Best of Seven Jul 2022

Best of 7

Two teams face off in a best of 7 series.

I provide analysis and plots.

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 let’s say the probability that Team A wins each game is P. The probability that Team B wins each game is now therefore Q = 1 – P, and the sum of the probabilities P + Q = 1.0 as expected.

For example, if the probability that Team A wins each game is P = 0.81, then the probability that Team B wins each game is Q = 1 – P = 0.19.

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 wins?

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

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

Explanation:

$({P^4}\cdot{Q})$ is the one-time probability of Team A winning 4 games and Team B winning one game.
${_4{{C}_1}}$ 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

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

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

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

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

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

$_{(\mathbf{0}+3)}{{C}_\mathbf{0} = \prescript{_3}{{C}_{0}} = \frac {3!}{{3!}\cdot{0!}} = 1$

$_{(\mathbf{1}+3)}{{C}_\mathbf{1} = \prescript{_4}{{C}_{1}} = \frac {4!}{{3!}\cdot{1!}} = 4$

$_{(\mathbf{2}+3)}{{C}_\mathbf{2} = \prescript{_5}{{C}_{2}} = \frac {5!}{{3!}\cdot{2!}} = 10$

$_{(\mathbf{3}+3)}{{C}_\mathbf{3} = \prescript{_6}{{C}_{3}} = \frac {6!}{{3!}\cdot{3!}} = 20$

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

Using PLOTLY live plots

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Static Plots