#!/usr/bin/env ruby

# This is another search problem.  We're not entirely 
# sure if the relative gap between the  pro and the
# father matters, and we're not even sure that the son
# can beat the father, much less the pro. Alternately, 
# he might beat them both handily.
#
# So, we'll take sample points: we'll give the son 11 
# different chances against the pro: 0% chance of 
# winning, 10% chance of winning, ... , 90% chance, 
# 100%.  Then, for each of these, we'll give him all
# the same chances up to this chance against his father.
# At each of these, we run a number of trials with both
# orders.
#
# Yes, this gets expensive: it's the upper triangle of a
# 11x11 matrix, so we're going to do 
#    11x11/2*2*TRIALS = 121*TRIALS trials.  
# But, hey, cycles are cheap.

TRIALS=10000
RANGE=[0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0]

n_dpd = 0
n_pdp = 0

def win(series)
  did_win = 0 # 1 is "won last", 2 is "won bet"

  series.each do |p_now|
    if p_now > rand()
      did_win += 1
      return true if did_win == 2
    else
      did_win = 0
    end
  end

  return false
end

RANGE.each do |p_pro|
  RANGE.each do |p_dad|
    break if p_dad > p_pro # we'll let dad == pro through for now
      TRIALS.times do
        n_dpd += 1 if (win([p_dad, p_pro, p_dad]))
        n_pdp += 1 if (win([p_pro, p_dad, p_pro]))
      end
  end
end

puts "Total Wins: dad/pro/dad=#{n_dpd}, pro/dad/pro=#{n_pdp}"