/* 

  Card problem in Picat.

  Second assignment of the CPLINT course:
  https://edu.swi-prolog.org/mod/assign/view.php?id=243
  """
  http://cplint.eu/p/cards.swinb
  Cards

  Suppose you have two decks of poker cards (52 cards, 4 suits, 13 ranks: A, K, Q, J, 10, 9, 8, 7, 6, 5, 4, 3, 2).

  Suppose you draw a card from the first deck and one from the second.

  Write a program to compute the probability that in this way you obtain one pair (two cards of the same rank).

  Can you do it with a single probabilistic clause?

  Add code to compute the probability that you draw at least one ace.
  """

  Note that the two cards are from two different deck of cards!

  This is a port of my Gamble model gamble_cards.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.

main => go.

/*

  var : pair
  Probabilities:
  false: 0.92500000000000004 (37 / 40)
  true: 0.07500000000000000 (3 / 40)
  mean = [false = 0.925,true = 0.075]

  var : at least one ace
  Probabilities:
  false: 0.85029999999999994 (8503 / 10000)
  true: 0.14970000000000000 (1497 / 10000)
  mean = [false = 0.8503,true = 0.1497]

  var : exactly one ace
  Probabilities:
  false: 0.85670000000000002 (8567 / 10000)
  true: 0.14330000000000001 (1433 / 10000)
  mean = [false = 0.8567,true = 0.1433]

  var : two aces
  Probabilities:
  false: 0.99360000000000004 (621 / 625)
  true: 0.00640000000000000 (4 / 625)
  mean = [false = 0.9936,true = 0.0064]

  var : a king or queen in spades
  Probabilities:
  false: 0.92279999999999995 (2307 / 2500)
  true: 0.07720000000000000 (193 / 2500)
  mean = [false = 0.9228,true = 0.0772]


*/
go ?=>
  reset_store(),
  run_model(10_000,$model,[show_probs_rat,mean,
                           presentation=["pair","at least one ace","exactly one ace",
                                         "two aces","a king or queen in spades"]]),
  nl.
go => true.

suit(C) = C.first.
rank(C) = C.second.

model() =>
  Suits = ["hearts","spades","clubs","diamonds"],
  Ranks = ["ace","king","queen","jack","v10","v9","v8","v7","v6","v5","v4","v3","v2"],

  C1 = [uniform_draw(Suits),uniform_draw(Ranks)],
  C2 = [uniform_draw(Suits),uniform_draw(Ranks)],
  
  % Probability of a pair
  Pair = (rank(C1) == rank(C2)).check,

  % At least an ace
  AtLeastOneAce = (rank(C1) == "ace" ; rank(C2) == "ace").check,

  % Exactly one ace
  % ExactlyOneAce = cond( [C1.rank,C2.rank].remove_dups==["ace"],true,false),
  ExactlyOneAce = xor(C1.rank=="ace",C2.rank=="ace"),

  % Two aces
  TwoAces = (rank(C1) == "ace",rank(C2) == "ace").check,

  % A King or a Queen in spades
  AKingOrQueenInSpades = cond(( (C1.suit == "spades", membchk(C1.rank,["king","queen"]))
                                 ;
                                (C2.suit == "spades", membchk(C2.rank,["king","queen"]))
                                ),true,false),

  add_all([ ["pair",Pair],
            ["at least one ace",AtLeastOneAce],
            ["exactly one ace",ExactlyOneAce],
            ["two aces",TwoAces],
            ["a king or queen in spades",AKingOrQueenInSpades]
          ]).