#!/usr/bin/env python

import collections, itertools

gray_die  = [0,1,1,1,2,3]
black_die = [0,2,2,2,3,4]

row_template = """\
|-
| style="text-align:center;"| {}
| style="text-align:right;" | {}
| style="text-align:right;" | {}
| style="text-align:right;" | {}\
"""
def print_row(attribute):
  # to calculate the plain odds, count how many of the 36 outcomes are favorable
  hitogram = collections.Counter(sum(x) for x in itertools.product(gray_die, black_die))
  chance = sum(frequency for (roll,frequency) in hitogram.items() if roll <= attribute) / 36.0
  # lucky charm lets you reroll if you fail, so it's the inverse probability of losing twice in a row
  lucky_chance = 1 - (1 - chance)**2

  # dark fortune makes things complicated. we need to pick one of the dice to reroll
  fortune_numerator = 0
  for (g, b) in itertools.product(gray_die, black_die):
    if g + b <= attribute:
      # no need to use dark fortune. effectively the reroll is guaranteed success
      fortune_numerator += 6
    else:
      # see what would happen if we rerolled either one, and choose the best outcome
      new_chances = 0
      for keep, reroll_die in [(b, gray_die), (g, black_die)]:
        new_chances = max(new_chances, sum(1 for new_value in reroll_die if keep + new_value <= attribute))
      fortune_numerator += new_chances
  fortune_chance = fortune_numerator / (36*6.0)
  print(row_template.format(attribute,
      format_percent(chance),
      format_percent(lucky_chance),
      format_percent(fortune_chance)))
def format_percent(x):
  if x == 1: return "100%"
  return "{:.1%}".format(x)

for n in range(8):
  print_row(n)