# What is the chance that a rabbit won't fall off a table if you put it somewhere and it moves.

If you would put a rabbit randomly on a circular table with radius $r= 1$ meter and it moves $1$ meter in a random direction, what is the chance it won't fall off?

I tried to do this using integrals, but then I noticed you need a double integral or something and since I'm in the 5th form I don't know how that works.

-
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? – 5xum Feb 22 at 8:56
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add the upvote. – 5xum Feb 22 at 8:56
Is the table circular? – Lovsovs Feb 22 at 9:09
The "Oberlehrers" stroke again. This is an excellent question. Above all: It is an authentic question by someone who saw a mathematical situation in his surroundings – not an artificial question as in homework. – Christian Blatter Feb 22 at 11:08
I absolutely love this question because it enhances the likelihood that some day I may encounter a textbook on probability and randomness titled "How Dumb Is the Rabbit?" – A. I. Breveleri Feb 22 at 19:41

If the rabbit is placed at distance $x>0$ from the center, and moves straightforward in an uniformly distributed direction, his chances of not falling off are $$p(x)={2\arccos{x\over2}\over 2\pi}\ .$$ Assuming that his starting point is uniformly distributed with respect to area the probability $P$ that he will not fall of is therefore given by $$P={1\over\pi}\int_0^1 p(x)\cdot 2\pi x\>dx={2\over\pi}\int_0^1\arccos{x\over2}\>x\>dx\ .$$ Substituting $x:=2\cos t$ and partial integration then produce $$P={2\over3}-{\sqrt{3}\over 2\pi}\doteq0.391\ .$$

-
Wow that looks awesome! Thank you so much for your help! – Koen van Wesel Feb 22 at 18:25
I will try to make a JAVA programme and see if it is close. – N.S.JOHN Feb 26 at 2:56

Hint: this problem is angularly symmetric. Therefore, only the starting distance from the center matters.

Consider a disk of radius $1$ centered at the origin. For a rabbit at $(x,0)$, for which angles does it fall off the table (do a little trig)? Now, use the symmetry and integrate.

To avoid the double integral, you can integrate along the positive radius and multiply each value you get by $2\pi r$ (the circumference of the circle with that radius). (This $2\pi r$ also comes out of the double integral because it's angularly symmetric.)

-
It might be added that a reasonable assumed uniform distribution would be with regard to area rather than distance from the centre. So the random variable $X$ representing the distance from the centre is square-root-uniform: namely, $$\frac{\mathrm P[x\leqslant X\leqslant x+\mathrm dx]}{\mathrm dx}=2x$$ for $0\leqslant x\leqslant 1$. – John Bentin Feb 22 at 10:29
I'm assuming that the distribution on the table is angularly symmetric (not necessarily uniform) and the directions of travel are uniformly distributed and independent of starting position. – Michael Burr Feb 22 at 10:40
I'm pretty sure that thanks to the symmetry you can just assume without loss of generality that the rabbit moves up/north/whatever, and look at what proportion of the table is within 1m of the top. Can't you? In other words yes, it's a double integral, but once you transform the location into co-ordinates based on the direction of motion, the direction integral becomes trivial. – Steve Jessop Feb 22 at 12:40
@SteveJessop: That sounds a lot easier. Make that an answer! – Henning Makholm Feb 22 at 14:47
In addition to SteveJessop's comment, if the original distribution of location is evenly by area (as opposed to evenly-chosen angle and distance from centre), the problem resolves to geometry of intersection of two unit circles. – Neil Slater Feb 22 at 15:49

Let Roger Rabbit's location on the circular table be defined by the point $(X,Y)$, where random variables $X$ and $Y$ have a bivariate Uniform distribution inside the unit circle, with joint pdf $f(x,y)$:

By symmetry, we can assume, without loss of generality, that the rabbit jumps say 1m to the east (as Steve Jessop has noted above). Then, the probability that Roger Rabbit still lands on the table is $P[(X+1)^2 + Y^2 \leq 1]$:

All done.

Notes

• The Prob function used above is from the mathStatica package for Mathematica. As disclosure, I should add that I am one of the authors.
-