Probability of all 12 Cupless teams winning by 2050 (Simplified for plausibility)

[SnowblindNYR]

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

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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?

Click to read more...

 

[HockeyGuy73]

huh?

 

[SnowblindNYR]

huh?

LOL. I was asking if the probability that all of the 12 teams that haven't won the cup yet will win the cup at least once by 2050, if we were to assume (I know this is false) that each team's chances are 1/30. It's a probability problem and I was asking if I did it right from someone that knows probability theory.

 

[Patman]

The expected value for all teams is

\sum{\frac{n}{n-i+1}}_i^n

My guess is if you only use the last 12 terms you'd get your answer.

Working backwards... 30+15+10+7.5=62.5

So the result is higher than that... 30/12=10/4=2.5 so it's at least 82.5 years...

Obviously this is not the variance which is a touch harder to calculate. The real question is "how desperately unlikely is it?"

 

[SnowblindNYR]

The expected value for all teams is

\sum{\frac{n}{n-i+1}}_i^n

My guess is if you only use the last 12 terms you'd get your answer.

Working backwards... 30+15+10+7.5=62.5

So the result is higher than that... 30/12=10/4=2.5 so it's at least 82.5 years...

Obviously this is not the variance which is a touch harder to calculate. The real question is "how desperately unlikely is it?"

Interesting. That's a bit over my head, haha. I'm a simple probability guy.

 

[Bear of Bad News]

I typically simulate this stuff since it's easier and close enough.

I assumed independence and a lack of correlation from season to season. I also assumed no additional expansion or contraction.

TEAMS WITH CUP|PERCENT
0|0.000%
1|0.000%
2|0.002%
3|0.034%
4|0.290%
5|1.668%
6|6.463%
7|15.875%
8|25.603%
9|26.544%
10|16.774%
11|5.901%
12|0.845%

So under these assumptions, there's about a 0.8% chance of all twelve teams winning the Cup in the 36-year period between 2015 and 2050.

 

[Patman]

I just cooked up the variance formulation... Wasn't the worst. The advantage is that the spacing between each term is itself a geometric distribution. I proved that on a lonely whiteboard. I never cared to finish it out.

Same form as the above essentially.... Swapping the variance form for the expectation part

\frac{n^2(i-1)} openbrace(n-i+1)^2}

IMO, stuff like this is why I'm generally against league expansion

Edit: hmm... Won't allow me to use an opening brace where it says openbrace... I wish vB would embed TeX

 

[66871]

It might be worth noting for this discussion that over the past eleven seasons, non-cup winning teams have underperformed in terms of getting into the playoffs (even though four teams have 'joined the club').

Specifically, if playoff berths were a random draw, 80 playoff berths would be expected to have gone to non-winners In reality, only 70 did. So perhaps instead of n/30 as the odds for a given year (where n is the number of teams yet to win a cup), a better number is 0.875 * n / 30.

That said, as time has gone on since the lockout, teams yet to win seem to be doing better against the expected value for berths. So perhaps another way of looking at it is to just assume that a cup winner always makes the playoffs the following season. So instead of the above, you get 0.9375 * n / 30.