Faulty reasoning in two person card game problem
Suppose Mike plays a randomised strategy of playing "0" with probability $\dfrac58$ and playing "1" with probability $\dfrac38$

Then @Herr K.'s table becomes (in expectation) 
$$\begin{array}{|c|c|c|}\hline&amp;\text{'0'}&amp;\text{'1'}\\\hline\text{Mike's }\frac58,\frac38 \text{ combo}&amp;\frac18,-\frac18&amp;\frac18,-\frac18\\\hline\end{array}$$

and there is nothing Cheng can do to get a better result than an expected loss of $\frac18$.  So in that sense the game is advantageous to Mike if Mike takes this approach

Incidentally, here Cheng can restrict expected losses to this amount by playing the same mixed strategy, and both players doing so is a Nash equilibrium