Bertrand’s Paradox
Pick Up Sticks
Shuffle Board
Like
a Rolling Stone
Throwing
Darts Blinfolded
Back to
the Ball in the Box
The
Distance along the Surface
The
Straight-Line Distance
The L Distribution
The Best Bet
Consider the following problem. A game is played by marking a point A on the surface of a spherical ball of radius a. The ball is then shaken in a closed box and when the box is opened the straight line distance, L, from the point of tangency on the floor of the box to A is determined. Near what value of L is the result of this experiment most likely to occur? The surprising (at least to us) answer is close to the diameter of the ball!
does indeed give the most probable value of the central angle from the
point of tangency to A, but not the most probable value of L. Thus, the
question, "What is the most probable position for A?" has no unambiguous
answer. In general, to answer such questions in "Geometric
Probability" one must carefully define the experiment, i.e., the relevant
random variables and their probability distributions. The details in the
solution of the Sphere Game will be presented after examining a better
known (1, 2) two-dimensional example which vividly illustrates
this need for clarity.
An equilateral triangle is inscribed in a circle of radius a. The length of a side of the triangle, S, is related to the radius as shown in the following figure.
Now consider the question: What is the probability that a "random" chord of the circle will exceed S ? Each of the following experimental scenarios generates a "random" chord, but with different probabilities of exceeding S. Thus, as it is originally stated the problem is inherently ambiguous.
Imagine a large set of calibrated rods with lengths uniformly distributed in the interval [0, 2a]. If one of these rods is picked at random and laid across the circle, it makes a chord of length c. The probability that this cord exceeds S is given by
.
Let r stand for the radial distance from the circle’s
center to the chord of length c. Then 
.
So if
.
Let x stand for the position of the chord from the center O. If the chord is to the left of O, x is negative. The degenerate case of two tangent lines rather than a single chord is designated by x = a. Thus, x is a random variable uniformly distributed in (-a, a]. The radial distance r is the absolute value of x. Thus, the probability that the chord length exceeds S is given by
.Imagine a circular disk of radius a placed with
its axis parallel to a plane. Let P mark a point of vertical tangency on
the rim of the disk. Now allow the disk to roll without slipping for a
random distance in plane. The point P will now make an angle 
with respect to its original position. This angle will be a random variable
uniformly distributed in
. Drop a segment from P perpendicular to the plane. This generates the
"random" chord PQ.
The length of this chord is given by 
. Thus, the probability that the chord length exceeds S is given
by 
Imagine picking a point M at random from the interior of the circle.Through M construct the chord AB perpendicular to the radial segment OM. For c = 2MB to be larger than S requires the following:
. So
The Distance along the Surface
Place the origin of a three dimensional coordinate system at the center
of the sphere. If A’s position is uniformly distributed over the surface
of the sphere, then using the standard spherical coordinate angles
, the probability of A being at a particular position is proportional to
the element of surface area 
, so the probability density is
.
is independent
of , so the
probability density function for
is simply computed as
.
is calculated as follows:
.Since the distribution for
is unimodal and symmetric, the average value of
is equal to both the median and the most probable value of
: 
.
Let
be the central
angle from the point of tangency to point A. This is just the supplement
to .
.
It follows that the mean for this outside distance is the same as its median
value is the same as its most probable value: 
.
The straight-line distance from 
is given by
This same result can be obtained using the law of cosines: 
.
The value of L on the "equator" is given by
. This last result is of course obvious from the Pythagorean theorem.
For 
Hence,
the median value of L, 
, corresponds to the length associated with the most probable value of
.
However, computing the average value of L gives the following result:
.
The surface area of the band of width ds is
This demonstrates the familiar, yet still surprising result, that the lateral surface area ("crust") of a slice of a sphere depends only on the slice’s thickness and not its location. The probability density function gives the probability that point A is to be found in a slice of thickness dz . It is given by
. This result
can also be obtained from the standard fundamental transformation formula(3,
4, 5):
.
In terms of z the straight-line distance is given by the following expression.
.
Now
So it is
again confirmed that
. In terms of z the average value of L is computed as 
,
which agrees with the calculation on the distribution of
.
The most straightforward way to analyze the Sphere Game is to use the probability distribution for L. The key step is to again use the fundamental transformation formula.
.This formula is the special case for n = 3 of a general result derived by Lord (6) for a sphere in n dimensions. The equation for three dimensions has been used in the analysis of protein folding (7).
The graph of the linear L distribution is plotted ibelow. Since this
is a right triangle, the median value of L with equal areas on each side
can be found at 
the base length from the left vertex. Thus, 
.
The mean value of L is computed rather simply as 
. The probability of measuring a value of L greater than the mean is given
by
.Finally, the most important and intuition-defying result of the Sphere Game is now readily apparent. The most probable value of L is at the right end point, i.e., Mode of the L distribution = 2a. In analogy with Bertrand’s paradox, defining the experiment in terms of the length L yields a different result for the “most probable location of point A” than using the outside distance or the central angle. In a way this result does make geometric sense. As L increases, the area on the surface of the sphere in a band generated by an interval dL also increases due to its longer “reach”. It should be remembered that dL and dz are not equivalent. In fact,
.So for small but fixed 
. The area generated by a given 
is 
(this result
gives an alternate derivation for the probability density of L; 
). Hence, for small but fixed
the most area is generated by the largest value of L, namely the diameter
of the sphere.
Since L is a continuous random variable a non-zero probability requires
a non-zero width in the range of L-values. Suppose for a "win" the game
requires that the measured value of L must be within a stated tolerance, 
, of the "guessed" value. Since
is an increasing function, the maximum probability of a win occurs for
a "bet" of 
, which wins whenever L is within the interval 
. This maximum probability is given by
, which corresponds to the position of the "best bet" for the distance
on the outside of the ball, would be
Thus betting on a value of L nearly equal to the diameter of the ball
is about
times more
likely to win than a bet on the "more intuitive" choice of 
.
Betting on the mean value of L, 
, gives an even smaller value for the probability of a win.
) has a probability of winning equal to
.
) will win more often than any other value of L you could choose!
Conclusion
The median value,
, would seem to be the obvious answer to the question, “What is the most
probable straight line distance between a fixed point and a second point
picked at random from the surface of a sphere of radius a?”. The rather
humorous and surprising result is that not only is our intuition wrong,
but the answer is the largest possible value of L. This peculiarity is
a result of initially stating the problem in terms of the Euclidean distance
rather than the central angle or the exterior arclength. Like Bertrand’s
Paradox in two dimensions it’s a reminder that answers to problems in probability
depend critically on what’s being asked!
References:
We would welcome any comments, criticisms and/or suggestions.
Please feel free to contact either one of us at the addresses shown
below.
| Al Lehnen | Gary E Wesenberg |
| Mathematics Department | Department of Biochemistry |
| Madison Area Technical College | University of Wisconsin |
| 3550 Anderson Street | 6606B Biochemistry |
| Madison, WI 53704 | 433 Babcock Drive |
| (608) 246-6567 | Madison, WI 53706 |
| alehnen@matcmadison.edu | (608) 263-5923 |
| http://my.execpc.com/~aplehnen/al | gary@biochem.wisc.edu |