Why is the answer to this $2$ dice and a coin probability question not $\frac{1}{2}$? I was trying to solve the following homework probability question which has the following setup:

We have $2$ dice: $A$ and $B$. Die $A$ has $4$ red faces and $2$ white faces, whereas die $B$ has $2$ red faces and $4$ white faces. On each turn, a fair coin is tossed. If the coin lands on heads then die $A$ is thrown, but if the coin lands on tails then die $B$ is thrown. After this the turn ends, and on the next turn the coin is once again tossed to determine the die (i.e. we throw the coin and the corresponding die on each turn).

From this game I'm asked to answer $2$ questions:


*

*Show that the probability of obtaining a red face on any $n$-th throw is $\frac{1}{2}$.

*If the first $2$ consecutive die throws result in red faces, what is the probability that the third throw is also red?



To answer part $1$ I used the law of total probability. Denoting obtaining a red face on the $n$-th die throw as $P(R_n)$ I get
\begin{align*}
P(R_n) &= P(R_n \vert A) P(A) + P(R_n \vert B) P( B) \\
& = \left(\frac{4}{6}\right)\left(\frac{1}{2}\right) + \left(\frac{2}{6}\right)\left(\frac{1}{2}\right)\\
& = \frac{1}{2}
\end{align*}
But on question $2$ is where I started running into trouble. Using the same notation, what I want to calculate is $P(R_3 \vert R_2 R_1)$. And recalling that for events $E_1$ and $E_2$ we can say that
$$
P(E_2 \vert E_1) = \frac{P(E_2E_1)}{P(E_1)}
$$
I get that
$$
P(R_3 \vert R_2 R_1) = \frac{P(R_3 R_2 R_1)}{P(R_2 R_1)} \tag{1}
$$
and from here I obtain 2 different solutions using $2$ distinct methods:
Answer 1
Using that on the first part of the question we showed that $P(R_n) = \frac{1}{2}$, and noticing that the die throws are independent since what I threw before does not affect how I throw the next coin toss or how I roll the next die, from equation $(1)$ I get
\begin{align*}
P(R_3 \vert R_2 R_1) &= \frac{P(R_3 R_2 R_1)}{P(R_2 R_1)}\\
&= \frac{P(R_3) P(R_2) P(R_1)}{P(R_2) P(R_1)}\\
& = P(R_3) = P(R_n) = \frac{1}{2}
\end{align*}
Answer 2
Using the law of total probability on $(1)$ I get
\begin{align*}
P(R_3 \vert R_2 R_1) & = \frac{P(R_1 R_2  R_3 \vert A)P(A)+ P(R_1  R_2  R_3 \vert B)P(B)}{P(R_1  R_2 \vert A ) P(A)+ P(R_1  R_2 \vert B) P(B)}\\
&= \frac{\left[\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)\right]\frac{1}{2}+ \left[\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)\right]\frac{1}{2}}{\left[\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)\right]\frac{1}{2}+\left[\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)\right]\frac{1}{2}}\\
& = \frac{3}{5}
\end{align*}

To me both of the previous answers seem to be following coherent logic, but since I didn't get the same answer I knew one of them was wrong. I decided to write a program to simulate the game and I found out that the correct solution was $P(R_3 \vert R_2 R_1) = \frac{3}{5}$. But even though I verified this answer to be correct I couldn't seem to understand what part of my analysis is wrong on Answer 1.
So my question is, why is
$$
P(R_3 \vert R_2 R_1) \neq \frac{1}{2}\quad ?
$$
 A: The flaw in your second answer is that it is not necessarily the same die that is thrown after each coin toss.  The way you have written the law of total probability is acceptable in your first answer, but not in the second, because it is only in the second answer that you conditioned on the events $A$ and $B$, which represent the outcomes of the coin toss.  Consequently, the result is incorrect because it corresponds to a model in which the coin is tossed once, and then the corresponding die is rolled three times.
First, let us do the calculation the proper way.  We want $$\Pr[R_3 \mid R_1, R_2] = \frac{\Pr[R_1, R_2, R_3]}{\Pr[R_1, R_2]}$$  as you wrote above.  Now we must condition on all possible outcomes of the coin tosses, of which there are eight:
$$\begin{align}
\Pr[R_1, R_2, R_3] &= \Pr[R_1, R_2, R_3 \mid A_1, A_2, A_3]\Pr[A_1, A_2, A_3] \\ &+ \Pr[R_1, R_2, R_3 \mid A_1, A_2, B_3]\Pr[A_1, A_2, B_3] \\
&+ \Pr[R_1, R_2, R_3 \mid A_1, B_2, A_3]\Pr[A_1, B_2, A_3] \\ &+ \Pr[R_1, R_2, R_3 \mid A_1, B_2, B_3]\Pr[A_1, B_2, B_3] \\
&+\Pr[R_1, R_2, R_3 \mid B_1, A_2, A_3]\Pr[B_1, A_2, A_3] \\ &+ \Pr[R_1, R_2, R_3 \mid B_1, A_2, B_3]\Pr[B_1, A_2, B_3] \\
&+\Pr[R_1, R_2, R_3 \mid B_1, B_2, A_3]\Pr[B_1, B_2, A_3] \\ &+ \Pr[R_1, R_2, R_3 \mid B_1, B_2, B_3]\Pr[B_1, B_2, B_3] \\
\end{align}$$
and since each of the $2^3 = 8$ triplets of ordered coin tosses has equal probability of $1/8$ of occurring,
$$\Pr[R_1, R_2, R_3] = \tfrac{1}{8}\left((\tfrac{2}{3})^3 + 3(\tfrac{2}{3})^2(\tfrac{1}{3}) + 3(\tfrac{2}{3})(\tfrac{1}{3})^2 + (\tfrac{1}{3})^3\right) 
= \tfrac{1}{8}.$$  A similar (but simpler) calculation for the denominator yields $1/4$, and the result follows.
Of course, none of this is necessary; it is only shown here to illustrate how the calculation would be done if it were to be done along the lines of your second answer.
A: With the  second approach:
To have first two die throws resulting in red, there are 2x2 paths: AA,AB,BA, BB $\to$
$P(R1\cap R2)=(\frac{1}{2})^2(\frac{2}{3})^2$+2$(\frac{1}{2})^2\frac{2}{3}\frac{1}{3}$+$(\frac{1}{2})^2(\frac{1}{3})^2$ =$\frac{1}{4}$
To have three  die throws resulting in red, there are 2x2x2 paths:AAA,BBB,AAB,ABA,BAA,BBA,BAB,ABB $\to$
$P(R1\cap R2\cap R3)= (\frac{1}{2})^3(\frac{2}{3})^3$+3$(\frac{1}{2})^3(\frac{2}{3})^2\frac{1}{3}$+3$(\frac{1}{2})^3(\frac{1}{3})^2\frac{2}{3}$+ $(\frac{1}{2})^3(\frac{1}{3})^3$ =$\frac{1}{8}$
$\frac{P(R1\cap R2\cap R3)}{P(R1\cap R2)}$ =$\frac{1}{2}$
In this way, you get consistent result compared to first approacch.
