/* 

  All Girls world puzzle in Picat.

  From https://brainstellar.com/puzzles/probability/6
  """
  All Girls World?

  In a world where everyone wants a girl child, each family continues having 
  babies till they have a girl. What do you think will the boy-to-girl ratio 
  be eventually?

  Assuming probability of having a boy or a girl is the same and there is no 
  other gender at the time of birth.

  
  Answer:
  Suppose there are N couples. First time, N/2 girls and N/2 boys are born. N/2 
  couples retire, and rest half try for another child.

  Next time, N/4 couples give birth to N/4 girls and N/4 boys. Thus, even in the 
  second iteration, the ratio is 1:1. It can now be seen that this ratio will 
  always remain the same, no matter how many times people try to give birth to a 
  favored gender.
  """

  Note: I added a limit of the maximum number of children in a family (20 for
        the current model) since otherwise it might go on indefinitely. 

  This is a port of my Gamble model gamble_all_girls_world.rkt which also includes
  more exact probabilities.

  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 : len
  Probabilities:
  1: 0.50053333333333339 (1877 / 3750)
  2: 0.24990000000000001 (2499 / 10000)
  3: 0.12220000000000000 (611 / 5000)
  4: 0.06130000000000000 (613 / 10000)
  5: 0.03310000000000000 (331 / 10000)
  6: 0.01676666666666667 (503 / 30000)
  7: 0.00880000000000000 (11 / 1250)
  8: 0.00376666666666667 (113 / 30000)
  9: 0.00180000000000000 (9 / 5000)
  10: 0.00096666666666667 (29 / 30000)
  11: 0.00043333333333333 (13 / 30000)
  12: 0.00020000000000000 (1 / 5000)
  14: 0.00010000000000000 (1 / 10000)
  15: 0.00006666666666667 (1 / 15000)
  13: 0.00006666666666667 (1 / 15000)
  mean = 2.00627

  var : ratio
  Probabilities:
  0: 0.50053333333333339 (1877 / 3750)
  1: 0.24990000000000001 (2499 / 10000)
  2: 0.12220000000000000 (611 / 5000)
  3: 0.06130000000000000 (613 / 10000)
  4: 0.03310000000000000 (331 / 10000)
  5: 0.01676666666666667 (503 / 30000)
  6: 0.00880000000000000 (11 / 1250)
  7: 0.00376666666666667 (113 / 30000)
  8: 0.00180000000000000 (9 / 5000)
  9: 0.00096666666666667 (29 / 30000)
  10: 0.00043333333333333 (13 / 30000)
  11: 0.00020000000000000 (1 / 5000)
  13: 0.00010000000000000 (1 / 10000)
  14: 0.00006666666666667 (1 / 15000)
  12: 0.00006666666666667 (1 / 15000)
  mean = 1.00627


  The ratio is thus ~ 1 (with a possible limit of 20 children per family).

  Cf the geometric distribution:
  Picat> X=[1-geometric_dist_cdf(1/2,I) :  I in 0..10]
X = [0.5,0.25,0.125,0.0625,0.03125,0.015625,0.0078125,0.00390625,0.001953125,0.0009765625,0.00048828125]

  A more exact probability:

  Picat> X=[1-geometric_dist_cdf(1/2,I) :  I in 0..10].sum
  X = 0.99951171875

  Picat> X=[1-geometric_dist_cdf(1/2,I) :  I in 0..100].sum
  X = 1.0

*/
go ?=>
  reset_store,
  run_model(10,$model,[show_probs_rat,mean,presentation=["len","ratio"]]),
  % fail,
  nl.
go => true.

family(L,Limit) = Ret =>
  C = uniform_draw(["boy","girl"]),
  L2 = L ++ [C],
  if C == "girl" ; L2.len >= Limit then
    Ret = L2
  else
    Ret = family(L2,Limit)
  end.
    
model() =>
  Limit = 20,

  ThisFamily = family([],Limit),
  Len = ThisFamily.len,
  Ratio = Len - 1,
  add("len",Len),
  add("ratio",Ratio).