# 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 :)

• This may be relevant: math.stackexchange.com/questions/73758/… – joriki Oct 23 '11 at 13:46
• This is a duplicate of this question of mine. You have $n$ Bernoulli trials with $p=\frac{1}{3}$ for having a consonant. You are seeking to find a probability of having no runs of length $\ell \ge 2$. – Sasha Oct 23 '11 at 13:47
• @Sasha: Well spotted; but I'm not voting to close as a duplicate since the $m=2$ case is very difficult to extract from Byron's general answer and this question asks specifically for that case, which has a relatively simple answer. – joriki Oct 23 '11 at 14:01

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.)

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:= 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= True


Building on @Mike's answer to this question, with $2$ of $6$ sides labeled with consonants, we have $p=\frac13$ and $q=\frac23$, so the probability to have the first occurrence of a double success at the $k$-th throw is

$$\begin{eqnarray} P_k &=& \frac{\left(\frac13\right)^2\left(\left(\frac23 + \sqrt{\left(\frac23\right)^2+4\cdot\frac13\cdot\frac23}\right)^{k-1} - \left(\frac23 - \sqrt{\left(\frac23\right)^2+4\cdot\frac13\cdot\frac23}\right)^{k-1}\right)}{2^{k-1}\sqrt{\left(\frac23\right)^2+4\cdot\frac13\cdot\frac23}} \\ &=& \frac49 \left( \left(\frac23\right)^{k-1}-\left(-\frac13\right)^{k-1}\right)\;. \end{eqnarray}$$

What we need in the present case is the sum of these probabilities up to and including $n$:

$$\begin{eqnarray} \sum_{k=1}^n \frac49 \left( \left(\frac23\right)^{k-1}-\left(-\frac13\right)^{k-1}\right) &=& \frac49 \left(\frac{1-\left(\frac23\right)^n}{1-\frac23}-\frac{1-\left(-\frac13\right)^n}{1-\left(-\frac13\right)}\right) \\ &=& 1-\frac13\left(4\left(\frac23\right)^n-\left(-\frac13\right)^n\right) \end{eqnarray}$$

Subtracting this from $1$ to get the probability of at least one such double success yields

$$\frac{4\cdot2^n-(-1)^n}{3^{n+1}}$$