Probability Analysis of the Head-To-Head Tie Breaker in a Round-Robin Tournament
This blog is a detailed probability analysis that supports the previous entry titled "I Hate the Head-To-Head." Seriously, its pretty technical. Its an example of what happens when you cross a nerd with a football fan. If you don't like math, I recommend you read "I Hate the Head-To-Head" instead.
This post originally appeared on the blog site Irrational Musings in December 2010.
I developed the analysis when I had a lot of time on my hands. I have since cleaned it up and posted it here. I hope you enjoy it.
Using probability analysis, I will show that the head-to-head tie
breaker is equally likely to select the 2nd best team as it is to select
the best team. I relied on definitions from Wikipedia
for this analysis. The primary result to take from that wiki page is that the ratio of likelihoods of two system parameter given a single observed event can be found from ratio of the probabilities of the observed event given each system parameter.
L(Q1|A)/L(Q2|A) = P(A|Q1)/P(A|Q2)
(I recognize that the assumptions made here may be a substantial leap and may not be reflective of the probability of a given team to defeat another, or of the variability of such probabilities between different teams. When a team's strengths line up directly with another's weaknesses, the probability of defeat is greatly affected.)
Suppose we have an n-team round-robin tournament in which two teams, TeamA and TeamB, are tied for the championship with one loss a piece at the end of the tournament. For the purposes of calculating probabilities, I will make the assumption that there exists a clear superior team that has a better than 50% chance of beating any other team in the conference. My problem then becomes one of determining which is the superior team, TeamA or TeamB.
I will define strength between 0 and 1 such that the superior team has a strength of 1 and each of the other teams has a strength that is less by some increment 0<epsilon<1.
strength(TeamI) = 1-epsilonI, where epsilonI is between 0 and 1 (0<epsilonI<1)Since the strength of the superior team is 1, that means its epsilon must be zero. I assume that strength is constant regardless of the team faced or the venue of the game. The probability that team TeamI defeats TeamJ is calculated as the strength of TeamI divided by the sum of their strengths.
P (TeamI defeats TeamJ ) = (1-epsilonI)/( (1-epsilonI) + (1-epsilonJ) )
As an example, if epsilonI = 1/3, then the probability that the best team
(whose strength is 1) beats TeamI is P(best Team defeats TeamI) = 60%. Similarly, the probability that TeamI defeats the best team is 40%.
Let me assign a few observation events to probability constants that establish TeamA and TeamB as tied with one loss at the end of the round-robin tournament during which TeamA defeated TeamB but lost to inferior TeamK.
A' - TeamA defeats TeamB
A'' - TeamK defeats TeamA
A''' - TeamB defeats TeamK
A - all three events, such that P(A) = P(A')P(A'')P(A''')
and a few system parameter conditions
Q1-TeamA is the superior team (epsilonA = 0)
TeamB is the 2nd best team (epsilonB = epsilon2, where 0 < epsilon2 < epsilonK)
TeamK is an inferior team (epsilonK > epsilon2)
Q2-TeamB is the superior team (epsilonB = 0)
TeamA is the 2nd best team (epsilonA = epsilon2, where 0 < epsilon2 < epsilonK)
TeamK is an inferior team (epsilonK > epsilon2)
The HTH tiebreaker assumes that since Team A defeated TeamB, then condition Q1 is true (ie, TeamA is the superior team). However, this analysis shows that the two conditions are equally likely.
Given the sequence of events that TeamA beat TeamB beat TeamK beat TeamA, I wish to calculate the likelihood that TeamA is the superior team versus the likelihood that TeamB is the superior team. This is the likelihood of Q1 given A, L(Q1|A), as compared to the likelihood of Q2 given A, L(Q2|A). The equation from above applies for this purpose.
The probability of A' given Q1 is
P(A'|Q1) = 1/(2 - epsilon2)
P(A''|Q1) = (1 - epsilonK)/(2 - epsilonK)
P(A'''|Q1) = (1 - epsilon2)/(2 - epsilon2 - epsilonK)
P(A|Q1) = P(A'Q1)P(A''Q1)P(A''Q1) = (1 - epsilonK)(1 - epsilon2)/( (2 - epsilonJ)(2 - epsilon2 - epsilonK)(2 - epsilon2) )
P(A'|Q2) = (1 - epsilon2)/(2 - epsilon2)
P(A''|Q2) = (1 - epsilonK)/(2 - epsilon2 - epsilonK)
P(A'''|Q2) = 1/(2 - epsilonK)
P(A|Q2) = P(A'|Q2)P(A''|Q2)P(A'''|Q2) = (1 - epsilonK)(1 - epsilon2)/( (2 - epsilonK)(2 - epsilon2 - epsilonK)(2 - epsilon2) )
L(Q1|A)/L(Q2|A) = P(A|Q1)/P(A|Q2) = 1
Given the assumptions made here, it is clear that, regardless of the increments epsilon2 and epsilonK, the likelihood that TeamA is the best team is equal to the likelihood that TeamB is the best team, meaning that in the event of two teams tied with one-loss at the completion of a round-robin tournament, the loser of the head-to-head game is just as likely to be the superior team as the winner. The head-to-head tie breaker is equivalent to flipping a coin.