Randomly cutting a sandwich

Suppose the sandwich has side length 1. Let $t \in [0,1]$ be location of the first point (measured from a corner). We have three cases to consider:

1. If the second point is on the same side (probability $\frac{1}{4}$), the area of the resulting smallest piece will be zero.
2. If the second point is on an adjacent side (probability $\frac{1}{2}$), the area of the resulting smallest piece will be $\frac{1}{2}tr$, where $r \in [0,1]$ is the location of the second point on the adjacent side.
3. If the second point is on the opposite side (probability $\frac{1}{4}$), the area of the resulting piece will be the smaller of the two resulting quadrilaterals. Specifically, it will be $\min(\frac{t+r}{2},1-\frac{t+r}{2})$, where $r \in [0,1]$ is the location of the second point on the opposite side.

We will solve the general case, which is to find the probability that the smallest slice will have a fraction at least $p$ of the total area. the problem asks for $p=\frac{1}{4}$. Since the total area of the square is $1$, we are effectively asking: “what is the probability that the smallest area is at least $p$?”.

Case 1. If we are in the first case (points on the same side), the probability that the area is greater than $p$ is zero, since the area is always zero. So we conclude that:
\[
a_1 = 0
\]

Case 2. If we are in the second case (points on an adjacent sides), we want the probability that $\frac{1}{2}tr > p$, where $r$ and $t$ are chosen uniformly at random in $[0,1]$. This amounts to computing:
\begin{align}
a_2 &= \int_0^1 \int_0^1 \mathbf{1}\bigl( \tfrac{1}{2}tr \gt p \bigr)\,\mathrm{d}r\,\mathrm{d}t \\
&= \int_{2p}^1 \int_0^1 \mathbf{1}\bigl(r \gt \tfrac{2p}{t} \bigr)\,\mathrm{d}r\,\mathrm{d}t \\
&= \int_{2p}^1 \left( 1-\frac{2p}{t}\right)\,\mathrm{d}t \\
&= 1-2p+2p\log(2p)
\end{align}Note that we had to change the lower bound of the integral for $t$ because when $t \lt 2p$, we have $\tfrac{2p}{t}\gt 1$, so the integrand is zero. Geometrically, this is the fraction of the region $(t,r)\in[0,1]^2$ where $tr\gt p^2$, sketched below.

Case 3. If we are in the third case (points on opposite sides), we perform a similar computation with the area of the quadrilateral:
\begin{align}
a_3 &= \int_0^1 \int_0^1 \mathbf{1}\bigl( \min(\tfrac{t+r}{2},1-\tfrac{t+r}{2}) \gt p \bigr)\,\mathrm{d}r\,\mathrm{d}t \\
&= \int_0^1 \int_0^1 \mathbf{1}\bigl( \tfrac{t+r}{2}\gt p\text{ and } 1-\tfrac{t+r}{2} \gt p \bigr) \,\mathrm{d}r\,\mathrm{d}t \\
&= \int_0^1 \int_0^1 \mathbf{1}\bigl( p \lt \tfrac{t+r}{2} \lt 1-p\bigr) \,\mathrm{d}r\,\mathrm{d}t \\
&= \int_0^1 \int_0^1 \mathbf{1}\bigl( 2p \lt t+r \lt 2-2p\bigr) \,\mathrm{d}r\,\mathrm{d}t \\
&= 1-4p^2
\end{align}A trick to evaluating the above integral is to do it geometrically. Sketching the region for which $2p\lt t+r\lt 2-2p$, we see that it is the square of area $1$ with the two corners cut off, whose area is $(2p)^2$, so the result is $1-4p^2$. See below for an illustration.

Putting everything together, the probability that the area of the smallest piece is at least $p$ is given by sum of the probabilities we already calculated, each weighted by their likelihood of occurrence, so $f(p) = \frac{1}{4}a_1 + \frac{1}{2}a_2 + \frac{1}{4}a_3$. Plugging in the results from above and simplifying, we obtain our final answer:

$\displaystyle
f(p) = \frac{3}{4}-p-p^2+p \log (2 p)
$

Here is what the function looks like:

Naturally, the plot only extends to $p=\frac{1}{2}$ because the area of the smaller piece cannot be more than half of the total area. When $p=\frac{1}{4}$ (what the original problem asked about), we obtain $f(\frac{1}{4}) = \frac{7}{16}-\frac{\log (2)}{4}\approx 0.264213$. So the probability that the smaller piece is at least one quarter of the whole area is about 26.4%.