Faulty reasoning in two person card game problem I was asked this question in an interview: There are two people Mike and Cheng with one blank card each. Both write either a '0' or a '1' on it without showing the other while writing. They then simultaneously show their cards to each other. If the payoffs are as follows depending on the cards shown:

1,1: Mike pays Cheng 3

0,0: Mike pays Cheng 1

1,0: Mike receives 2 from Cheng

0,1: Mike receives 2 from Cheng

Is this game fair?

Using game theory I reason: Cheng will always play 1 because his payoff from 1 is (-2 or 3) as opposed to his payoff from playing 0 which is (1 or -2). Knowing that Cheng will prefer to play 1, Mike always plays 0 to get a payoff of 2. So the game is unfair to Cheng.

But I also reason: Mike will never play 1 because he has a payoff (2 or -3) vs playing 0 which has a payoff (-1 or 2). So Mike will always prefer to play 0 (this is also inline with the above reasoning). Knowing that Mike will always prefer 0, Cheng always plays 0 so Mike gets -1 every round. So this game is unfair to Mike.

Regardless if the game is played one round or many rounds, I am unable to conclude which is correct.

Can anyone help explain why I am getting conflicting results?

PS: Later the interviewer told me that the game is unfair and Mike always wins.

PPS: I apologize in advance if I have made any mistakes in the posting - this is my first post.
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