Suppose the matching pennies game :
$$\begin{array}[ccc]1 & q & 1-q\\p&(1,-1)&(-1,1)\\1-p&(-1,1)&(1,-1)\end{array}$$
Suppose we were player 2 and choose our strategy in function of 1s choices : $q=f(p)$
The gain is $$G2(p)=(1-2p)(2f(p)-1)\\ G2'(p)=-2(2f(p)-1)+(1-2p)2f'(p)=0\\ \Rightarrow \frac{f'}{2f-1}=\frac{1}{1-2p}\Rightarrow f(p)=\frac{1}{2}\left(1+\frac{A}{1-2p}\right)$$
$$\Rightarrow G2max=A, p\neq 1/2$$
So player 2 could expect a gain of A , if A were chosen small but non zero, then player 2 could do a statistic on 1s choices, put in the formula and get the probability to make his own choices, and this could induce a bias so that player 2 could have an average gain different than zero on the long term ?
Edit : in fact the question more precisely is : $f(p)$ has to remain in [0,1]. What is more close to reality, setting $f(p)=\cos(g(p))^2$ from the onset, giving $q=1-p$ and $G2=(1-2p)^2$ or bounding it in the above calculation ?
(Anyhow both calculations lead to unrealistic results.)
Suppose we know both $p$ and $q$. Then the expected gain of the first player is $(pq + (1-p)(1-q)) - (p(1 - q) + q(1 - p)) = 2pq - p - q + 1 + 2pq - p - q = 1 - 2p - 2q + 4pq = (1 - 2p)(1 - 2q)$. And for the second player it is $-(1 - 2p)(1 - 2q)$, because the game is a zero sum one. For a given $p$ the expected gain of the second player reaches maximum, when
$$q = \begin{cases} 0 & \quad p > \frac{1}{2} \\ 1 & \quad p < \frac{1}{2} \end{cases}$$
If $p = \frac{1}{2}$, then the expectation does not depend on $q$ at all.
That means, that the largest possible expected gain of the second player is $|1 - 2p|$. Because it is a zero sum game, the first player will do their best to minimise it.
Thus, the only mixed Nash equilibrium here is when $p = q = \frac{1}{2}$. In this case everyones expected gain will be $0$.
From that we can conclude, that the game is unbiased.
And if you want to find a game that is similar, but biased, try applying the same reasoning to the following variant of matching pennies (known as "prison poker"):
$$\begin{array}[ccc]1 & q & 1-q\\p&(2,-2)&(-3,3)\\1-p&(-3,3)&(4,-4)\end{array}$$