Problem 43 of Monte Carlo solutions to Fifty Challenging Problems...

(This is another part of the Fifty Problems series, a set of example applications of Monte Carlo methods. In each post, I present source code which answers a probabilistic question using simulated models of the underlying system.)

Problem 43: A bar is broken at random in two places. Find the average length of the shortest, middle-est, and longest pieces.

#!/usr/bin/env ruby

TRIALS=100000

# Much like 42, this is pretty trivial stuff.

cum_short = 0.0
cum_med = 0.0
cum_long = 0.0

TRIALS.times {
  b1 = rand()
  b2 = rand()

  # put them on a stick reasonably...
  if (b1 > b2)
    x = b1
    b1 = b2
    b2 = x
  end

  ls = [b1, b2-b1, 1.0-b2].sort

  cum_short += ls[0]
  cum_med += ls[1]
  cum_long += ls[2]
}

m_short = cum_short/TRIALS
m_med = cum_med/TRIALS
m_long = cum_long/TRIALS

puts "After #{TRIALS} trials, average lengths:"
puts " #{m_short}, #{m_med}, #{m_long}"

I've been coding my way through Fifty Challenging Problems in Statistics with Solutions. This post is a part of the Fifty Challenging Problems series.

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