ags@pucc-i (Seaman) (02/27/84)
Here is another way to solve the chain puzzle which allows more opportunity
for generalization. As before, we assume that the required lengths are x
and y, which are independent random variables described by a positive
probability density function p. This means that p(x) >= 0 for all x,
p(x) = 0 for x < 0, and
{integral from 0 to infinity} p(x) dx = 1.
Therefore the probability that the length is between a and b is given by
{integral from a to b} p(x) dx.
The available chain is cut into lengths t and L-t, where 0 <= t <= L/2.
We do not know which of x or y is the larger, but by symmetry we may
compute the probability on the assumption that x <= y, and multiply by 2.
The probability that the two pieces are long enough is twice the double
integral of p(x)p(y) dx dy over the region
x <= y <= L-t (inner integral over y)
0 <= x <= t (outer integral over x).
In the special case where p(x) = a * exp(-a * t), this double integral
evaluates to the same result I gave previously and gives a maximum at
t = L/3. Note that by changing the density function p(x) it is possible
to come up with any answer you like. We cannot hope for an answer that
is totally independent of the underlying probability distribution.
It IS rather remarkable that the t=L/3 answer works for ALL exponential
distributions, regardless of the expected value. It is as if the chain
lengths were determined by the radioactive decay times of two atoms of
an unknown substance, whose half-life might be measured in nanoseconds
or in millenia!
Let's briefly consider what happens if you need three pieces. I have
neither the time nor the patience to carry this through, but here is
how it might be done: Compute the triple integral of p(x)p(y)p(z) dx dy dz
over the region
y <= z <= L-s-t (inner integral over z)
x <= y <= t (middle integral over y)
0 <= x <= s (outer integral over x)
and multiply the result by 3! = 6, to allow for all possible rearrangements
of x, y and z. This gives you a function f(s,t) expressing the probability
of success when you cut lengths of s, t, and L-s-t. Finally, maximize this
function over s and t. We are assuming s <= t <= L-s-t.
--
Dave Seaman
..!pur-ee!pucc-i:ags
"Against people who give vent to their loquacity
by extraneous bombastic circumlocution."