Suppose the team wins $37\%$ of the time.

In order to "win", they must win best of 7 (or 4 games).

Answer the following questions to determine the probability that the team would have won a best of 7 playoff series (i.e., won 4 games) had they made the playoffs last season.

Rephrase this question in terms of sequences of 0s and 1s. What is the shortest length of a sequence? What is the longest length of a sequence?

If we let 1 represent a win, and 0 a loss, then the shortest length of wins are 1111(4) and longest length of win is 0001111(7).

What are the number of sequences that correspond to them winning the series?

$\textbf{I don't know if this is right (my best guess)}$

If they win in 4 games, that's $\binom{4}{4}$ = 1.

If they win in 5 games, that's $\binom{5}{4}$ = 5.

If they win in 6 games, that's $\binom{6}{4}$ = 15.

If they win in 7 games, that's $\binom{7}{4}$ = 35.

Adding 1 + 5 + 20 + 35 = 56.

Calculate the number of sequences as they relate to this problem. (The answer is $not$ $2^7$ as not all series would last 7 games.)


Calculate the probability that they will win the series.

I need help on the last two, and is my 56 right?

  • $\begingroup$ Your counting is flawed. You can't say that there are $\binom 64$ "ways to win in six games" because one of those ways would be $111100$ which would actually be a win in four games. If you want to do it this way, you need to require that the last game played is a win. Also, you can't just add these together because some strings are more probable than others. $\endgroup$ – lulu Dec 11 '16 at 19:10
  • $\begingroup$ Whats the best way to do it? $\endgroup$ – K Split X Dec 11 '16 at 19:11
  • $\begingroup$ Well, you can do it your way if you correct for the errors I mentioned. Personally, I prefer to imagine that all seven games are played regardless of the fact that the winner is probably decided before game $7$. The winner of the series then wins at least $4$ of the games and you can count $4,5,6,7$ wins separately. $\endgroup$ – lulu Dec 11 '16 at 19:13
  • $\begingroup$ But the question specically says 4 games, if you count 4,5,6,7 thats another question , is it not? $\endgroup$ – K Split X Dec 11 '16 at 19:15
  • $\begingroup$ No. The team that wins $5$ or more teams is the only team that gets $4$ (the other team only gets $2$ in this scenario). That's the beauty of the method...the winner is correctly determined by simple majority. $\endgroup$ – lulu Dec 11 '16 at 19:18

I'd argue that the clearest way to picture this is to imagine that all $7$ games are played out, despite the fact the winner may well have been decided before game $7$. Indeed, with seven games played exactly one team will have won at least $4$ games and that team is, clearly, the first to have won $4$ games in the series.

With that in mind, the probability is easy to compute. Assume $p$ is the probability that team $A$ wins any particular game (here $p=.37$) and let $p_i$ be the probability that $A$ wins exactly $i$ games in the series. We see that $$p_i=\binom 7i p^i(1-p)^{7-i}$$ and we deduce that the probability that $A$ wins the series is $$\sum_{i=4}^7 p_i=\sum_{i=4}^7 \binom 7i p^i(1-p)^{7-i}$$

Worth mentioning this question on a similar theme, wherein some useful approximation methods are discussed (relevant if $7$ is replaced by a large number).


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