The rectangle prefers to live without the dots!

With $x_1 \le y_1$ and $x_2 \le y_2$ (WLOG by reflection), the probability that we want to minimize is $1 - x_1y_1(1-x_1)(1-y_1) - x_2y_2(1-x_2)(1-y_2) + x_1y_1(1-x_2)(1-y_2).$ I am not sure if there is a clever way to do this minimization; I certainly did it uncleverly. I used my favorite computer algebra package to compute the solution set of the simultaneous set of equations given by the vanishing of the four partial derivatives of this polynomial, and it returned $\begin{aligned} \{ (0, 0, 0, 0), (0, 0, 0, 1), (0, 0, 1/2, 1/2), (0, 0, 1, 0), (0, 0, 1, 1), & (0, 3/4, 1/2, 1/2), (0, 1, 0, 1), (0, 1, 1, 0), (0, 1, 1, 1), (1/2, 1/2, 1/4, 1), \\ (1/2, 1/2, 1, 1/4), (1/2, 1/2, 1, 1), (3/4, 0, 1/2, 1/2), & (1, 0, 0, 1), (1, 0, 1, 0), (1, 0, 1, 1), (1, 1, 1, 1), \\ (-w + 1, -w + 1, w, w), & (w + 2, w + 2, -w - 1, -w - 1) \} \end{aligned}$ where $w = \frac{\sqrt{5}-1}2.$ We can throw out all the solutions where one of the points is on the border of the square, and the last solution has coordinates that are too large, so that leaves $\left( \frac{3-\sqrt{5}}2, \frac{3-\sqrt{5}}2, \frac{\sqrt{5}-1}2, \frac{\sqrt{5}-1}2 \right)$ which gives a probability of $4-5w = \frac{13-5\sqrt{5}}2 \approx 0.90983.$

The distance between $(1-w,1-w)$ and $(w,w)$ is $\sqrt{2}(2w-1) = \sqrt{10}-\sqrt{8} \approx \fbox{0.33385}.$

Let's consider a square in the $xy$ -plane which vertices are $(0,0)$ , $(0,1)$ , $(1,1)$ , $(1,0)$ . Let's also place two points $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$ randomly and uniformly in the square such that $0\leq x_i \leq 1$ and $0\leq y_i \leq 1$ for $i \in \{0,1\}$ . Also $x_1 \geq x_2$ and $y_1 \geq y_2$ . We call $E$ the event "at least one of the points is in the rectangle". If $A$ is the event "point $P_1$ is in the rectangle" and $B$ is the event "point $P_2$ is in the rectangle", than $E$ will be

$\displaystyle E=(A \cup B)\setminus (A\cap B)$

If event $A$ happens, i.e. if $P_1$ is in the rectangle, than, given that $b \geq a$ and $d \geq c$ , we'll have

$\displaystyle \begin{cases} 0 \leq a \leq x_1 \\ x_1 \leq b \leq 1 \\ 0 \leq c \leq y_1 \\ y_1 \leq d \leq 1 \end{cases} \hspace{5pt} \Longrightarrow \hspace{5pt} \mathbb{P}(A)=(1-x_1)x_1(1-y_1)y_1$

Similarly, for event $B$

$\displaystyle \begin{cases} 0 \leq a \leq x_2 \\ x_2 \leq b \leq 1 \\ 0 \leq c \leq y_2 \\ y_2 \leq d \leq 1 \end{cases} \hspace{5pt} \Longrightarrow \hspace{5pt} \mathbb{P}(B)=(1-x_2)x_2(1-y_2)y_2$

If the rectangle contains both points, i.e. if the event $A \cap B$ occurs, than, given that $b \geq a$ and $d \geq c$ , we'll have

$\displaystyle \begin{cases} 0 \leq a \leq x_1 \\ x_2 \leq b \leq 1 \\ 0 \leq c \leq y_2 \\ y_1 \leq d \leq 1 \end{cases} \hspace{5pt} \Longrightarrow \hspace{5pt} \mathbb{P}(A \cap B)=(1-x_2)x_1(1-y_1)y_2$

Hence

$\displaystyle \mathbb{P}(E)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A \cap B)=(1-x_1)x_1(1-y_1)y_1+(1-x_2)x_2(1-y_2)y_2-(1-x_2)x_1(1-y_1)y_2$ .

Now, in order to minimize the probability that "neither $P_1$ nor $P_2$ are in the rectangle", we can maximize the probability that "at least one of the two point is in the rectangle". In other words we can maximize $\mathbb{P}(E)$ . This leads to the following problem

$\displaystyle \begin{cases} \text{max} \mathbb{P}(E) \\ 0 \leq x_i \leq 1 \\ 0 \leq y_i \leq 1 \end{cases}$

for $i \in \{0,1\}$ . I solved it via graphical methods and found a couple of point for which the object fuction reached its highest value, than I evaluated the distance. I suspect there's at least a couple of two points $(P_1,P_2)$ for which $\mathbb{P}(E)$ is maximum, due to the symmetry of the square, but I don't have any proof of this. It would be also interesting if someone could provide an analytical solution for the non-linear optimization problem (if there is any).

Draw the graph of the equations.You will get the solution on your own.I am happy to be the first solver of this question. Hint:Use the meaning of the statement to minimize the probability of neither points in the rectangle and analyse this beautiful Cartesian system.Cheers.