What is the probability that A will win atleast three of the next four games? In a set of five games in tennis between two players A and B,the probability of a player winning a game is $\frac{2}{3}$ who has won the earlier game.A wins the first game.What is the probability that A will win atleast three of the next four games?

Since the probability of $A$ winning atleast three of the next four games is asked.This is possible if $A$ wins 2nd,3rd,4th games or 2nd,3rd,5th games or 3rd,4th,5th games or 2nd,4th,5th games or 2nd,3rd,4th,5th games.
Let the probability of a player winning a game is $p$ if he has not won the earlier game.If he has won the earlier game,then the probability of a player winning a game is $\frac{2}{3}$ as given in the question.
So the required probability is $\frac{2}{3}\times\frac{2}{3}\times\frac{2}{3}+\frac{2}{3}\times\frac{2}{3}\times p+p\times\frac{2}{3}\times\frac{2}{3}+\frac{2}{3}\times p \times \frac{2}{3}+\frac{2}{3}\times\frac{2}{3}\times \frac{2}{3}\times\frac{2}{3}$
but i dont know what is the value of $p$,since it is not given in the question.I put $p=\frac{1}{2}$ but it is giving wrong answer.
Correct answer is $\frac{4}{9}$,i dont know where have i made the mistake or my logic is not correct?Please help me.Thanks.

$p$ is just the probability that the other player doesn't win, which is 1/3.
Also you need to multiply all the terms except the last by another 1/3, since B is winning one game, the probability of which is 1/3.
