  We construct a Parrondo's game using discrete time quantum walks. Two lossing
games are represented by two different coin operators. By mixing the two coin
operators $U_{A}(\alpha_{A},\beta_{A},\gamma_{A})$ and
$U_{B}(\alpha_{B},\beta_{B},\gamma_{B})$, we may win the game. Here we mix the
two games in position instead of time. With a number of selections of the
parameters, we can win the game with sequences ABB, ABBB, \emph{et al}. If we
set $\beta_{A}=45^{\circ},\gamma_{A}=0,\alpha_{B}=0,\beta_{B}=88^{\circ}$, we
find the game 1\emph{}with {\normalsize
$U_{A}^{S}=U^{S}(-51^{\circ},45^{\circ},0)$,
$U_{B}^{S}=U^{S}(0,88^{\circ},-16^{\circ})$ will win and get the most
profit.}If we set
$\alpha_{A}=0,\beta_{A}=45^{\circ},\alpha_{B}=0,\beta_{B}=88^{\circ}$
and{\normalsize{} the game 2 with $U_{A}^{S}=U^{S}(0,45^{\circ},-51^{\circ})$,
$U_{B}^{S}=U^{S}(0,88^{\circ},-67^{\circ})$, will win most. And}game
1\emph{}{\normalsize is equivalent to the}game\emph{}2\emph{}with the changes
of sequences and steps. But at a large enough steps, the game will loss at
last.
