If a player tosses a 6, he wins.
If a 4 or 5 is tossed the player throws the dice again.
If a 1, 2 or 3 is tossed, the dice passes to the other player.
Is the game biased?
We can construct the table:
| A's Turn | B's Turn | A wins | B wins | |
| A's Turn | 1/3 | 1/2 | 1/6 | 0 |
| B's Turn | 1/2 | 1/3 | 0 | 1/6 |
| A Wins | 0 | 0 | 1 | 0 |
| B Wins | 0 | 0 | 0 | 1 |
\[P= \left( \begin{array}{cccc} 1/3 & 1/2 & 1/6 & 0 \\ 1/2 & 1/3 & 0 & 1/6 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)\]
.This matrix is of the form
\[P= \left( \begin{array}{cc} Q & R \\ O & I \end{array} \right)\]
where\[Q=\left( \begin{array}{cc} 1/3 & 1/2 \\ 1/2 & 1/3 \end{array} \right), \; Q= \left( \begin{array}{cc} 1/6 & 0 \\ 0 & 1/6 \end{array} \right), O= \left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right), \; I= \left( \begin{array}{cccc} 1 & 0 \\ 0 & 1 \end{array} \right)\]
\[P^n= \left( \begin{array}{cc} Q^n & (Q^{n-1}+Q^{n-2}+...+Q+I)R \\ O & I \end{array} \right)\]
But
\[(Q^{n-1}+Q^{n-2}+...+Q+I)(I-Q)=I-Q^n \rightarrow Q^{n-1}+Q^{n-2}+...+Q+I=(I-Q^n)(I-Q)^{-1}\]
.We can write
\[P^n= \left( \begin{array}{cc} Q^n & (I-Q^n)(I-Q)^{-1}R \\ O & I \end{array} \right)\]
The eigenvalues of
\[Q\]
are 5/6, -1/6, so \[Q^n \rightarrow \left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right) \]
as \[n \rightarrow \infty\]
.Hence
\[P^n= \left( \begin{array}{cc} 0 & (I-Q)^{-1}R \\ O & I \end{array} \right)\]
as \[n \rightarrow \infty\]
.\[(I-Q)^{-1}R= \left( \begin{array}{cc} 4/7 & 3/7 \\ 3/7 & 4/7 \end{array} \right)\]
is the part of the matrix \[P^n\]
concerned with the probabilities of making a transition from the various possible starting states to the various possible winning states. If A starts, the probability of him winning provided the game goes on long term is 4/7, so the game is biased in his favour.