/* 

  Sleeping beauty problem in Picat.

  From https://medium.com/@theo.seeds/why-probability-isnt-real-df50b72a073e
  """
  There’s a probability paradox called the "Sleeping Beauty Paradox". It seems 
  simple at first, but philosophers and statisticians still haven’t figured it 
  out.

  Here’s how it goes. An experimenter flips a coin on Sunday. Then, he puts 
  his experiment subject, Sleeping Beauty, to sleep. When Sleeping Beauty 
  wakes up, she won’t remember if the coin was heads or tails, and she won't 
  remember if she’s been woken up before.

  If the coin’s heads, then the scientist wakes up Sleeping Beauty on Monday.

  If the coin’s tails, then the scientist wakes up Sleeping Beauty on Monday, 
  puts her back to sleep, and then wakes her up again on Tuesday.

  So, suppose you’re Sleeping Beauty, and you wake up. You have no idea how 
  the coin flip went and you have no idea what day it is. What’s the 
  probability that the coin came up heads?

  The paradox

  There are two ways of looking at this problem. They both make sense when 
  you explain them. They both seem equally valid. And they both give you completely 
  different answers.

  One way of looking at the problem is looking at the coin flip. The coin had 
  a 50/50 chance of being heads. And you’re gonna wake up either way — the 
  fact that you’re awake now gives you no new information about the coin flip. 
  So the coin is still 50% likely to have come up heads.

  Another way of looking at the problem is by looking at the fact you’ve 
  woken up. Suppose you ran this experiment 100 times. Sleeping Beauty would 
  wake up 100 times in a world where the coin was heads, and 200 times in a 
  world where the coin was tails. So if you’re Sleeping Beauty and you’re 
  waking up, there’s only a 1/3 probability that the coin was heads.

  Obviously, the probability can’t be 1/2 and 1/3. It’s a paradox. So what 
  the hell’s going on?
  """

  From https://en.wikipedia.org/wiki/Sleeping_Beauty_problem
  """
  This has become the canonical form of the problem:

    Sleeping Beauty volunteers to undergo the following experiment and is told all of the 
    following details: On Sunday she will be put to sleep. Once or twice, during the 
    experiment, Sleeping Beauty will be awakened, interviewed, and put back to sleep with an 
    amnesia-inducing drug that makes her forget that awakening. A fair coin will be tossed 
    to determine which experimental procedure to undertake:

    - If the coin comes up heads, Sleeping Beauty will be awakened and interviewed on Monday only.
    - If the coin comes up tails, she will be awakened and interviewed on Monday and Tuesday.

    In either case, she will be awakened on Wednesday without interview and the experiment ends.

    Any time Sleeping Beauty is awakened and interviewed she will not be able to tell which 
    day it is or whether she has been awakened before. During the interview Sleeping Beauty 
    is asked: "What is your credence now for the proposition that the coin landed heads?"
  """


  As mentioned in both these texts, there is two different approaches with different
  answers on the probability of head:
  - 1/2
  - 1/3

  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, ppl_common_utils.
import util.
% import ordset.

main => go.

/*
  Here we show two different approaches of modelling this problem
  of the probability of heads: either 1/2 or 1/3.

  var : sleeping beauty wakeups
  Probabilities:
  tails: 0.6663110162710056
  heads: 0.3336889837289944
  mean = []

  var : p
  Probabilities:
  true: 0.5004000000000000
  false: 0.4996000000000000
  mean = 0.5004

*/
go ?=>
  reset_store,
  run_model(20_000,$model,[show_probs_trunc,mean]),
  nl,
  nl.
go => true.
  
  
model() =>
  Coin = categorical([1/2,1/2],[heads,tails]), 
  SB = new_list(2), % The Sleeping Beauty

  if Coin = heads then
    SB[1] := wake_up,
    add("sleeping beauty wakeups",heads), % 1/3 solution
  else
    SB[1] := wake_up_and_sleep,
    SB[2] := wake_up,
    add("sleeping beauty wakeups",tails),
    add("sleeping beauty wakeups",tails),
  end,
  
  % 1/2 solution:
  % I.e. monday was a wake up but tuesday was not
  P = check( (SB[1] == wake_up ; SB[2] != wake_up)),

  add("p",P).