Best of Seven Jul 2022

Best of 7
Two teams face off in a best of 7 series.

I provide analysis and plots.

Click for Best of Seven Live Chart


Evenly Matched Teams

If the two teams are evenly matched, then we say the probability that Team A wins each game is 0.50. The probability that Team B wins each game is also 0.50, and the sum of these two probabilities is 1.0 as expected.

The question here is, what is the likelihood of each possible playoff outcome:

  • 4-0 Team A wins
  • 4-1 Team A wins
  • 4-2 Team A wins
  • 4-3 Team A wins and
  • 4-0 Team B wins
  • 4-1 Team B wins
  • 4-2 Team B wins
  • 4-3 Team B wins

Team A is Stronger

If the two teams are not evenly matched, then we might say the probability that Team A wins each game is P = 0.81, for example. The probability that Team B wins each game is now therefore Q = 1 – P = 0.19, and the sum of the probabilities P + Q = 1.0 as expected.

Again, the question is, what is the likelihood of each possible playoff outcome:

  • 4-0 Team A wins,
  • 4-1 Team A wins,
  • 4-2 Team A wins,
  • 4-3 Team A wins, and
  • 4-0 Team B wins,
  • 4-1 Team B wins,
  • 4-2 Team B wins,
  • 4-3 Team B win?

Discussion

This answer to this problem is relatively easy to discuss yet somewhat complex to calculate, Fortunately, it can be broken down into several similar parts.


Series Goes 4-0 or 0-4

An easy start: if Team A wins in 4 straight, then the probability is simply

    \[{P^4} = 0.81^4 = 0.430\]

And similarly: if Team B wins in 4 straight, then the probability is simply

    \[{Q^4} = 0.19^4 = 0.001\]

in this example very unlikely.


Series Goes 4-1 or 1-4

Now, if Team A is to win the series 4-1, then after 4 games, they must be up 3-1 just before the 5th game in the series. That means there is one loss in the 4 games played and there are only 4 ways that can happen.

Therefore the probability is

    \[({P^4}\cdot{Q})\cdot{\prescript{_4}{{C}_1}} = ({0.81^4}\cdot {0.19})\cdot {4} = 0.327\]

Explanation:

  • ({P^4}\cdot{Q}) is the one-time probability of Team A winning 4 games and Team B winning one game.
  • {\prescript{_4}{{C}_1}} is the number of combinations in which one team wins 3 games and the other wins only one game.
    Click for explanation of combinations

Similarly, if Team B is to win the series 4-1, then after 4 games, they must be up 3-1 just before the 5th game in the series. Therefore the probability is

    \[({Q^4}\cdot{P})\cdot{\prescript{_4}{{C}_1}} = ({0.19^4}\cdot {0.81})\cdot {4} = 0.004\]


Series Goes 4-2 or 2-4

If Team A is to win the series 4-2, then after 5 games, they must be up 3-2 just before the 6th game in the series. That means there are 2 losses in the 5 games played and there are only 10 ways that can happen.

So we have established the pattern, if Team A is to win the series 4-\mathbf{X}, the probability is

    \[({P^4}\cdot{Q})\cdot{\prescript{_{(\mathbf{X}+3)}}{{C}_\mathbf{X}}}\]

Likewise, if Team B is to win the series \mathbf{X}-4, the probability is

    \[({Q^4}\cdot{P})\cdot{\prescript{_{(\mathbf{X}+3)}}{{C}_\mathbf{X}}}\]


    \[\prescript{_{(\mathbf{0}+3)}}{{C}_\mathbf{0} = \prescript{_{{3}}}{{C}_{0}} = \frac {3!}{{3!}\cdot{0!}} = 1\]

    \[\prescript{_{(\mathbf{1}+3)}}{{C}_\mathbf{1} = \prescript{_{{4}}}{{C}_{1}} = \frac {4!}{{3!}\cdot{1!}} = 4\]

    \[\prescript{_{(\mathbf{2}+3)}}{{C}_\mathbf{2} = \prescript{_{{5}}}{{C}_{2}} = \frac {5!}{{3!}\cdot{2!}} = 10\]

    \[\prescript{_{(\mathbf{3}+3)}}{{C}_\mathbf{3} = \prescript{_{{6}}}{{C}_{3}} = \frac {6!}{{3!}\cdot{3!}} = 20\]


Observation

Notice that compared to a single game, a seven game series increases the probability that the strong team wins the series. For example, with P = 0.81, as shown above, the probability that Team A wins the series is calculated at 0.972. Note further that the probability that Team B wins the series decreases from 0.19 to only 0.028.

Using Odds:

The single game had odds of 81:19 or about 4:1 whereas the series has odds of 972:28 or about 35:1.




Plots

See the plots of all possible outcomes for all probabilities:

    \[P \mid 0 < P < 1\]

Example probabilities are given for the specific case where P = 0.81.

Live Plot

Click for Best of Seven Live Chart

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All Outcomes
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Static Plots

All Outcomes
Winning Outcomes


Reference

Best of Seven Outcome Probability Dec 2020

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