Chance of two teams being "inseparable"

[MagareyLegend]

Just on that though, one day two teams on a precarious position on the ladder are going to be on the same points for & against or the same percentage at the end of the season & the same win/loss & I bet there is no rule for it. Imagine the situation, what do you do - toss a coin?

[spell_check]

Just on that though, one day two teams on a precarious position on the ladder are going to be on the same points for & against or the same percentage at the end of the season & the same win/loss & I bet there is no rule for it. Imagine the situation, what do you do - toss a coin?

I don't know about that, this was discussed last year (see viewtopic.php?f=1&t=3676) and no one else knows on here either.

[Mr66]

Just on that though, one day two teams on a precarious position on the ladder are going to be on the same points for & against or the same percentage at the end of the season & the same win/loss & I bet there is no rule for it. Imagine the situation, what do you do - toss a coin?

I don't know about that, this was discussed last year (see viewtopic.php?f=1&t=3676) and no one else knows on here either.

If this is the case, then the clubs head-to-head meetings are the next tie-breaker.
If they are level on wins, then the % between the two games is considered, ie, if North beats
Norwood by 10 pts then Norwood wins by 5 pts in the return, North goes ahead.
I THINK the AFL has adopted this procedure, not sure about other leagues.
The statistical odds of two teams having the same wins/losses AND % is astronomical.
I don't it has ever happened regardless of win/loss ratio in any league across Australia.

[spell_check]

Just on that though, one day two teams on a precarious position on the ladder are going to be on the same points for & against or the same percentage at the end of the season & the same win/loss & I bet there is no rule for it. Imagine the situation, what do you do - toss a coin?

I don't know about that, this was discussed last year (see viewtopic.php?f=1&t=3676) and no one else knows on here either.

If this is the case, then the clubs head-to-head meetings are the next tie-breaker.
If they are level on wins, then the % between the two games is considered, ie, if North beats
Norwood by 10 pts then Norwood wins by 5 pts in the return, North goes ahead.
I THINK the AFL has adopted this procedure, not sure about other leagues.
The statistical odds of two teams having the same wins/losses AND % is astronomical.
I don't it has ever happened regardless of win/loss ratio in any league across Australia.

This would be the most sensible approach to this, but I reakon it would be nice to know for certain, if only for curiosity sake.

[Hondo]

2 comments:

1 That caller what a classic! I wonder if that morning he realised it was as far to walk to his letter-box as it was to walk back?

2 The chances of two teams finishing on identical % and points would be infestimal .... any stats guys out there able to work out the odds of that happening?

[spell_check]

2 comments:

1 That caller what a classic! I wonder if that morning he realised it was as far to walk to his letter-box as it was to walk back?

2 The chances of two teams finishing on identical % and points would be infestimal .... any stats guys out there able to work out the odds of that happening?

Ecky would be better at working out odds like that, but I have a feeling that it would depend on how many points were scored in that season. You would have to base it on a particular season to get a figure.

[Grahaml]

I would think that you would be able to apply normal distribution rules to work out the odds. I'm not good enough to do it myself but maybe if you really want to know contact a uni department. I have a feeling though that you can't work it out to an exact number like a lot of odds calculations because the odds of scoring 1000 points for a season is different to scoring 1 point for the season.

[Ecky]

I would think that you would be able to apply normal distribution rules to work out the odds. I'm not good enough to do it myself but maybe if you really want to know contact a uni department.

And chances are they would forward you onto me, as most of the uni stats lecturers would know that I am the only statistician in South Australia crazy enough to care about such things...

I'll have a go at it... just give me a bit of time...

[pipers]

most of the uni stats lecturers would know that I am the only statistician in South Australia crazy enough to care about such things..

A crazy statistician?

I've heard it all now.

Go mental, Ecky!

