What is the probability?
A playoff series between two teams proceeds one game at a time until one team has won 5 games. What is the probability that the series lasts 9 games if each team is equally likely to win each game?
The answer is 0.2734375.
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3 solutions
You don't have to count the last game, as it doesn't matter who wins it as long as it happens.
What all series that last the full 9 games have in common is that their first 8 games have 4 wins for each of the teams. What then happens in the ninth game is of no consequence, (mathematically speaking).
Now there are C(8,4) = 70 possibilities where the first 8 games result in 4 wins for each team, out of 2 8 = 256 possibilities without restrictions.
This gives us a probability of 70/256 = 0 . 2 7 3 .
How can total cases here be 256? Let tge teams be A and B... It includes the following cases:
AAAAABBB AAAAABAB AAAAABBA AAAAAABB AAAAAAAB AAAAAABA AAAAABAA AAAAAAAA
But according to the condition, the series ends as soon as one of the teams win 5 matches. Hence, all the mentioned cases will end after 5 matches, which means they are all the same. Si how can we count all of them as different? Note: there are a lot of other cases that should not be counted similarly.
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Technically speaking, he is not counting the actual number of cases, but rather accounting for the probability that each case occurs.
For example, in a lottery, you either win or lose, and so there are 2 cases. Does that mean that your chance of winning is 50%? Not really, because there is only 1 ticket that wins and many tickets that lose. The proper way is to account for cases in terms of the number of tickets, each of which have the same probability of winning.
In a similar way, the probability of playing out to ABABABAB is going to be 2 8 1 , while the probability of playing out to AAAAA is 2 5 1 . If you try and count cases like @Nicolas Bryenton , then you will still need to account for the varying probability as in the lottery win/lose scenario.
What @brian charlesworth is doing, is to 'pad' with extra cases of the form AAAAAAAA, AAAAAAAB, AAAAAABB, etc, such that we can attribute each of them with the same probability of 2 8 1 chance of occuring, and there are 2 3 such ways. This would give us that AAAAA has 2 8 2 3 = 2 5 1 probability of occuring, which is what we want.
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Great explanation, Calvin. Sometimes my quest for brevity results in the absence of certain important observations. I'll try to be more expansive, when necessary, in the future. :)
Thanks for clarifying!
I believe you are on to something. This question must be incorrect. The true solution would be 70/116, provided my quick casework is correct (this was actually my original solution, but it was wrong, so I tried to find a different (albeit problematic) solution.
Probability of A winning a game, p=0.5
Probability of a losing a game, q=0.5
Out of 8 games, probability of A winning 4 and losing the rest is:
8 C 4 p 4 q 8 − 4
=70/256=0.273
In order for the series to end in 9 games, one team must win exactly 4 out of 8 games (in any order) and then go on to win the 9th game.
So that is 8C4 (1/2)^9 + 8 C4 (1/2)^9 = 140/512 = 0.273