Question

The National Basketball Association championship series is a best of $7$ series, meaning that the first team to win $4$ games is declared the champion. In its history, no team has ever come back to win the championship series after being behind $3$ games to $1$. Assuming that each of the games played in this year’s series is equally likely to be won by either team, independent of the results of earlier games, what is the probability that the upcoming championship series will result in a team coming back from a $3$ games to $1$ deficit to win the series?

Short answer

The probability that the upcoming championship series will result in a team coming back from a $3$games to $1$deficit to win the series is$6.25\%$.

Step-by-step solution

Given Information

$\mathrm{L}=$loss for the team

$W=$win for the team

Substitution

Since each group is assumed to be equally likely to win, the probability of success and the probability of a failure for a team is both $\frac{1}{2}$.

$P\left(W\right)=P\left(L\right)=\frac{1}{2}$

When the first four games resulted in $3$failures and $1$victory for the team, then there are $\left(\begin{array}{l}4\\ 1\end{array}\right)=4$possible games that could have resulted in the win.

Calculation

When the first four games resulted in $3$failures and $1$victory for the team, while the team can still reach back to win, then the final three games all require to have been won for the team.

$P(3$games to 1 deficit and come back for one team $)$

$=\left(\begin{array}{l}4\\ 1\end{array}\right)\times P\left(L\right)\times P\left(L\right)\times P\left(L\right)\times P\left(W\right)\times P\left(W\right)\times P\left(W\right)\times P\left(W\right)$

$=4\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}$

$=\frac{1}{32}$

Calculation

However, either of the two teams could have $3$games to $1$deficit and come back, while the probability is $\frac{1}{32}$for each of the two teams.

$P(3$games to $1$deficit and come back $)$

$=P(3$games to $1$deficit and come back for first team )

$+P(3$games to 1 deficit and come back for second team)

$=\frac{1}{32}+\frac{1}{32}$

Adding the given expression

$=\frac{2}{32}$

Diving$2$both sides,

$=\frac{1}{16}$

$=0.0625$

We get,

$=6.25\%$

Final answer

The probability that the upcoming championship series will result in a team coming back from a $3$games to $1$deficit to win the series

$6.25\%$.