# Balls in an urn

Suppose that there are 6 balls in an urn, 2 red and 4 white. There are two players. The first player draws a ball at random. If the ball is white then it is replaced and the other player draws, and so forth, continuing until a red ball is drawn. The player drawing the first red ball loses.

a) What is the probability that the player who goes first wins? b) What is the probability that the player who goes first wins if after each draw of a white ball it is replaced in the urn by a red ball?

a) Denote $q$ the probability that the player who goes first loses. Then you have that he can either lose directly (i.e. draw a red ball with probability $2/6$) or he can draw a white ball ($4/6$), then player draw a white ball as well ($4/6$) and then the game practically starts over, so he will lose with probability $q$ from then on (renewal argument). Thus $$q=\frac26+\frac46\cdot\frac46\cdot q$$ which gives $$q=\frac35$$ So the probability that the first player wins is $p=1-q=\frac25$.
b) Now the probability $q$ that he loses is $$q=\frac26+\frac46\cdot\frac36\cdot\left[\frac46+\frac26\cdot\frac16\cdot 1\right]=\frac{31}{54}$$ and the required probability is $p=1-q=\frac{23}{54}$.