I am holding a best of 3 series against two basketball teams: $Team\,A$ and $Team\,B$. Either team must win or lose the series; a tie is impossible. I must calculate the probability of either team winning the series.
During their exhibition games (outside of the 3-game series), $Team\,A$ beat $Team\,B$ in $3$ out of $5$ games.
Through experimental probability, $Team\,A$ has a $3/5$ chance of winning a game against $Team\, B$.
Therefore, $Team\,B$ must have a $2/5$ chance of winning a game against $Team\, A$.
Using these values, I determined the binomial probability of $Team\, A$ winning 2 out of 3 series games ($P_{2_A}$) as follows, assuming that losing is the only other outcome:
$P_{2_A}={3\choose2} \times \left(\frac{3}{5}\right)^2 \times \left(1- \frac{3}{5}\right)^{3-2}$
$P_{2_A} = 3\times \frac{9}{25}\times \frac{2}{5}$
$P_{2_A} = \frac{54}{125}$
I also did the same math for $Team\, B$ ($P_{2_B}$), using the assumptions as prior:
$P_{2_B}={3\choose2} \times \left(\frac{2}{5}\right)^2 \times \left(1- \frac{2}{5}\right)^{3-2}$
$P_{2_B} = 3\times \frac{4}{25}\times \frac{3}{5}$
$P_{2_B} = \frac{36}{125}$
When adding $P_{2_A}$ and $P_{2_B}$ together, the probabilties do not add to $1$. I do not believe this should happen as there are only two possible endgames: $Team\, A$ wins or $Team\, B$ wins.
$\frac{54}{125} + \frac{36}{125} = \frac{90}{125} \, or \, \frac{18}{25}$
Is it incorrect to assume that all binomial probabilties must add to 1? Could binomial distribution be the incorrect approach for this type of probability scenario?
Best Answer
You missed the possibility that one team would win all three games. You assumed that the series would end two games to one. The easiest way to compute these possibilities is to assume they play all three games, but you can also condition on whether one team wins the first two games. Yes, the probabilities will sum to $1$ if you get all of them.