Smallest number of games per match to get $0.8$ chance to win the match. If the ﬁrst person to win $n$ games wins the match, what is the smallest value of $n$ such that $A$ has a better than $0.8$ chance of winning the match?
For $A$ having a probability of $0.70$, I get smallest $n = 5$ (Meaning there must be $5$ games per match for $A$ to have a $0.8$ chance to win.) I got this by doing $1 - 0.7^5 = 0.832$. Although I would have thought it would have been lower.

Using the same method as before, with A having a probability of winning a game, the probabilities of A winning the match are about $0.7$ for $n=1$, $0.784 $ for $n=2$, $0.837$ for $n=3$, $0.874$ for $n=4$ and $0.901$ for $n=5$.
So the answer is $n=3$ to exceed $0.8$.
$1−0.7^5$ is the answer to the question "What is the probability B wins at least one game before A wins 5 games?"
