Distribution of throws of die rigged to never produce twice in a row the same result A die is “fixed” so that each time it is rolled the score cannot be the same as the preceding score, all other scores having probablity 1/5. If the first score is 6, what is the probability that the nth score is 6 and what is the probability that the nth score is 1?
 A: The Markov Chain has 6 states labelled 1 through 6. For each state, the transition probability to each of the other states is $1/5$. So the probability transition matrix is $M=\dfrac{1}{5}(J-I)$ where $J$ is the all ones matrix, and $I$ is the identity matrix.
The distribution vector for the $n$th score is given by $(1,0,0,0,0,0)M^n$, i.e. by the first row of $M^n$.
A: Let $p_n$ be the probability the $n$-th toss is a $6$. Note that $p_1=1$. We obtain a recurrence for $p_n$. 
The probability the $n$-th toss is not a $6$ is  $1-p_n$. Given that, the probability the $n$-th toss is a $6$ is $\frac{1}{5}$. 
Thus
$$p_{n+1}=\frac{1}{5}(1-p_n)=\frac{1}{5}-\frac{1}{5}p_n.$$
The homogeneous recurrence $p_{n+1}=-\frac{1}{5}p_n$ has general solution $A(-1/5)^n$. A particular solution of the non-homogeneous recurrence is $1/6$. 
Thus the general solution of our recurrence is $\frac{1}{6}+\frac{A(-1)^n}{5^n}$. Since $p(1)=1$, we have $A=-\frac{25}{6}$ and therefore 
$$p_n=\frac{1}{6}\left(1+\frac{(-1)^{n-1}}{5^{n-2}}    \right).$$
Now that we know the probability of a $6$, given the first is a $6$, we can easily compute the probability of $1$, since by symmetry $1$ to $5$ are equally likely.
