Consider a stick, cut twice, probability the smallest is 1/5 Consider a stick of unit length, you take two points (uniform independent), and then break the stick on these two points. You now got 3 segment. What is the probability the smallest segment  is less than 1/5?
$X, Y$ be the two cuts.
$P(X < 1/5  \text{ and } Y-X < 1/5 \text{ and } (1-Y-X)<1/5). $
 A: Let $P(S)$ be the probability that the smallest segment is less than $\frac{1}{5}$; let $X$ be the location of the first cut, and $Y$ be the location of the second cut.
If $X < \frac{1}{5}$ or $X > \frac{4}{5}$, then you already know that one of the segments has length less than $\frac{1}{5}$, so the conditional probability $P\big(S \vert X \in [0,\frac{1}{5}) \cup (\frac{4}{5},1]\big)=1$.
If $\frac{2}{5} < X < \frac{3}{5}$, then the possible locations where the second cut could facilitate $S$ are the regions of length $\frac{1}{5}$ on either side of $X$, and at either end of the stick, all of which are non-overlapping.
Therefore, the conditional probability $P(S \vert X \in (\frac{2}{5},\frac{3}{5}]) = \frac{4}{5}$.
Moving $X$ left of $\frac{2}{5}$, you linearly lose area in which $Y$ can land to facilitate $S$, until right before $X=\frac{1}{5}$, the good region is only length $\frac{3}{5}$. The same argument follows on the other side.
Combining all of these results, I end up with a piecewise function for the conditional probability of $P(S|X)$:
$$
P(S \vert X) = \begin{cases}
1 & 0 \leq X < \frac{1}{5} \\
\frac{2}{5} + X & \frac{1}{5} \leq X < \frac{2}{5} \\
\frac{4}{5} & \frac{2}{5} \leq X < \frac{3}{5} \\
\frac{7}{5} - X & \frac{3}{5} \leq X < \frac{4}{5} \\
1 & \frac{3}{5} \leq X \leq 1
\end{cases}
$$
and by integrating that function over $X$, I find that $P(S)=\frac{21}{25}=84\%$.
