/* 

  Coupon collector's problem in Picat.

  From 
  ## https://www.maa.org/frank-morgans-math-chat-rolling-a-die-until-every-face-has-appeared
  https://sites.williams.edu/Morgan/math-chat-archives/rolling-a-die-until-every-face-has-appeared/
  """
  LD CHALLENGE (Steve Jabloner). How many times do you think you need to roll a
  normal die to be 90% sure that each of the six faces has appeared at least once? Why?

  ANSWER. It takes 23 rolls, as Al Zimmerman discovered in a computer experiment 
  rolling 5000 dice until 90% of them had shown each face. In a less accurate 
  experiment with just 300 dice, Eric Brahinsky wrongly concluded that 22 rolls 
  provide a 90.7% probability. 
  """

  This is a port of my Gamble model gamble_coupon_collectors3.rkt

  Note: See ppl_geometric_cereal_box.pi for a more extensive test
        of the coupon collector's problem.

  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.

main => go.

/*
  Here we just show the HPD interval of 90%

  var : v

  HPD intervals:
  HPD interval (0.9): 6.00000000000000..23.00000000000000

*/
go ?=>
  reset_store,
  N = 6,
  % run_model(10_000,$model(Map,N),Map,[show_probs,mean]),
  % We don't show anything automatically
  run_model(10_000,$model(N),[]), 
  show_hpd_intervals(get_store().get("v"),[0.90]),
  nl.
go => true.

model(N) =>
  % From ppl_geometric_cereal_box.pi
  Vs = [ 1+geometric_dist(1 - (I / N)) : I in 0..N-1],
  add("v",Vs.sum).