SudsMeister
This seems to be intuitively correct.
Bunuel, if possible, could you point out why is this wrong?
Thank you.
rkrish7
The correct answer is 75%.
You have a total of eight different possibilities:
4-0 in favor of Team A
4-1 in favor of Team A
4-2 in favor of Team A
4-3 in favor of Team A
4-0 in favor of Team B
4-1 in favor of Team B
4-2 in favor of Team B
4-3 in favor of Team B
So the total number of choices for the game to end with less than 7 games = 6 different possibilities
Therefore % less than 7 games = 6/8 = 75%
This approach assumes each of the scenarios is equally weighted, when they are not.
For example:
4 and 0 can occur only 1 way: the winning team wins the first 4 games. Total probability: (1/2)^4 = 1/16 for a given team.
4 and 1 has to have the last game end in a win, so the first four games contain 3 wins and 1 loss.
3 wins and 1 loss can be accomplished 4!/3! = 4 ways because the 1 loss can occur anywhere among the 4 games.
So the total probability of that scenario is: 4*(1/2)^4 *(1/2) = 4/32
The simplistic equal weight approach ignores the different ways each scenario can be accomplished.
Just because one can count the number of scenarios doesn't mean they are equally likely.
For example:
Scenario 1: The Sun rises tomorrow
Scenario 2: The Sun doesn't rise tomorrow
Captures all possibilities.
Hopefully each is not a 50% probability