I believe $\left ( 1 - 2a \right )$ should be the right answer.
Partition will divide original array into two sub arrays.
I am using a random variable $X$ to denote the fractional length of smaller partition of the original array with respect to the length of original array.
Consequently, the fractional length of the larger partition will be $\left ( 1 - X \right )$ with respect to the length of original array.
For example suppose, $X = 0.2$ then it will indicate that our smaller sub array (among the two sub arrays that we got after partitioning) is $0.2$ time the original array.
and hence the larger sub array is $\left ( 1 - 0.2 \right )$ or $0.8$ times the original array.
Since $X$ is denoting the fractional length of smaller sub array after partition, its value can not exceed more than $0.5$ times the original array otherwise it will no longer be the smaller sub array.
Hence the domain of random variable $X$ is $\left ( 0, \frac{1}{2} \right ]$, that is $X$ can take any values between $0$ and $\frac{1}{2}$.
It can be inferred that as $X$ approaches to $\frac{1}{2}$, $\left ( 1 - X \right )$ also approaches to $\frac{1}{2}$, that is the sizes of two sub arrays are being equal or in other words the partition becomes more balanced.
$X = \frac{1}{2}$ means we have got two sub arrays of equal size,and both of them are $0.5$ times the original array, this is the most desirable situation from the efficiency's point of view.
As $X$ approaches to $0$, the partition becomes more unbalanced.
Now if we want a split more balanced than $a\mid \left ( 1 - a \right )$ where $a\ \epsilon \left ( 0, \frac{1}{2} \right ]$, we have to set our random variable $X's$ value larger than $a$, so that our random variable is more nearer to $\frac{1}{2}$ , then the $a$ is (as a value nearer to $\frac{1}{2}$ implies a more balanced partition).
So to get a partition more balanced than $a\mid \left ( 1 - a \right )$, $X$ must lie in the range $\left ( a, \frac{1}{2} \right ]$.
Hence the probability of getting a more balanced split than $a\mid \left ( 1 - a \right )$ is same as $P\left ( X > a \right )$,
And $P\left ( X > a \right ) = \frac{\text{length of favourable domain of X}}{\text{length of complete domain of X}}$.
$\Rightarrow P\left ( X > a \right ) = \frac{\left ( \frac{1}{2} - a \right )}{\left ( \frac{1}{2} - 0 \right )}$
$\Rightarrow P\left ( X > a \right ) = \left ( 1 - 2a \right )$.
It can also be observed that using the above formula when $a$ becomes $ \frac{1}{2} $ there is $0$ probability of getting a more balanced split, and when $a$ tends to $0$, there is almost full probability of getting a more balanced split.
So derived formula for probability is consistent with our intuition.