[Ecky]

OK, here is my solution to the question:
What is the chance that two teams in the SANFL have the exact same points for, points against and premiership points at the end of the minor round?
I've made a number of assumptions/approximations that I believe won't have a significant effect on the calculation
Notation:
PF = points for of a team
PA = points against of a team

Assumptions:
PF and PA are independent, both between the same team and between different teams.
the average (expected) score of a team in a game is 90 points
there are 20 rounds in a season
PF has a binomial distribution with n=3600 and p=0.5. (3600 = 20*90*2 and can be interpreted as the total number of points scored in games involving a particular team. p=0.5 as I have assumed that each team has an equal chance of scoring. Obviously not quite correct, but it becomes too complicated if you assume a more complicated distribution. A further reason for choosing p=0.5 is that pairs of teams which are likely to be in this unique situation are likely to be around the 50% mark.
The probability that two teams have the same premiership points, given that they have the same PF and PA is 1/3. Premiership points and PF,PA are highly correlated, so this is hard to work out. If anyone reckons this approximation is way out, let me know. I basically assumed that there was an equal chance of the three events:
- team A is higher
- team B is higher
- they have the same premiership points

Now,
Let X~B(3600,0.5) be the PF of team A.
let Y~B(3600,0.5) be the PF of team B.

We want to calculate P(X=Y), the chance that teams A and B have the same PF. To do this I used some code in the stats package "R" that simply summed up:
P(X=0)P(Y=0)+P(X=1)P(Y=1)+...+P(X=3600)P(Y=3600)
This probability was calculated to be pf=0.0094028
(Note that if you change the average score in a game from 90 to 100 it becomes 0.00892, so not that different)
Now there are 36 pairs of teams in a 9 team competition. So the chance that PF=PA for any pair of teams in a given year is:
36*pf^2 = 0.0031829
Hence the chance that two teams have the same PF, PA and premiership points is
(36*pf^2)/3 = 0.001060959
Hence this is likely to happen once every 942 years, or in other words the odds of this happening in any given year is about 1000/1. Quite unlikely, but not impossible.

Let me know if you think I have made a mistake, or I haven't explained some of these steps clearly enough, or you have a suggestion for a better way to calculate this approximation.

[spell_check]

What would the odds look like if you made it that a team would have 12 wins and 8 losses with this scenario? Or perhaps 8 wins and 12 losses?

[Ecky]

What would the odds look like if you made it that a team would have 12 wins and 8 losses with this scenario? Or perhaps 8 wins and 12 losses?

Not sure if I understand your question. Are you saying that GIVEN that two teams have both finished on 12-8, what is the chance they have the same PF and PA?

[spell_check]

What would the odds look like if you made it that a team would have 12 wins and 8 losses with this scenario? Or perhaps 8 wins and 12 losses?

Not sure if I understand your question. Are you saying that GIVEN that two teams have both finished on 12-8, what is the chance they have the same PF and PA?

Yep. I realised I put team instead of both teams, lol.

[Ecky]

Well, it would basically be the same as the above calculation, without the dividing by 3 factor. Hence the chance of this would be approx 0.00318. or a 313/1 chance. It is more likely because you have already assumed what the teams wins and losses are.

You could do something like assuming p=0.55 for PF and p=0.45 for PA, since they are better than average teams, but this would make bugger all difference.

[spell_check]

Well, it would basically be the same as the above calculation, without the dividing by 3 factor. Hence the chance of this would be approx 0.00318. or a 313/1 chance. It is more likely because you have already assumed what the teams wins and losses are.

You could do something like assuming p=0.55 for PF and p=0.45 for PA, since they are better than average teams, but this would make bugger all difference.

Interesting though, I doubt if anyone has covered this before in a maths equation. Or maybe they have, but not publicly.

[scott]

My brain hurts.

[bulldogproud]

Darn Ecky, you gave away one of the questions on the Victorian Pure Maths exam!!!!!!!!!! *grins*

[Hondo]

Does it impact the calc when you take in to account that you can have different PF and PA but still have an identical %

ie, 1000 v 1000
same as 900 v 900

[Pseudo]

Darn Ecky, you gave away one of the questions on the Victorian Pure Maths exam!!!!!!!!!! *grins*

Not quite. The Vics use PF and PA differently

[Ecky]

Does it impact the calc when you take in to account that you can have different PF and PA but still have an identical %

ie, 1000 v 1000
same as 900 v 900

edit now that I read what you wrote properly

Actually no I didn't allow for this. If you want to include these cases, the chance it would happen would increase a little, but not by that much since there aren't many cases (besides when PF=PA) that you can get different PF and PA and still get exactly the same percentage.