0
$\begingroup$

A basketball playoff series consists of a best-of-seven (first team to win 4 games). While commentating the match, announcers stated that in the past 20 years, 12 times the team that won game 1 went on to win the series, suggesting that winning a series 60% of the time was surprisingly high.

Show if it is considered high. In other words, assuming both teams are equally likely to win a game and games are independent, what is the probability that the team that won game 1 wins the entire series?

I am completely lost in this question and do not even know how to begin. If anyone could help it would be much appreciated! Thank you.

$\endgroup$
5
  • 4
    $\begingroup$ Hint: Consider the remaining $6$ games (imagine that they are all played even if the match is settled prior to game $7$). The team that won the first game wins iff it wins at least $3$ of those games. $\endgroup$
    – lulu
    Oct 12, 2019 at 20:22
  • $\begingroup$ @lulu I'm sorry I have attempted multiple ways to approach the question and still cannot seem to wrap my head around it. So i understand that in the remaining 6 games, the team that won game 1 has to win 3 more before the second can win 4 but i have no clue where to go from there $\endgroup$ Oct 12, 2019 at 23:02
  • $\begingroup$ The posted solution is correct and, I'd say, optimal. $\endgroup$
    – lulu
    Oct 12, 2019 at 23:04
  • $\begingroup$ @lulu could you please explain why to divide by 64. I understand the sum of the choose part just not why to divide by 64. Thanks $\endgroup$ Oct 12, 2019 at 23:48
  • $\begingroup$ We divide by $64 = 2^6$ since there are two equally likely outcomes for each of the six games. $\endgroup$ Oct 13, 2019 at 17:58

2 Answers 2

1
$\begingroup$

Assuming the teams are evenly matched, the binomial theorem can be used. For simplicity assume all six games are played, then the probability of winning at least three games is $(\sum_{k=3}^6 \binom{6}{k})/64=0.65625$.

$\endgroup$
3
  • $\begingroup$ Is there a way to attempt this question without binomial theorem because we are not allowed to use it in our solution and are supposed to use an alternative method. $\endgroup$ Oct 12, 2019 at 21:52
  • $\begingroup$ could you please explain the solution a little more much appreciated $\endgroup$ Oct 12, 2019 at 23:52
  • $\begingroup$ For each k, the binomial term represents the number of possible ways of winning k games out of 6. 64 is the total number of possible outcomes. I can't think of an alternative method except deriving the binomial rather than using it. For example, k=5 means losing 1 and winning 5 - there are 6 ways depending on which game is lost. $\endgroup$ Oct 13, 2019 at 18:25
0
$\begingroup$

Since you are not familiar with the Binomial Theorem, we can consider cases.

The team that wins the first game wins the series in four games: For this to occur, the team that won the first game must win the next three games. Since each team is equally likely to win each game, the probability that the team that won the first game wins the series in four games is $$\Pr(\text{wins series in four games} \mid \text{won the first game}) = \left(\frac{1}{2}\right)^3$$

The team that wins the first game wins the series in five games: For this to occur, the team that won the first game must win two of the next three games and the fifth game.

There are two equally likely outcomes for each of the next three games. There are $\binom{3}{2}$ ways for the team which won the first game to win exactly two of the next three games. Thus, the probability that the team which won the first game wins exactly two of the next three games is $$\frac{\binom{3}{2}}{2^3}$$ The probability that the team which won the first game then wins the fifth game is $1/2$. Hence, the probability that the team which won the first game wins the series in five games is $$\Pr(\text{wins series in five games} \mid \text{won the first game}) = \frac{\binom{3}{2}}{2^3} \cdot \frac{1}{2} = \binom{3}{2}\left(\frac{1}{2}\right)^4$$

The teams that wins the first game wins the series in six games: For this to occur, the team which won the first game must win exactly two of the next four games and then win the sixth game.

There are two equally likely outcomes for each of the next four games. There are $\binom{4}{2}$ ways for the team which won the first game to win exactly two of these four games. The probability that it then wins the sixth game is $1/2$. Hence, the probability that the team which won the first game wins the series in six games is $$\Pr(\text{wins the series in six games} \mid \text{won the first game}) = \frac{\binom{4}{2}}{2^4} \cdot \frac{1}{2} = \binom{4}{2}\left(\frac{1}{2}\right)^5$$

The team that wins the first game wins the series in seven games: For this to occur, the team which won the first game must win exactly two of the next five games and then win the seventh game.

I will leave it to you to compute this probability.

Add the conditional probabilities of these four mutually exclusive cases to find the probability that the team which won the first game wins the series.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .