Consider $a$ be the number of die $A$ and$b$ be the number of die
$B$.

It is given that $a\mathit{+}b$ is even.

Therefore,$a$ and $b$ are both even or both odd.

Thus, the probability of both even or both odd is,

$P=\frac{1}{2}$

Thus, for ${100}^{\text{th }}$ time, the required probability is ${\left(\frac{1}{4}\right)}^{100}$.