# A game is played by rolling a six sided die which has four red faces and two blue faces. One turn co

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 $\frac{4}{9}$
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Caiden Barrett
When Beryl throws the die once, Adnan can throw it 2,3 or 4 times. The required probability is
$\underset{\text{Beryl=1}}{\underset{⏟}{\frac{1}{3}}}\left(1-\underset{\text{Adnan=1}}{\underset{⏟}{\frac{1}{3}}}\right)$
Similarly, when Beryl throws the die 2 times, Adnan may throw 3 or 4 times, giving the required probability
$\underset{\text{Beryl=2}}{\underset{⏟}{\frac{2}{3}\frac{1}{3}}}\left(1-\left(\underset{\text{Adnan=1}}{\underset{⏟}{\frac{1}{3}}}+\underset{\text{Adnan=2}}{\underset{⏟}{\frac{2}{3}\frac{1}{3}}}\right)\right)$
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
$\underset{\text{Beryl=3}}{\underset{⏟}{\frac{2}{3}\frac{2}{3}\frac{1}{3}}}\left(\underset{\text{Adnan=4}}{\underset{⏟}{\frac{2}{3}\frac{2}{3}\frac{2}{3}\left[\frac{2}{3}+\frac{1}{3}\right]}}\right)$
The sum of these terms yields the required answer.
###### Not exactly what you’re looking for?
Ellen Chang
Well, A throws it more often than B in the following cases:
B throws 1, A throws 2,3,4
B throws 2, A throws 3,4
B throws 3, A throws 4

And now you go and check for the all-together probability of these events. For example:B throws 1, A throws 2,3,4:
$\mathbb{P}\left(B=1\right)=\frac{2}{6}$ and $\mathbb{P}\left(A\in \left\{2,3,4\right\}\right)=\mathbb{P}\left(A\ne 1\right)=\frac{4}{6}$. And finally $\mathbb{P}\left(B=1\cap A\ne 1\right)=\mathbb{P}\left(B=1\right)\cdot \mathbb{P}\left(A\ne 1\right)=\frac{2}{9}$. Note that I used that the events are independent.