/* 

  Winning gold medals in Picat.

  From https://www.reddit.com/r/Probability/comments/1e5lhgf/probability_problem_w_no_solution/
  """
  Probability Problem w/ no solution?

  The problem:
  "olympic athletes from the US and China are among those most likely to win a gold metal. 
  For a random athlete from either of these two countries, you have the following information: 
  the probability of winning a gold medal if they practiced at least five hours per day = 0.6 
  the probability of winning a gold meal if they practiced less than five hours per day = 0.3 
  if an american athlete wins a gold medal, what is the probability that they practiced at 
  least five hours per day?"
  Select only one:
  1/3, 2/5, 1/2, 3/5

  I suspect that the problem as stated can't have exact answer but instead a possible range. 
  But by doing the calculations and using Bayes, I think that the problem statement is 
  totally wrong...
  """

  We don't care about the country since it seems to be irrelevant.
  For this model, it depends on how many possible hours an athlete can train per day.

  Cf my Gamble model gamble_winning_gold_medals.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.

/*

  maxHoursPerDay = 10
  var : p
  Probabilities:
  true: 0.6644113667117727
  false: 0.3355886332882274
  mean = [true = 0.664411,false = 0.335589]

  var : practice time per day
  Probabilities:
  9: 0.1423094271538115
  7: 0.1368967072620658
  8: 0.1305818673883627
  5: 0.1290031574199368
  6: 0.1256202074875959
  0: 0.0746504285069914
  4: 0.0683355886332882
  3: 0.0667568786648624
  1: 0.0645015787099684
  2: 0.0613441587731168
  mean = 5.34326


  maxHoursPerDay = 20
  var : p
  Probabilities:
  true: 0.8516104294478528
  false: 0.1483895705521472
  mean = [true = 0.85161,false = 0.14839]

  var : practice time per day
  Probabilities:
  10: 0.0611579754601227
  9: 0.0601993865030675
  19: 0.0590490797546012
  18: 0.0588573619631902
  16: 0.0588573619631902
  6: 0.0584739263803681
  12: 0.0580904907975460
  11: 0.0580904907975460
  14: 0.0565567484662577
  15: 0.0557898773006135
  5: 0.0552147239263804
  13: 0.0542561349693252
  7: 0.0542561349693252
  17: 0.0531058282208589
  8: 0.0496549079754601
  2: 0.0312500000000000
  4: 0.0306748466257669
  1: 0.0300996932515337
  3: 0.0283742331288344
  0: 0.0279907975460123
  mean = 10.5537


  maxHoursPerDay = 12
  var : p
  Probabilities:
  true: 0.7276916631755341
  false: 0.2723083368244659
  mean = [true = 0.727692,false = 0.272308]

  var : practice time per day
  Probabilities:
  10: 0.1068286552157520
  11: 0.1059907834101382
  9: 0.1057813154587348
  8: 0.1041055718475073
  6: 0.1032677000418936
  5: 0.1011730205278592
  7: 0.1005446166736489
  2: 0.0561374109761207
  1: 0.0561374109761207
  3: 0.0540427314620863
  4: 0.0532048596564726
  0: 0.0527859237536657
  mean = 6.39171


*/
go ?=>
  member(MaxHoursPerDay,[10,20,12]),
  println(maxHoursPerDay=MaxHoursPerDay),
  reset_store,
  run_model(10_000,$model(MaxHoursPerDay),[show_probs,mean]),
  nl,
  % show_store_lengths,
  fail,
  nl.
go => true.
  
  
model(MaxHoursPerDay) =>

  
  PracticeTimePerDay = random_integer(MaxHoursPerDay),
  % Using uniform dist - unsurprisingly - gives about the same result.
  % PracticeTimePerDay = uniform_dist(0,MaxHoursPerDay),

  GoldMedal = cases([[PracticeTimePerDay < 5, flip(0.3)], 
                     [PracticeTimePerDay >= 5, flip(0.6)]]),

  % """
  % if an american athlete wins a gold medal, what is the probability
  % that they practiced at least five hours per day?
  % """
  P = check(PracticeTimePerDay >= 5),

  observe(GoldMedal == true),
  
  if observed_ok then
    add("practice time per day",PracticeTimePerDay),
    % add("gold medal",GoldMedal),
    add("p",P)
  end.