There Just Might Be A Gremlin In Your House.

I used a Markov chain to model this problem. There are 21 states: $n = 0, 1, 2, ..., 20$ . Of these, 19 states are transient (their chances decrease to zero in the limit) and 2 states are permanent (for $n = 0$ and $n = 20$ ).

To model a state change, you can skip the situation where the chosen Gremlin or Mogwai chooses one of its own (nothing changes). Then the state changes become simple: in one of the transient states, $n$ decreases with probability $p$ or $n$ increases with probability $1-p=q$ .

Let $A$ be the 19 x 19 matrix showing the probabilities to move from one of the 19 transient states: $1 \leq n \leq 19$ . Then $A_{i+1,i}=q$ (here $q=\frac{19}{36}$ ); $A_{i,i+1}=p$ (here $p=\frac{17}{36}$ ) for $i=1,2,...,18$ and $A_{i,j}=0$ for all other $1 \leq i, j \leq 19$ . Let $Q$ be the 2 x 19 matrix showing the probabilities to reach one of the two permanent states (n=0 or n=20). The matrix $R$ is the 2 x 2 matrix showing the probabilities for the permanent states. Here $R = I_{2,2}$ , the identity matrix of order 2 x 2.

Furthermore, let $b=\begin{pmatrix}b_1 \\ b_2 \end{pmatrix}$ be the vector with the initial probabilities of the 19 transient states ( $b_1$ ) and the two permanent states ( $b_2$ ). Here, $b=\begin{pmatrix}1 \\ 0 \\ 0 \\ . \\ . \\ . \\ 0 \\ 0 \\ 0\end{pmatrix}$ .

The asked probabilities are governed by $M=\begin{pmatrix}A & 0 \\ Q & R \end{pmatrix}$ , where $0$ is the 19 x 2 matrix with all zeroes.

The probabilities can be computed one step at a time: we start with $b$ . After the first conversion, we have $Mb$ . After the 8th conversion we have $M^8b$ . You can see:

$M^k =$

$=\begin{pmatrix}A^k & 0 \\ QA^{k-1} + RQA^{k-2} + R^2 QA^{k-3} + ... + R^{k-1}Q & R^k \end{pmatrix}$

$= \begin{pmatrix} A^k & 0 \\ Q(A^{k-1}+A^{k-2}+...+I) & R^k \end{pmatrix}$

In the limit, $A^k \rightarrow 0$ and the series of 2 x 19 matrices in the lower left corner approach $Q(I-A)^{-1}$ . Now, we don't need to know the complete inverse of the matrix $I-A$ , but only its first column, let's call that vector $(c_i)_i$ , since $b_1=e_1$ , the first unit vector of dimension 19.

It can be shown that there are two solutions for $c_i$ to have all but rows 1 and 19 of $(I-A)^{-1}c$ become zero: $c_i = a.1^i$ and $c_i = b.(\frac p q)^i$ for real constants $a$ and $b$ . Carefully choosing $a$ and $b$ so that the last row becomes $0$ also and row $1$ becomes exactly $1$ , gives us: $c_i=\frac{(1-f^{20-i})}{q.(1-f^{20})}$ , for $i=1, 2, ..., 19$ and $f=\frac p q$ .

Now, only what is left, is to calculate only the bottom (second) row of $Q(I-A)^{-1}b=Qc$ , or, $Q_{2,19}.c_{19}=q.\frac{(1-f)}{q.(1-f^{20})}=\boxed{ \frac{1-f}{1-f^{20}} = \frac{ 1-\frac{17}{19} }{ 1-(\frac{17}{19})^{20} } \approx 0.118 }$ .

This problem is just like what's given in the following link that uses Markov Chain. Gambler's Ruin Problem

The answer is, therefore, equal to = {1 - (17/19)}/{1 - (17/19)^{20}} = 0.118024.