**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."