/* 

  Flipping coins until HTTTH or HTHTHT in Picat.

  From "Perplexing the Web, One Probability Puzzle at a Time" 
  https://www.quantamagazine.org/perplexing-the-web-one-probability-puzzle-at-a-time-20240829/?mc_cid=94caee8978
  (about Daniel Litt)
  """
  Puzzle 2
 
  You keep flipping a fair coins until you see five consecutive flips of the
  form HTTTH or HTHTH (where H means heads and T means tails). Then stop.
  Those last five flips are more likely to have been
  - HTTTH
  - HTHTH
  - equally likely
  """

  The problem was originally stated in Daniel Litt's
  https://x.com/littmath/status/1786115416231641335 (May 2, 2024)
  """
  Keep flipping a fair coin until you see five consecutive flips of the form 
  HTTTH or HTHTH (where H means heads and T means tails), then stop. 
  Your last 5 flips are more likely to have been:
  HTTTH  15.2%
  HTHTH  14.5%
  Equally likely 53.5%
  Don't know/see results 16.7%
  """

  Here we code H -> 1, T -> 0 


  Cf my Gamble model gamble_flipping_coins_until_pattern.rkt

  This program was created by Hakan Kjellerstrand, hakank@gmail.com
  See also my Picat page: http://www.hakank.org/picat/

*/

import ppl_distributions, ppl_utils.
import util.

main => go.

/*

  patterns = [[1,0,0,0,1],[1,0,1,0,1]]
  var : len
  Probabilities (truncated):
  6: 0.0597000000000000
  5: 0.0593000000000000
  8: 0.0566000000000000
  7: 0.0555000000000000
  .........
  101: 0.0001000000000000
  96: 0.0001000000000000
  95: 0.0001000000000000
  91: 0.0001000000000000
  mean = 19.7193

  var : pattern
  Probabilities:
  [1,0,0,0,1]: 0.5568000000000000
  [1,0,1,0,1]: 0.4432000000000000
  mean = [[1,0,0,0,1] = 0.5568,[1,0,1,0,1] = 0.4432]


  patterns = [[1,0,1],[1,1,1]]
  var : len
  Probabilities (truncated):
  3: 0.2549000000000000
  4: 0.1834000000000000
  5: 0.0914000000000000
  6: 0.0794000000000000
  .........
  39: 0.0001000000000000
  38: 0.0001000000000000
  35: 0.0001000000000000
  32: 0.0001000000000000
  mean = 6.8642

  var : pattern
  Probabilities:
  [1,0,1]: 0.5899000000000000
  [1,1,1]: 0.4101000000000000
  mean = [[1,0,1] = 0.5899,[1,1,1] = 0.4101]


  patterns = [[1,0,1,1],[1,1,1,0],[1,1],[0,0]]
  var : len
  Probabilities (truncated):
  2: 0.5044000000000000
  3: 0.2488000000000000
  4: 0.1251000000000000
  5: 0.0588000000000000
  .........
  11: 0.0010000000000000
  12: 0.0004000000000000
  13: 0.0003000000000000
  14: 0.0001000000000000
  mean = 2.9917

  var : pattern
  Probabilities:
  [0,0]: 0.5026000000000000
  [1,1]: 0.3718000000000000
  [1,0,1,1]: 0.1256000000000000
  mean = [[0,0] = 0.5026,[1,1] = 0.3718,[1,0,1,1] = 0.1256]


  patterns = [[1,1],[1,0]]
  var : len
  Probabilities (truncated):
  2: 0.5068000000000000
  3: 0.2455000000000000
  4: 0.1228000000000000
  5: 0.0638000000000000
  .........
  11: 0.0007000000000000
  13: 0.0005000000000000
  14: 0.0003000000000000
  15: 0.0001000000000000
  mean = 2.9916

  var : pattern
  Probabilities:
  [1,1]: 0.5023000000000000
  [1,0]: 0.4977000000000000
  mean = [[1,1] = 0.5023,[1,0] = 0.4977]

*/
go ?=>
  PatternsList = [
                   [
                     [1,0,0,0,1], % HTTTH
                     [1,0,1,0,1]  % HTHTH
                   ],
                   [
                     [1,0,1], % HTH
                     [1,1,1]  % HHH
                   ],

                   [
                     [1,0,1,1], % HTHH
                     [1,1,1,0], % HHHT This is not reached
                     [1,1],     % HH
                     [0,0]      % TT
                   ],

                   [
                     [1,1], % HH
                     [1,0]  % HT
                   ]
              ],
  member(Patterns,PatternsList),
  println(patterns=Patterns),
  reset_store,
  run_model(10_000,$model(Patterns),[show_probs_trunc,mean]),
  nl,
  % show_store_lengths,
  fail,
  nl.
go => true.
  
model(Patterns) =>

  % [Found,Pattern] = flip_until_pattern(Patterns,[]),
  [Found,Pattern] = draw_until_one_pattern([],Patterns,[0,1]),
  % add("found",Found),
  add("pattern",Pattern),
  add("len",Found.len).