/30)*(11/30)*)(10/30)*(9/30)*(8/30)*(7/30)*(6/30)*(5/30)*(4/30)*(3/30)*(2/30)*(1/30)*(12/30)*(18/30)^24
12 first time cup winners and 2 repeats of those 12 first time cup winners and 23 current cup teams:
(12/30)*(11/30)*)(10/30)*(9/30)*(8/30)*(7/30)*(6/30)*(5/30)*(4/30)*(3/30)*(2/30)*(1/30)*(12/30)^2*(18/30)^23
.
.
.
12 first time cup winners and 25 current cupless teams repeating:
(12/30)*(11/30)*)(10/30)*(9/30)*(8/30)*(7/30)*(6/30)*(5/30)*(4/30)*(3/30)*(2/30)*(1/30)*(12/30)^25
Each one I did a combination of to see what the combo was of current cupless wins out of all 37 years. If it's only 12 current cupless teams I did (37!)/[(25!)(12!)]. If it's 12 current cupless teams, 1 repeat of a cupless teams, and 24 current cup teams I did a combination of 13 out of 37, or (37!)/[(24!)(13!)]. I kept going until I had 37 current cupless teams with 12 unique cupless teams and the rest as repeats of those current cupless teams, which was the combination of 37 out of 37, or 1. For each combination I multiplied each probability up top and added everything together.
Result: Probability is 0.0001047584 and odds are 1 in about 9,546.
Does that seem about right?
1) I assume a 1/30 probability for all teams to win a cup in a given season, even though I know that's silly to assume that, it's probably worse odds that it'll be that simple given how long it takes bottom feeders (e.g. Buffalo and Florida) to become cup contenders. Really I don't think you can possibly get this probability.
2) I used a sample of 37 years even though the playoffs already started this year. Just much easier.
So without further ado I'd be interested if someone that understand probability well can tell me if my methodology is correct:
1) There's a possibility of all 12 playoff-less teams winning once to satisfy this query. However there's also a chance that there will be all 12 playoff-less teams winning once and then another winning a second time, with 24 teams that currently have cups occupying the other years. This can go on to 12 current cupless teams winning once and 25 current cupless years winning in the rest of the 25 years. So 12 current cupless teams and 25 current cup winners to 12 current cupless teams and the rest being repeats of any of those cupless teams. So these are the possibilities for me:
12 first time cup winners and 25 current cup teams:
(12/30)*(11/30)*)(10/30)*(9/30)*(8/30)*(7/30)*(6/30)*(5/30)*(4/30)*(3/30)*(2/30)*(1/30)*(18/30)^25
12 first time cup winners and 1 repeat of those 12 first time cup winners and 24 current cup teams:
(12/30)*(11/30)*)(10/30)*(9/30)*(8/30)*(7/30)*(6/30)*(5/30)*(4/30)*(3/30)*(2/30)*(1/30)*(12/30)*(18/30)^24
12 first time cup winners and 2 repeats of those 12 first time cup winners and 23 current cup teams:
(12/30)*(11/30)*)(10/30)*(9/30)*(8/30)*(7/30)*(6/30)*(5/30)*(4/30)*(3/30)*(2/30)*(1/30)*(12/30)^2*(18/30)^23
.
.
.
12 first time cup winners and 25 current cupless teams repeating:
(12/30)*(11/30)*)(10/30)*(9/30)*(8/30)*(7/30)*(6/30)*(5/30)*(4/30)*(3/30)*(2/30)*(1/30)*(12/30)^25
Each one I did a combination of to see what the combo was of current cupless wins out of all 37 years. If it's only 12 current cupless teams I did (37!)/[(25!)(12!)]. If it's 12 current cupless teams, 1 repeat of a cupless teams, and 24 current cup teams I did a combination of 13 out of 37, or (37!)/[(24!)(13!)]. I kept going until I had 37 current cupless teams with 12 unique cupless teams and the rest as repeats of those current cupless teams, which was the combination of 37 out of 37, or 1. For each combination I multiplied each probability up top and added everything together.
Result: Probability is 0.0001047584 and odds are 1 in about 9,546.
Does that seem about right?