I would like to know how to find the answer to this probability problem.
Two players, $A$ and $B$, are playing an arbitrary game (no draw is possible). The winner is the player who wins two consecutive games. Player $A$ has $2/3$ chances of winning a single game and player $B$ $1/3$.
Example: Player $A$ loses the first game, but wins the two next games, so he wins the overall game.
What is the probability that Player $A$ wins the overall game?
Best Answer
Let $E$ be the probability that $A$ wins the game finishes in an even number of games, and $O$ the probability that $A$ wins the game in a odd number of games. We have $\mathbb{P}(A \text{ wins})=E+O.$
Let us have a look at $E$. Denots $a$ a victory for $A$ and $b$ a victory for $B$ on each round.
If the game finishes in $2$ games, then it is $aa$.
If the game finishes in $4$ games, then it is $abaa$.
If the game finishes in $6$ games, then it is $ababaa$.
Now we begin to see a pattern. The probability of winning in $2k$ matches, is $(2/3)^{k+1}(1/3)^{k-1}$ This gives $$ E=\sum_{k=1}^{\infty} (2/3)^{k+1}(1/3)^{k-1}=\frac{4}{7} $$ I leave the odd part to you, but you should find $O=\frac{4}{21}$ for a total of $\frac{16}{21}$