3
$\begingroup$

Given the size of the cube(L) and the size of a stick (r), and supposing the first head of the stick fell on some random point of the cube, what is the probability that the second head will also be inside of the cube.

I know that if first head of the stick is inside the box and farther than r away from the border, then both heads are inside box. But after that, some sticks may have one extreme outside of the box depending on the orientation. Also, if the stick is on a corner of the cube, the probability that the second head is outside the cube, would be larger, than if it was on the side. How can I calculate the probability that the second head is outside the box given that the first head is inside?

$\endgroup$
3
  • $\begingroup$ Take a white sphere and paint one hemisphere of it green. Consider now a spherical cap on this sphere. If you can figure out how much of the cap is green, you are on your way to solving this problem. While I suspect finding how much green has an elementary solution, I am not seeing it yet. Gerhard "Maybe If I Change Color..." Paseman, 2019.11.06. $\endgroup$ Nov 6, 2019 at 21:16
  • 3
    $\begingroup$ How are you dropping the stick? Are you taking some big region, uniformly picking a location for one end, and uniformly picking an orientation on the sphere? Are you doing something else? The answer may depend on your probability measure. $\endgroup$ Nov 7, 2019 at 3:37
  • $\begingroup$ Sticks fell in a infinite area of space with a known frequency of sticks per volume. I'm selecting a cubical volume on said space. I care about the sticks heads (not about the body). If neither head is in the box, I consider that the stick is outside the box. $\endgroup$
    – Carlos
    Nov 19, 2019 at 2:20

1 Answer 1

4
$\begingroup$

I assume one end of the stick is dropped uniformly in the cube and the other end is chosen uniformly in a sphere of radius $r$ around the first. If $r < L$ then conditionally on the orientation ${\bf x}=(x_1,x_2,x_3)$ with $|{\bf x}| = r$ of the stick the probability that it is inside the cube is $\left(1-\frac{|x_1|}{L}\right)\left(1-\frac{|x_2|}{L}\right)\left(1-\frac{|x_3|}{L}\right)$. Integrating this over the sphere of radius $r$ gives a probability $$ 1 - \frac{3}{2}\frac{r}{L} + \frac{2}{\pi} \frac{r^2}{L^2} - \frac{1}{4\pi} \frac{r^3}{L^3} $$ for the stick to fall completely inside the cube. The formula for $L < r < \sqrt{3} L$ can be computed similarly.

$\endgroup$
2
  • $\begingroup$ Thank you, this solution is correct, I also checked it against a simulation and it agrees with this equation. I understand that if r>L this solution does not make sense, since sometimes |x1|/L >1, but I'm satisfied that this solution is exact for r<L. $\endgroup$
    – Carlos
    Nov 7, 2019 at 20:17
  • $\begingroup$ Also just for completion, this is the probability that the second head will fall inside the cube given that the first head already fell inside the cube (P(A|B)=P(B|A)). Since the question was about the probability that the stick will fall inside, it may fall with the first head first or the second head first. Then given that any head felt inside the box, the probability that the other head will also fall inside can be computed from your answer: P(A U B)/(P(A U B)+P(A U ~B)+P(~A U B))=P(A|B)/(2-P(A|B)) $\endgroup$
    – Carlos
    Nov 7, 2019 at 21:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.