Fractal Probability Explorer
Many famous fractals are built by a repeated “keep some parts, remove others” rule.
If you pick a point uniformly from the original region (triangle, segment, square, cube),
a natural probability question is:
What is the chance the point lies in the remaining set after \(n\) iterations?
For these classical constructions, the answer is closely tied to how the fractal scales and how
much measure (length/area/volume) remains each step.
This calculator focuses on that remaining-measure probability and connects it to the
self-similarity dimension.
Measure as probability
Suppose you start with a region \(R\) of finite measure (length in 1D, area in 2D, volume in 3D).
You build an approximation \(F_n \subseteq R\) after \(n\) iterations.
If a point \(X\) is chosen uniformly from \(R\), then:
\[
P(X \in F_n)=\frac{\mu(F_n)}{\mu(R)},
\]
where \(\mu\) is the appropriate measure (length/area/volume).
So “probability of landing in the remaining set” is literally the remaining fraction of measure.
Self-similar scaling and the per-step fraction \(q\)
Many self-similar removal fractals share this pattern:
- Each iteration keeps \(N\) smaller copies of the previous shape.
- Each copy is scaled by a linear factor \(r\) (e.g., \(r=\tfrac12\) or \(r=\tfrac13\)).
- The construction lives in an embedding space \(\mathbb{R}^d\) (so measure scales like \(r^d\)).
Because lengths scale by \(r\), areas scale by \(r^2\), and volumes scale by \(r^3\),
the remaining measure fraction after one step is:
\[
q = N\cdot r^{d}.
\]
If \(0<q<1\), the remaining measure shrinks each iteration, and after \(n\) steps:
\[
P(\text{remaining after }n)=q^{n}.
\]
Classic examples
Sierpinski triangle
Start with an equilateral triangle. Each step divides it into 4 congruent triangles (scale \(r=\tfrac12\))
and removes the middle one, leaving \(N=3\) triangles.
Since \(d=2\) (area):
\[
q = 3\left(\tfrac12\right)^2=\tfrac34,
\qquad
P = \left(\tfrac34\right)^n.
\]
Example: \(n=3\) gives \(P=(3/4)^3=27/64\approx 0.4219\). The removed fraction is \(1-P\approx 0.5781\).
Cantor set
Start with a line segment \([0,1]\). Each step removes the middle third and keeps two segments
of length \(1/3\), so \(N=2\), \(r=\tfrac13\), and \(d=1\):
\[
q = 2\left(\tfrac13\right)=\tfrac23,
\qquad
P = \left(\tfrac23\right)^n.
\]
Sierpinski carpet
Start with a square. Divide into a \(3\times 3\) grid (scale \(r=\tfrac13\)), remove the center square,
leaving \(N=8\) squares in 2D:
\[
q = 8\left(\tfrac13\right)^2=\tfrac89,
\qquad
P = \left(\tfrac89\right)^n.
\]
Menger sponge
Start with a cube. Divide into \(3\times 3\times 3\) (scale \(r=\tfrac13\)) and remove the central cube and the
six face-center cubes, leaving \(N=20\) cubes in 3D:
\[
q = 20\left(\tfrac13\right)^3=\tfrac{20}{27},
\qquad
P = \left(\tfrac{20}{27}\right)^n.
\]
Dimension tie-in: why fractals are “in between”
A standard self-similarity (scaling) dimension for many classical fractals is:
\[
D=\frac{\log N}{\log(1/r)}.
\]
This dimension measures how the number of pieces grows relative to the scaling factor.
For example:
- Sierpinski triangle: \(D=\log 3 / \log 2 \approx 1.585\) (between a curve and a surface).
- Cantor set: \(D=\log 2 / \log 3 \approx 0.631\) (smaller than a line).
- Sierpinski carpet: \(D=\log 8 / \log 3 \approx 1.893\).
- Menger sponge: \(D=\log 20 / \log 3 \approx 2.727\).
Important: \(D\) describes scaling complexity, but the probability in this calculator comes from the
embedding-space measure fraction \(q^n\). In many removal constructions, \(q<1\), so
\(\mu(F_n)\to 0\) as \(n\to\infty\) even though the fractal remains infinitely detailed.
University note: measure zero vs. “still infinite”
It can feel paradoxical that a fractal can have infinitely many points but zero area/volume in the limit.
The resolution is that “number of points” (cardinality) is not the same as measure.
Many fractals are uncountably infinite but still have measure zero.
Chaos game connection (intuition)
For the Sierpinski triangle, the “chaos game” generates points by repeatedly moving halfway toward a random vertex.
The points eventually concentrate on the Sierpinski set, visually reinforcing that the removed region becomes more and more dominant
in measure (since the remaining area fraction is \((3/4)^n\)), while the remaining set stays structured.
