$\star$ Consider modifying the

`PARTITION`

procedure by randomly picking three elements from array $A$ and partitioning about their median (the middle value of the three elements). Approximate the probability of getting at worst an $\alpha$-to-$(1-\alpha)$ split, as a function of $\alpha$ in the range $0 < \alpha < 1$.

First, for simplicity's sake, let's assume that we can pick the same element twice. Let's also assume that $0 < \alpha \le 1/2$.

In order to get such a split, two out of three elements need need to be in the smallest $\alpha n$ elements. The probability of having one is $\alpha n / n = \alpha$. The probability of having exactly two is $\alpha^2 - \alpha^3$. There are three ways in which two elements can be in the smallest $\alpha n$ and one way in which all three can be in the smallest $\alpha n$ so the probability of getting such a median is $3\alpha^2 - 2\alpha^3$. We will get the same split if the median is in the largest $\alpha n$. Since the two events are mutually exclusive, the probability is:

$$ \Pr\{\text{OK split}\} = 6\alpha^2 - 4\alpha^3 = 2\alpha^2(3 - 2\alpha) $$