/* 

  100 coin tosses in sequence in Picat.

  What is the probability that in 100 recorded coin tosses there are
  first 50 heads and then 50 tails?

  For 10 coin tosses:
  - the probability that the first 5 are 1s and the last 5 are 0s is    
    1/(2**10) = 1/1024 (0.0009765625)
  - the probability of exactly 2 runs is  9/512 (0.017578125)

  For 100 coin tosses
  - the probability that the first 50 are 1s and the last 50 are 0s is 
    1/(2**100) = 1/1267650600228229401496703205376 = very small
    and too small to simulated

  Cf my Gamble model gamble_coins_in_sequence.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.

/*

  For 10 coins:

  var : p
  Probabilities:
  0: 0.9992000000000000
  1: 0.0008000000000000
  mean = 0.0008
  [len = 10000,min = 0,mean = 0.0008,median = 0.0,max = 1,variance = 0.00079936,stdev = 0.028273]

  var : p2
  Probabilities:
  0: 0.9823000000000000
  1: 0.0177000000000000
  mean = 0.0177
  [len = 10000,min = 0,mean = 0.0177,median = 0.0,max = 1,variance = 0.0173867,stdev = 0.131859]

*/
go ?=>
  reset_store,
  run_model(10_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 = 10,
  M = N // 2,

  X = random_integer_n(2,N),


  % first m are 1s and last m are 0s
  P = check1( (X[1..M].remove_dups == [1], X[M+1..N].remove_dups == [0])), 

  Runs = get_runs(X),
  % Exactly two runs
  P2 = check1(Runs.len == 2),

  add("p",P),
  add("p2",P2).