Probability related, dice problem

Question

A dice is turned randomly from one face to one of four adjoined faces. Suppose we start from the position with the number 6 on top. Find the probability $P_n$ that after $n$ such turns the dice will show the number $6$ again on top. Find the limit $\lim_{n\rightarrow \infty} P_n$ .

Answer 1

After the first step the $6$ will be on the side of the dice (neither up or down). After that due to symmetry it will be as probable that it is up that it is down.

We have that if the $6$ is on top or bottom the next step it will be on the side certainly. OTOH if it's on the side it will be $1/4$ chance that it will be on top next step and $1/4$ chance that it will be at the bottom at the next step. In total this means that when on side it will be $1/2$ chance that it will remain on the side and $1/2$ chance that it will be at top or bottom next step.

This means that the probability of the $6$ being on top or bottom is half the probability that it was at the side the previous step which is the complementary event that it's at top or bottom. Let $S_n$ be the event that the $6$ is on the side and $T_n$ is that it is not (ie top or bottom). We've see that $\Pr(T_{n+1}|T_n) = 0$ and $\Pr(T_{n+1}|S_n) = 1/2$. Let $Q_n$ be the probability that it's at top or bottom we have:

$$Q_{n+1} = \Pr(T_{n+1}) = \Pr(T_{n+1}\cap T_n) + \Pr(T{n+1}\cap S_n) = \Pr(T_{n+1}|T_n)\Pr(T_n) + \Pr(T_{n+1}|S_n)\Pr(S_n) \\= {1\over 2}\Pr({S_n}) = {1\over 2}(1-\Pr(T_n)) = (1-Q_n)/2$$

And $Q_n = 2P_n$ by symmetry and therefore we have:

$$P_{n+1} = (1-2P_n)/4$$

or

$$P_n = {1+2\cdot (-2)^{-n}\over 6}$$

which can be proven by induction (with the initial condition $P_1 = 0$. Therefore we have $\lim P_n = 1/6$ as expected.

Answer 2

Let $Q_n$ denote the probability that the face opposite to face $6$ is on top after $n$ turns.

By symmetry we find that: $$P_n=Q_n\text{ for }n=1,2,\dots\tag1$$

Next to that we have: $$P_n=\frac14(1-P_{n-1}-Q_{n-1})\text{ for }n=1,2,\dots\tag2$$ and combining $(1)$ and $(2)$ we find: $$P_n=\frac14-\frac12P_{n-1}\text{ for }n=2,3,\dots$$

Of course we start with $P_0=1$ and $P_1=0$.

Can you take it from here?

Answer 3

This is Markov chain with state set $\{1,2,3,4,5,6\}$ and transition probability matrix

$$\begin{bmatrix} 0&\frac14&\frac14&\frac14&\frac14&0\\ \frac14&0&\frac14&\frac14&0&\frac14\\ \frac14&\frac14&0&0&\frac14&\frac14\\ \frac14&\frac14&0&0&\frac14&\frac14\\ \frac14&0&\frac14&\frac14&0&\frac14\\ 0&\frac14&\frac14&\frac14&\frac14&0 \end{bmatrix}.$$

(I don't know if my die is a standard one but I could get the neighboring faces by looking at my die.)

It is easy to check that the stationary probabilities are $\left[\frac16\ \frac16\ \frac16\ \frac16\ \frac16\ \frac16\right].$

So, $$\lim_{n\to \infty}\begin{bmatrix} 0&\frac14&\frac14&\frac14&\frac14&0\\ \frac14&0&\frac14&\frac14&0&\frac14\\ \frac14&\frac14&0&0&\frac14&\frac14\\ \frac14&\frac14&0&0&\frac14&\frac14\\ \frac14&0&\frac14&\frac14&0&\frac14\\ 0&\frac14&\frac14&\frac14&\frac14&0. \end{bmatrix}^n=\begin{bmatrix} \frac16&\frac16&\frac16&\frac16&\frac16&\frac16\\ \frac16&\frac16&\frac16&\frac16&\frac16&\frac16\\\frac16&\frac16&\frac16&\frac16&\frac16&\frac16\\\frac16&\frac16&\frac16&\frac16&\frac16&\frac16\\\frac16&\frac16&\frac16&\frac16&\frac16&\frac16 \\\frac16&\frac16&\frac16&\frac16&\frac16&\frac16 \end{bmatrix}.$$

From the sixth power of the state transition matrix we can tell that the probability of getting $6$ after six steps starting with six is:

$$P(S_6=6\mid S_1=6)=\frac{11}{64}\approx \frac16.$$