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Probability of a team winning

 
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Italian Stallion



Joined: 04 Mar 2009
Posts: 112
PostPosted: Mon Dec 06, 2010 10:29 am    Post subject: Probability of a team winning Reply with quote
Is there any way to calculate the approximate probability of a team winning based on the records of the two teams?

I can't seem to intuit the formula and math and don't know it.

Assuming everyone had an equal schedule.

If two .500 teams face each other, it would obviously be .500 and .500

If a. 600 team faced a .500 team, the .600 should win .600

But how the heck do you calculate a .400 team vs. a .600 team etc.... The .600 team should win more than .600 because of the below average competition, but I don't see the formula. (I think I'm getting old).
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DSMok1



Joined: 05 Aug 2009
Posts: 602
Location: Where the wind comes sweeping down the plains
PostPosted: Mon Dec 06, 2010 10:58 am    Post subject: Re: Probability of a team winning Reply with quote
Italian Stallion wrote:
Is there any way to calculate the approximate probability of a team winning based on the records of the two teams?

I can't seem to intuit the formula and math and don't know it.

Assuming everyone had an equal schedule.

If two .500 teams face each other, it would obviously be .500 and .500

If a. 600 team faced a .500 team, the .600 should win .600

But how the heck do you calculate a .400 team vs. a .600 team etc.... The .600 team should win more than .600 because of the below average competition, but I don't see the formula. (I think I'm getting old).


Convert to Z scores: Normsinv(0.6) = 0.253. Normsinv(0.4) = -0.253.
Normsdist(0.253-(-0.253)) = 0.693 or 69.3%.
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Italian Stallion



Joined: 04 Mar 2009
Posts: 112
PostPosted: Mon Dec 06, 2010 11:19 am    Post subject: Re: Probability of a team winning Reply with quote
DSMok1 wrote:
Italian Stallion wrote:
Is there any way to calculate the approximate probability of a team winning based on the records of the two teams?

I can't seem to intuit the formula and math and don't know it.

Assuming everyone had an equal schedule.

If two .500 teams face each other, it would obviously be .500 and .500

If a. 600 team faced a .500 team, the .600 should win .600

But how the heck do you calculate a .400 team vs. a .600 team etc.... The .600 team should win more than .600 because of the below average competition, but I don't see the formula. (I think I'm getting old).


Convert to Z scores: Normsinv(0.6) = 0.253. Normsinv(0.4) = -0.253.
Normsdist(0.253-(-0.253)) = 0.693 or 69.3%.



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bbstats



Joined: 25 Apr 2010
Posts: 38
PostPosted: Mon Dec 06, 2010 11:23 am    Post subject: Reply with quote
This gives you roughly the same result (+/- 1%)

Average Win%=(A-A*B) / (A+B-2*A*B)
where A=team 1 win% and team B=team 2 win%

or the easier estimate

Average Win%=.5 + A - B


(Obligatory)
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Ed Küpfer



Joined: 30 Dec 2004
Posts: 785
Location: Toronto
PostPosted: Mon Dec 06, 2010 1:43 pm    Post subject: Re: Probability of a team winning Reply with quote
Italian Stallion wrote:
But how the heck do you calculate a .400 team vs. a .600 team etc.... The .600 team should win more than .600 because of the below average competition, but I don't see the formula. (I think I'm getting old).


The poster above gave the log5 formula -- you can see there's a whole literature about it. I compared it to a regression model a long long time ago here.
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bbstats



Joined: 25 Apr 2010
Posts: 38
PostPosted: Mon Dec 06, 2010 10:50 pm    Post subject: Reply with quote
Hey, I'm givin' credit where credit is due -- I think you missed the link in my post there, Ed.
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Italian Stallion



Joined: 04 Mar 2009
Posts: 112
PostPosted: Tue Dec 07, 2010 12:48 am    Post subject: Reply with quote
Thanks guys.
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