Let's play a little game

Assume that at a given turn, A chooses label 0 and 1 with a probability of $p$ and $1-p$ respectively, likewisely B chooses label 0 and 1 with probability $q$ and $1-p$ , respectively.

We will start by calculating the expected value of money gained by B at this turn. Notice that the sum of the money A and B possesses does not change over the game, hence the sum of expected values must be 0 (so $E(A) = -E(B)$ ). This implies that if one of them manages to find a strategy that assures his expected value to be positive throughout the game, he is winning as $n$ goes to infinity.

It is known that the expected value equals to the weighted mean of the values of the possible outcomes with weights of corresponding probabilities, which means that

$E(B) = (-1) \cdot pq + 2 \cdot p(1-q) + 2 \cdot q(1-p) + (-a) \cdot (1-p)(1-q) = -(a+5)pq + (a+2)p + (a+2)q -a = -(a+5)[pq - \frac{a+2}{a+5}q \frac{a+2}{a+5}p + \frac{a}{a+5} = -(a+5)[(p-\frac{a+2}{a+5})(q-\frac{a+2}{a+5})-(\frac{a+2}{a+5})^2+\frac{a}{a+5}] = -(a+5)(p-\frac{a+2}{a+5})(q-\frac{a+2}{a+5}) + \frac{(a+2)^2}{a+5} -a = -(a+5)(p-\frac{a+2}{a+5})(q-\frac{a+2}{a+5}) + \frac{4-a}{a+5}$

Clearly B has a winning strategy if he manages to keep $E(B)$ positive at every turn. Note that if B chooses $q$ to be $\frac{a+2}{a+5}$ , the value $E(B)$ equals to $\frac{4-a}{a+5}$ , which is positive if $a<4$ . Thus we've concluded that if $a<4$ , with the strategy of choosing label 0 and 1 with probabilities $\frac{a+2}{a+5}$ and $1-\frac{a+2}{a+5}$ respectively, no matter what A's strategy is (nor the value of $p$ ), B is certainly winning as the number of games approaches infinity.

Similarly, if $a>4$ , with the strategy of pressing the button labelled with 0 and 1 with probability $\frac{a+2}{a+5}$ and $1-\frac{a+2}{a+5}$ respectively, A has got a winning strategy since $E(A)=-E(B)=-\frac{4-a}{a+5}>0$ .

If $a=4$ neither of them has a certain winning strategy, so the game can be called fair in this case.

Henceforth, our answer is 4.