Probability that no two consecutive throws of some (A,B,C,D,E,F)-die show up consonants 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 :)

Here is a solution different from the one given on the page @joriki links to. Call $c_n$ the probability that no two consecutive consonants appeared during the $n$ first throws and that the last throw produces a consonant. Call $b_n$ the probability that no two consecutive consonants appeared during the $n$ first throws and that the last throw did not produce a consonant. Thus one is looking for $p_n=c_n+b_n$.
For every $n\geqslant1$, $c_{n+1}=\frac23b_n$ (if the previous throw was a consonant, one cannot get a consonant now and if it was not, $\frac23$ is the probability to get a consonant now) and $b_{n+1}=\frac13b_n+\frac13c_n$ (if the present throw is not a consonant, one asks that two successive consonants were not produced before now). Furthermore $c_1=\frac23$ and $b_1=\frac13$. One asks for $p_n=3b_{n+1}$ and one knows that $9b_{n+2}=3b_{n+1}+3c_{n+1}=3b_{n+1}+2b_n$ for every $n\geqslant1$. The roots of the characteristic equation $9r^2-3r-2=0$ are $r_2=\frac23$ and $r_1=-\frac13$ hence $b_n=B_2r_2^n+B_1r_1^n$ for some $B_1$ and $B_2$. One can use $b_2=\frac13$ as second initial condition, this yields $B_2=\frac23$ and $B_1=\frac13$ hence $b_n=r_2^{n+1}-r_1^{n+1}$.
Finally, $p_n=3^{-n-1}\left(2^{n+2}-(-1)^{n}\right)$. (Sanity check: one can check that $p_0=p_1=1$, $p_2=\frac59$, and even with some courage that $p_3=\frac{11}{27}$, are the correct values.) 
