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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.

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It would have made a lot of sense to link to this closely related and already answered question of yours. – joriki Dec 14 '12 at 10:28
possible duplicate of Working out probability of winning a match – Did Dec 14 '12 at 12:36
It's strange that you used a wrong approach here more than half an hour after a correct answer to your other question had been provided. Had you actually read that? It would probably make more sense to ask about what you don't understand about that answer rather than introducing an additional complication by inverting the problem. – joriki Dec 14 '12 at 12:38
@did: I didn't vote to close as duplicate because his question asks about the smallest $n$. I don't think there's a closed form for it, so finding it probably just involves calculating the probability for various values of $n$, so in that sense you're right that the other question already contains everything that's needed here, but that's not a priori obvious. – joriki Dec 14 '12 at 12:48
@joriki As precise as ever... :-) You give in your previous-to-last comment my main reason to vote to close: as long as the OP sticks to this $1-0.7^5$ stuff, the rest seems pointless. – Did Dec 14 '12 at 13:07

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?"

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