TITLE: Probability that no two consecutive throws of some (A,B,C,D,E,F)-die show up consonants
QUESTION [9 upvotes]: I have a question on probability. I am looking people presenting different approaches on solving this. I already have one solution but I was not satisfied like a true mathematician ;).....so go ahead and take a dig.....if no one answers....I will post my solution....thanks!

There is an unbiased cubical die with its faces labeled as A, B, C, D, E and F. If the die is thrown $n$ times, what is the probability that no two consecutive throws show up consonants?

If someone has already asked a problem of this type then I will be grateful to be redirected :)

REPLY [6 votes]: The problem can be looked at from the Markov chain perspective. Let the state of the Markov chain be the number of consecutive consonants. We are only interested in there being no greater than 2 of them. So the state space is $0, 1, 2$. The transition matrix is then
$$
   P = \left( \begin{array}{ccc} 1-p & p & 0 \\ 1-p & 0 & p \\ 0 & 0 & 1 \end{array} \right)
$$
where $p$ is the probability if getting a consonant at a die throw, which is $p=\frac{4}{6}=\frac{2}{3}$.
With the starting state of zero consecutive consonants, the probability sought is $p_n = 1 - (P^n)_{13}$.
This gives the same result as given by joriki and Didier:
In[58]:= With[{p = 2/3}, 
 FullSimplify[(1 - 
      MatrixPower[{{1 - p, p, 0}, {1 - p, 0, p}, {0, 0, 1}}, n])[[1, 
     3]] == 3^(-1 - n) (2^(2 + n) - (-1)^n), 
  n \[Element] Integers && n >= 0]]

Out[58]= True