ags@pucc-i (Seaman) (02/25/84)
First, I blew it when I stated the exponential density function. It should be: p(t) = a * exp(-a * t) where the first "a" is needed to make the integral to come out to 1. The expected value of this distribution (I worked it out this time) is actually "1/a", not "a" as I stated. If we assume the required lengths of chain are x and y, which are independent random variables, each exponentially distributed with mean 1/a, then the probability that x>t for any given t>=0 is exp(-a * t), and similarly for y. Let L be the length of available chain. For 0 <= t <= L/2, suppose we cut the chain into two pieces of length t and L-t. The probability that these two pieces are long enough is the sum of the probabilities that: (1) x <= t and y <= t, (2) x <= t and t < y <= L-t, (3) y <= t and t < x <= L-t. The first probability is [1 - exp(-a * t)] ** 2. The second and third are each equal to [1 - exp(-a * t)] * [exp(-a * t) - exp(-a * (L-t))]. Therefore the probability we are trying to maximize turns out to be f(t) = 1 + 2*exp(-a*L) - exp(-2*a*t) - 2*exp(-a*(L-t)) whose derivative is f'(t) = 2 * a * [exp(-2*a*t) - exp(-a*(L-t))]. The probability turns out to have a maximum when -2*a*t = -a*(L-t), or t = L/3. The arbitrary constant "a" does not enter into the answer. It is quite possible to start with a different probability distribution, or even to assume that x and y are nonindependent, and get different answers. -- Dave Seaman ..!pur-ee!pucc-i:ags "Against people who give vent to their loquacity by extraneous bombastic circumlocution."