Qunatum Parrondo's games constructed by quantum random walk

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 $U_{A}(α_{A},β_{A},γ_{A})$ and $U_{B}(α_{B},β_{B},γ_{B})$, 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 $β_{A}=45^{\circ},γ_{A}=0,α_{B}=0,β_{B}=88^{\circ}$, 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 $α_{A}=0,β_{A}=45^{\circ},α_{B}=0,β_{B}=88^{\circ}$ 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.

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