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 05-12-2017, 11:47 AM #1 BirdsallSa grinder   Join Date: Oct 2014 Posts: 673 Probability of Winning a 7 Game Series I feel like the numbers I'm getting here can't be right. Assuming Team A has a 20% chance to win any given game and Team b is 80%: 1 Scenario of Series ending in 4 - (0.2)^4 = 0.16% 4 Scenarios of Series ending in 5 - (0.8)*(0.2)^4*4 = .48% 10 Scenarios of Series ending in 6 - (0.8)^2*(0.2)^4*10 = 1% 20 Scenarios of Series ending in 7 - (0.8)^3*(0.2)^4*20 = 1.6% So, by these numbers a team with an average 20% chance of winning a game only has a 3.24% chance of winning a 7 game series? This seems like it can't be right.
 05-12-2017, 11:51 AM #2 BirdsallSa grinder   Join Date: Oct 2014 Posts: 673 Re: Probability of Winning a 7 Game Series Actually I think I understand how this could be correct now. Winning this series is essentially asking how often a team with a 20% winning percentage will win at least 57% of its games against a team with an 80% winning percentage. So, this number really should be quite low.
 05-12-2017, 12:04 PM #3 statmanhal Pooh-Bah   Join Date: Jan 2009 Posts: 3,983 Re: Probability of Winning a 7 Game Series It's right except for a calculation error for 5 games. Should be .512%, not .48%. Total becomes 3.33%. Of course this assumes binomial distribution is applicable with a constant and independent win probability, which is probably as good as any other assumption.
05-12-2017, 08:49 PM   #4
David Sklansky

Join Date: Aug 2002
Posts: 13,961
Re: Probability of Winning a 7 Game Series

Quote:
 Originally Posted by statmanhal It's right except for a calculation error for 5 games. Should be .512%, not .48%. Total becomes 3.33%. Of course this assumes binomial distribution is applicable with a constant and independent win probability, which is probably as good as any other assumption.
Actually its not. Because home game advantage means that some games are more than 20% and others are less. And that is worse for the dog.

05-13-2017, 01:18 AM   #5
BirdsallSa
grinder

Join Date: Oct 2014
Posts: 673
Re: Probability of Winning a 7 Game Series

Quote:
 Originally Posted by David Sklansky Actually its not. Because home game advantage means that some games are more than 20% and others are less. And that is worse for the dog.
I realize that, but this was about getting the methodology right, so I used a simplified scenario where we assumed no home court advantage, and a standardized win percentage.

 05-19-2017, 03:55 PM #6 masque de Z Carpal \'Tunnel     Join Date: Aug 2009 Location: Stanford, CA USA Posts: 8,754 Re: Probability of Winning a 7 Game Series Do a tree diagram win-lose branches and account for order of games played to take care of home details etc. That likely requires a computer program to list the correct products because eg 4-1 can happen many different ways but not after 4-0 etc. You can do one on your own if you select a big piece of paper and start in the middle on the left and do it carefully branching each time in 2 outcomes (draw each branching to take place at given "columns" range eg 1-2 4-5 7-8 etc to account for home advantage also (so list the order first 2 games here then 2 games there etc). If the paper is big its doable. The problem of course is to think of a computer program to list it or to see a creative way to do it by hand faster. So challenge yourselves to discover it lol. If the teams have p1,p2 and q1=1-p1, q2=1-p2 depending where they play then Eg 4-0 can happen with p1^2*p2^2 but 4-1 can happen as p1^2*p2*q2*p1 or p1^2*q2*p2*p1 or p1*q1*p2*p2*p1 or q1*p1*p2*p2*p1 etc. See what i mean? Now try 4-2 outcomes and you will even more complicated sequences. A proper tree diagram will list all of them because you have less than 2^6~64 branches likely as many terminate earlier like 4-0 0-4 4-1, 1-4 etc . Last edited by masque de Z; 05-19-2017 at 04:12 PM.
 05-24-2017, 04:07 AM #7 masque de Z Carpal \'Tunnel     Join Date: Aug 2009 Location: Stanford, CA USA Posts: 8,754 Re: Probability of Winning a 7 Game Series Also if we ignore the difference playing home vs visiting we can see it this way too; To win you need to win the last of course. That means you need to win from 3-3,3-2,3-1,3-0 in the score before. Now 3-0 has p^3, 3-1 has 4*p^3*q, 3-2 has 5!/3!/2!*p^3*q^2=10*p^3*q^2 and 3-3 has 6!/3!/3!*p^3*q^3=20p^3*q^3 So overall you have a chance to win that is (p^4+4*p^4*q+10*p^4*q^2+20*p^4*q^2)= p^4*(1+4*q+10*q^2+20*q^3) (adding nothing new here just a summary of results for general p/q worth having) So for example 80-20 is 96.67% to win and 3.33% to lose. If its 55% vs 45% then the best team has 60.8%. Playoffs are a massive disgrace to the best team principle everywhere they are applied. They exist only to give inferior teams the chance to win everything and make an entire stellar season irrelevant because of near term volatility and lucky sob accidents or injuries and for ratings and stupid public. They are wasting an entire year of championship effort for almost nothing but a placement and some home advantage edge that is trivial tribute to a potentially massive difference in quality performance between teams. The only way a team deserves to win a championship is by winning against all the other teams in a year long effort that they play everyone perfectly symmetrically collecting more points. And a massive F OFF to everything else (when they play each other all year symmetrically anyway. And if they dont bloody make it so). Last edited by masque de Z; 05-24-2017 at 04:13 AM.

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