Assuming that you are talking abut the usual formulation of BB84 and Bob (the receiver) is supposed to choose either the basis $\{|0\rangle,|1\rangle\}$ or the basis $\{|+\rangle,|-\rangle\}$, then the probability is exactly 1/2 of each measurement when the wrong basis is chosen.
To see this, recall that
$$|+\rangle=\frac1{\sqrt2}|0\rangle+\frac1{\sqrt2}|1\rangle$$
$$|-\rangle=\frac1{\sqrt2}|0\rangle-\frac1{\sqrt2}|1\rangle$$
so that if for example we measure $|-\rangle$ in the $\{|0\rangle,|1\rangle\}$ basis, we obtain $|0\rangle$ with probability $(1/\sqrt2)^2=1/2$ and $|1\rangle$ with probability $(-1/\sqrt 2)^2=1/2$. Likewise
$$|0\rangle=\frac1{\sqrt2}|+\rangle+\frac1{\sqrt2}|-\rangle$$
$$|1\rangle=\frac1{\sqrt2}|+\rangle-\frac1{\sqrt2}|-\rangle$$
so that if for example we measure $|0\rangle$ in the $\{|+\rangle,|-\rangle\}$ basis, we obtain $|+\rangle$ with probability $(1/\sqrt2)^2=1/2$ and $|-\rangle$ with probability $(1/\sqrt 2)^2=1/2$.
In all cases with transmission and measurement bases mismatching, 1/2 the time we measure a state corresponding to 0 and 1/2 the time a state corresponding to 1. This probability is independent of the state transmitted.
