math (Probability)

romsek

Math Team
Sep 2015
2,958
1,673
USA
There is another very elegant way to solve this if you are interested.

We treat the whole thing as an absorbing Markov chain.

The state transition matrix is

\(\displaystyle T = \begin{pmatrix}
0 &\dfrac 1 2 &\dfrac 1 2 &0\\
\dfrac 5 6 &0 &0 &\dfrac 1 6\\
0 &0 &1 &0\\
0 &0 &0 &1
\end{pmatrix}\)

This breaks up into the $Q$ and $R$ matrices as

$Q=\begin{pmatrix}0 &\dfrac 1 2 \\\dfrac 5 6 &0\end{pmatrix}\\
R = \begin{pmatrix}\dfrac 1 2 &0 \\0 &\dfrac 1 6\end{pmatrix}\\
N = \left(I_2 - Q\right)^{-1} =
\begin{pmatrix}
\dfrac{12}{7} & \dfrac{6}{7} \\
\dfrac{10}{7} & \dfrac{12}{7} \\
\end{pmatrix}\\
B = NR =
\begin{pmatrix}
\dfrac{6}{7} & \dfrac{1}{7} \\
\dfrac{5}{7} & \dfrac{2}{7} \\
\end{pmatrix}$

From $B$ we can directly read off that $P[\text{MW|start with MP}] = \dfrac 6 7$ and $P[\text{MW|start with SP}] = \dfrac 5 7$
 
Mar 2020
20
0
Netherlands
There is another very elegant way to solve this if you are interested.

We treat the whole thing as an absorbing Markov chain.

The state transition matrix is

\(\displaystyle T = \begin{pmatrix}
0 &\dfrac 1 2 &\dfrac 1 2 &0\\
\dfrac 5 6 &0 &0 &\dfrac 1 6\\
0 &0 &1 &0\\
0 &0 &0 &1
\end{pmatrix}\)

This breaks up into the $Q$ and $R$ matrices as

$Q=\begin{pmatrix}0 &\dfrac 1 2 \\\dfrac 5 6 &0\end{pmatrix}\\
R = \begin{pmatrix}\dfrac 1 2 &0 \\0 &\dfrac 1 6\end{pmatrix}\\
N = \left(I_2 - Q\right)^{-1} =
\begin{pmatrix}
\dfrac{12}{7} & \dfrac{6}{7} \\
\dfrac{10}{7} & \dfrac{12}{7} \\
\end{pmatrix}\\
B = NR =
\begin{pmatrix}
\dfrac{6}{7} & \dfrac{1}{7} \\
\dfrac{5}{7} & \dfrac{2}{7} \\
\end{pmatrix}$

From $B$ we can directly read off that $P[\text{MW|start with MP}] = \dfrac 6 7$ and $P[\text{MW|start with SP}] = \dfrac 5 7$
OH thank you so much;)