Throw a coloured die until blue face is on top A game is played by rolling a six sided die which has four red faces and two blue faces. One turn consists of throwing the die repeatedly until a blue face is on top or the die has been thrown 4 times Adnan and Beryl each have one turn. Find the probability that Adnan throws the die more turns than Beryl I tried : Adnan throws two times and Beryl throws once = $\frac{2}{3}$ x $\frac{1}{3}$ Adnan throws three times and Beryl throws once =$\frac{4}{9}$ x $\frac{1}{3}$ Adnan throws three times and Beryl throws twice = $\frac{4}{9}$ x $\frac{2}{3}$ Adnan throws four times and Beryl throws once = $\frac{8}{27}$ x $\frac{1}{3}$ Adnan throws four times and Beryl throws twice = $\frac{8}{27}$ x $\frac{2}{3}$ Adnan throws four times and Beryl throws three times =$\frac{8}{27}$ x$\frac{4}{9}$ The answer says 0.365 Please help

When Beryl throws the die once, Adnan can throw it $2,3$ or $4$ times. The required probability is$$\underbrace{\frac13}_{\text{Beryl=1}}\left(1-\underbrace{\frac13}_{\text{Adnan=1}}\right)$$Similarly, when Beryl throws the die $2$ times, Adnan may throw $3$ or $4$ times, giving the required probability$$\underbrace{\frac23\frac13}_{\text{Beryl=2}}\left(1-\left(\underbrace{\frac13}_{\text{Adnan=1}}+\underbrace{\frac23\frac13}_{\text{Adnan=2}}\right)\right)$$and when Beryl throws it $3$ times, Adnan throws the die $4$ times. The last throw could result in a red face or a blue face. So the probability of this case is$$\underbrace{\frac23\frac23\frac13}_{\text{Beryl=3}}\left(\underbrace{\frac23\frac23\frac23\left[\frac23+\frac13\right]}_{\text{Adnan=4}}\right)$$The sum of these terms yields the required answer.