/* 

  Distinct six dice in Picat.

  From https://www.quora.com/What-are-the-most-interesting-or-popular-probability-puzzles-in-which-the-intuition-is-contrary-to-the-solution/answer/Justin-Rising
  """
  I'm going to flip six standard dice, and count the number of distinct values that come up. 
  So if the result is (6,5,1,2,4,6) I record a five, and if the result is (6,6,4,1,1,6) I 
  record a three. What's the probability that I record a four?

  I guarantee that unless you've worked out this puzzle before, your first answer is 
  going to be pretty far off. See the comments for the right answer.

  Answer:
  ... 
  And our final answer (including the extra factor from the first paragraph) is: 
  65×(64)×4!66≈0.501543209876543
  """

  Cf my Gamble model gamble_distinct_six_dice.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.

/*

  n = 6
  var : num unique
  Probabilities:
  4: 0.5001000000000000
  3: 0.2363000000000000
  5: 0.2295000000000000
  2: 0.0183000000000000
  6: 0.0157000000000000
  1: 0.0001000000000000
  mean = 3.9877

  var : p
  Probabilities:
  true: 0.5001000000000000
  false: 0.4999000000000000
  mean = [true = 0.5001,false = 0.4999]


  n = 10
  var : num unique
  Probabilities:
  6: 0.3513000000000000
  7: 0.3503000000000000
  8: 0.1324000000000000
  5: 0.1275000000000000
  4: 0.0187000000000000
  9: 0.0179000000000000
  3: 0.0016000000000000
  10: 0.0003000000000000
  mean = 6.5003

  var : p
  Probabilities:
  false: 0.9813000000000000
  true: 0.0187000000000000
  mean = [false = 0.9813,true = 0.0187]

*/
go ?=>
  member(N,[6,10]),
  println(n=N),
  reset_store,
  run_model(10_000,$model(N),[show_probs_trunc,mean]),
  nl,
  % show_store_lengths,nl,
  fail,
  nl.
go => true.
  
  
model(N) =>
  X = random_integer1_n(N,N),
  NumUnique = X.remove_dups.len,
  P = check(NumUnique == 4),
  
  add("num unique",NumUnique),
  add("p",P).