/* 

  Casino puzzle in Picat.

  From 
  Muhammad Zain Sarwar
  "The Casino Puzzle That Breaks Your Expectations!"
  https://medium.com/puzzle-sphere/youll-never-guess-how-long-it-takes-to-lose-all-your-money-3a7d57e7ac98
  """
  You find yourself at a casino in Las Vegas, and the dealer presents you with a game. 
  The rules are simple:

  You roll six dice at the same time and each dice is fair meaning every face 
  (1, 2, 3, 4, 5, or 6) has an equal chance of appearing.

  You count how many different numbers appear on the dice.

  If exactly four distinct numbers show up, then you win $1.

  Otherwise, you lose $1.

  You start with $100 and play until you run out of money.

  Each round takes one minute to play.

  Puzzle Statement

  How long do you think it will take for you to lose all your money?
  """

  Cf my Gamble model gamble_casino_puzzle.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.
% import ordset.

main => go.

/*

  var : d
  Probabilities:
  4: 0.5017000000000000
  3: 0.2328500000000000
  5: 0.2273500000000000
  2: 0.0227500000000000
  6: 0.0151500000000000
  1: 0.0002000000000000
  mean = 3.9787
  [len = 20000,min = 1,mean = 3.9787,median = 4.0,max = 6,variance = 0.613146,stdev = 0.783037]

  var : p
  Probabilities:
  true: 0.5017000000000000
  false: 0.4983000000000000
  mean = 0.5017
  show_simple_stats: no numeric data

  Since p > 0.5 the probability of loosing in the long run, is (on average) 0.

*/
go ?=>
  reset_store,
  run_model(20_000,$model,[show_probs_trunc,mean,
                           show_simple_stats % ,
                           % show_percentiles,
                           % show_hpd_intervals,hpd_intervals=[0.84,0.9,0.94,0.99,0.99999],
                           % show_histogram,
                           % min_accepted_samples=1000,show_accepted_samples=true
                          ]),
  nl,
  % show_store_lengths,nl,
  % fail,
  nl.
go => true.
  
  
model() =>
  N = 6,
  Game = random_integer1_n(6,N),

  D = Game.remove_dups.len,
  P = check(D == 4),
  
  add("d",D),
  add("p",P).



/*

  But let's simulate this by setting the two stop cases to 
  0 or 200 so we can see the average number of runs needed.

  var : v
  Probabilities:
  200: 0.6490000000000000
  0: 0.3510000000000000
  mean = 129.8

  var : c
  Probabilities (truncated):
  1970: 0.0020000000000000
  10404: 0.0015000000000000
  8872: 0.0015000000000000
  7396: 0.0015000000000000
  .........
  1026: 0.0005000000000000
  1000: 0.0005000000000000
  818: 0.0005000000000000
  760: 0.0005000000000000
  mean = 9385.34


  The probability of ending when reaching 200 in this setup is thus 
  about 0.65 and loosing is about 0.34.

  The mean number of runs is about 9 400 runs.


*/
go2 ?=>
  reset_store,
  run_model(2_000,$model2,[show_probs_trunc,mean]),
  nl,
  % show_store_lengths,nl,
  % fail,
  nl.
go2 => true.
  

throw(N) = random_integer1_n(6,N).remove_dups.len.

f(V,C,N) = Res =>
  if V <= 0 ; V >= 200 then
    Res = [V,C]
  else
    if throw(N) == 4 then
      Res = f(V+1,C+1,N)
    else
      Res = f(V-1,C+1,N)    
    end
  end.
  
model2() =>
  D = 6,
  [V,C] = f(100,0,D),
  
  add("v",V),
  add("c",C).