1. **Problem Statement:** We have six independent and identically distributed (IID) points in the unit square. The first three points form one triangle, and the last three points form another triangle. We want to find the probability that one of these triangles is entirely contained within the other. 2. **Understanding the Problem:** The problem involves geometric probability and the concept of one triangle being contained within another inside the unit square. This is a complex problem involving joint distributions of points and containment conditions. 3. **Key Insight:** The problem references a known closed-form expression for the probability that a given point $(x,y)$ lies within a randomly picked triangle formed by three IID points in the unit square, denoted $P(x,y)$. 4. **Given:** The function $P(x,y)$ is defined piecewise using $p_0(x,y)$ and symmetry relations, with $p_0(x,y)$ involving a complicated polynomial $f(x,y)$ and logarithmic terms. The integral over the unit square satisfies: $$\int_0^1 \int_0^1 P(x,y) \, dx \, dy = \frac{11}{144}.$$ This integral represents the average probability that a random point lies inside a random triangle formed by three IID points. 5. **Approach to the Containment Probability:** To find the probability that one triangle is contained within the other, consider the following: - The probability that a fixed point lies inside a random triangle is $\frac{11}{144}$. - For one triangle to be contained in another, all three vertices of the first triangle must lie inside the second triangle. - Since the points are IID and independent, the probability that all three vertices of the first triangle lie inside the second triangle is: $$\left(\frac{11}{144}\right)^3 = \frac{11^3}{144^3} = \frac{1331}{2985984}.$$ - Similarly, the probability that the second triangle is contained in the first is the same. - Since these two events are mutually exclusive (one triangle cannot be contained in the other simultaneously), the total probability is twice this value: $$2 \times \frac{1331}{2985984} = \frac{2662}{2985984}.$$ 6. **Simplify the fraction:** $$\frac{2662}{2985984} = \frac{1331}{1492992}.$$ 7. **Final answer:** The probability that one triangle is entirely contained within the other is: $$\boxed{\frac{1331}{1492992}}.$$