A fair dice tossed repeatedly. A wins if it is 1 or 2 on two consecutive tosses and B wins if it is 3, 4,5,6 on two consecutive tosses. The probability that A wins if the die is tossed indefinitely:
My attempt: let p be the probability of A winning.
P=(4/36)+(4/6)(2/6)*2(1-p) Which gives p=13/21.
But the answer is 5/21.
On
Another solution:
$$P = \dfrac{1}{9}+\dfrac{2}{3}\cdot \dfrac{1}{9} + \dfrac{1}{3}\cdot \dfrac{2}{3}\cdot \dfrac{1}{9} + \dfrac{2}{3}\cdot \dfrac{1}{3}\cdot \dfrac{2}{3}\cdot \dfrac{1}{9} + \cdots$$
Collecting even terms and odd terms into separate summations:
$$P = \dfrac{1}{9}\sum_{n\ge 0} \left(\dfrac{2}{9}\right)^n + \dfrac{1}{9}\cdot \dfrac{2}{3}\sum_{n\ge 0}\left(\dfrac{2}{9}\right)^n = \dfrac{5}{27}\cdot \dfrac{1}{1-\tfrac{2}{9}} = \dfrac{5}{27}\cdot \dfrac{9}{7} = \dfrac{5}{21}$$
Let $P_1$ be the probability of winning when the previous roll was $1,2$ (and the game is not over), and $P_2$ the probability of winning when the previous roll was $3,4,5,6$ and the game is not over.
So, we are looking for $\frac{P_1}3+\frac{2P_2}3$.
Note that when we just rolled $1,2$, we win if we roll $1$ or $2$, or if we roll otherwise and win from state $2$. So, $$P_1=\frac13+\frac23P_2$$ If we just rolled $3,4,5,6$, we need to roll $1,2$ to get to state $1$ to have any hope of winning. So, $$P_2=\frac13P_1$$ Hence, $P_1=\frac13+\frac29P_1\to\frac79P_1=\frac13$. Therefore, $P_1=\frac37$ and $P_2=\frac17$.
So, $$\frac{P_1}3+\frac{2P_2}3=\frac3{21}+\frac2{21}=\frac5{21}$$