## Testing a Game For Bias Example

Two players pay dice with the following rule.
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
The transition matrix is
$P= \left( \begin{array}{cccc} 1/3 & 1/2 & 1/6 & 0 \\ 1/2 & 1/3 & 0 & 1/6 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$
.
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.