Original question from AOPS:
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?
I researched and found some solutions on the internet (which also gives the same answer as with the official solution) giving 35/128, taking 2^9 as the total number of possibilities.
The official solution seems to be counting 000000000,000000001,000000010,…,111111110,111111111 total of 512
My question is shouldn’t it count only
00000, 11111, 000010, 000100, 001000, 010000, 100000, …
and NOT including sequences like 000000000 because such game is not possibly?
In other words 000000000,000000001,000000010,…000001111 should all be just counted as 00000.
My Calculation shows that the answer should be
$$
\frac{\binom{8}{4}}{ 1+\binom{5}{1} +\binom{6}{2} +\binom{7}{3} +\binom{8}{4} }
$$
Best Answer
In addition to the comments that show why outcomes you are counting are not equally probable, in the way you are thinking, it is easier to see how play doesn't last $9$ games.
That is either of the team wins in $5, 6, 7$ or $8$ games.
$P(W \leq 8) = \displaystyle \frac {2}{2^5} + \frac {2 \cdot{5 \choose 4} } {2^6} + \frac{2 \cdot{6 \choose 4}}{2^7} + \frac {2 \cdot{7 \choose 4}}{2^8} = \frac{93}{128}$
So the probability that the play lasts $9$ games is $\displaystyle \frac{35}{128}$
But it is of course unnecessary when you can simply use $~\displaystyle \frac{2 \cdot {8 \choose 4}}{2^9}$.