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 $\frac{1}{2}$

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 $\frac{1}{2}$

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

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 $\frac{1}{2}$

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 $\frac{1}{2}$

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