In a game with two equal teams, the home team wins any game with probability $p>\frac12$. In a best of three playoff series, a team with a home advantage has play at home, followed by a game away, followed by a home play if necessary. The series is over as soon as one team wins two games.
What is $P[T]$, the probability that the team with home advantage wins the series? Prove that $P[T]>p$ for all $p>\frac12$.
In the first sentence, I think there's a typo. I think it meant to say $p=\frac12$ (I don't know how you can get a number as your answer if its not equal to $\frac12$)
I drew a tree diagram as shown below, and calculated the probability that Team A wins.
$.5^2 + .5^3 + .5^3 = .5$
Best Answer
This is not a typo. $p$ is a variable and you are supposed to include it in your answer.
There are three sequences of results leading to the home advantage team winning, as you found out: WW ($p(1-p)$), WLW ($p^3$) and LWW ($(1-p)^2p$). Thus $$P[T]=p(1-p)+p(1-p)^2+p^3=(p-p^2)(2-p)+p^3=2p-3p^2+2p^3=p(2-3p+2p^2)$$ and if $p=\frac12+d$ with $d>0$ $$P[T]=p\left(2-3\left(\frac12+d\right)+2\left(\frac12+d\right)^2\right)$$ $$=p\left(2d^2-d+1\right)$$ $$=p\left(1+d(2d-1)\right)<p$$ where the last inequality is because $2d-1<0$ for $0<d<\frac12$. Indeed, the inequality is reversed compared to what is printed: conversely, $P[T]>p$ when $p<\frac12$.