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

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.

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:
$$\mathbb{P}(B=1) = \frac{2}{6}$$ and $$\mathbb{P}(A\in \{2,3,4\}) = \mathbb{P}(A \neq 1) = \frac{4}{6}$$. And finally $$\mathbb{P}(B=1 \cap A \neq 1) = \mathbb{P}(B=1) \cdot \mathbb{P}(A \neq 1) = \frac{2}{9}$$. Note that I used that the events are independent.

Let $$X_1, X_2\sim\text{Geo}(1/3)$$ be independent identically distributed geometric random variables with probability of success $$1/3$$ i.e. $$P(X_1=k)=(2/3)^{k-1}(1/3)$$ for $$k\geq 1$$.(Here $$X_1$$ is the minimum number of times a dice must be rolled for a blue to come up). Then $$Y_i=\min(X_i, 4)$$ is the number of rolls in a turn (say $$i=1$$ corresponds to Adnan and $$i=2$$ corresponds to Beryl.

We want to compute $$P(Y_1>Y_2)$$. To this end note the equality of events $$(Y_1>Y_2)=(X_1>Y_1, 4>Y_2)=(X_1>Y_1, X_2<4)=(X_1>X_2, X_2<4)$$ Using the law of total probability we can write $$P(Y_1>Y_2)=\sum_{k=1}^3P(X_1>k, X_2=k)\stackrel{(\star)}{=}\sum_{k=1}^3 \left(\frac{2}{3}\right)^k \left(\frac{2}{3}\right)^{k-1}\frac{1}{3}=\frac{266}{729}$$ which is approximately $$0.365$$. We used independence in $$(\star)$$.

The probability that Adnan throws twice and Beryl once should be $$(\frac{2}{3}*\frac{1}{3})*\frac{1}{3}$$, not what you've written. That's because the probability that A throws twice is that of getting red in the first roll, blue in the second, which is (2/3)*(1/3). Then the probability that B throws only once is a further 1/3. Correct this in your calculations, you should get the correct answer. Do keep in mind that when we write for 4 throws, the turn ends no matter a blue or a red face comes up, so multiply the probability of three consecutive reds with 1.

What I'm seeing here is that you're calculating the probabilities of needing exactly two, three or four throws incorrectly. Since this looks like a homework problem, I'm not going to do the full calculation for you, but consider instead a very similar problem. Let's take the same game, but with slightly different rules. Now we have an 8-sided die with 2 blue faces and 6 red, and each turn has a maximum of 5 throws. Let $$X$$ be the number of throws in a player's turn. Then, the probability of a player needing one throw is

$$\mathbb{P}(X = 1) = \mathbb{P}(roll~a~blue~face) = \frac{2}{8} = \frac{1}{4},$$

similarly to how you correctly calculated the same probability for your problem. However, if we consider the probability of having a turn with exactly two throws, we need the event "first throw is NOT blue AND second throw is blue, i.e.

\begin{align} \mathbb{P}(X = 2) &= \mathbb{P}(1^{st}~roll~is~red) \cdot \mathbb{P}(2^{nd}~roll~is~blue) \\ &= \frac{6}{8} \cdot \frac{2}{8} = \frac{3}{4} \cdot \frac{1}{4} = \frac{3}{16}. \end{align}

Similarly,

$$\mathbb{P}(X = 3) = \frac{6}{8} \cdot \frac{6}{8} \cdot \frac{2}{8} = \frac{9}{64} \\ \mathbb{P}(X = 4) = ... = \frac{27}{256}.$$

Now, the final case, $$\mathbb{P}(X = 5)$$ is slightly different. Because we are cutting things off at five throws, we have two possibilities together. Namely, either the first four throws are red and the fifth is blue or all five throws are red. So, we could add these probabilities together. But, the much better way to think about this problem is to use the Complement Rule (a.k.a. the Law of Total Probability, depending on the book):

$$\mathbb{P}(X = 5) = 1 - \mathbb{P}(X \neq 5) = 1 - \mathbb{P}(X = 1,~2,~3,~or~4).$$

Since the events $$\mathbb{P}(X=1)$$, $$\mathbb{P}(X=2)$$, $$\mathbb{P}(X=3)$$, and $$\mathbb{P}(X=4)$$ are independent, we can use the Special Addition Rule to find the last probability. Again, since this is clearly a homework problem, I will leave that last step to you. It is straightforward given everything I've already discussed here.

• Love downvotes with no explanation. If you disagree with my answer, at least tell me why. – Charles Beer Sep 3 at 20:38