/* 

  Consecutive numbers in Lotto ticket in Picat.

  From https://www.reddit.com/r/Probability/comments/1etgees/consecutive_numbers_in_lotto_random_pick_ticket/
  """
  Consecutive numbers in lotto 'random pick' ticket

  Just got this lotto ticket. Thought the number of consecutive numbers was 
  just way too many. Are these really random? 
  Or is this number of consecutive numbers normal? 
  """

  The Lotto described is 7 numbers in 1..40.

  Cf my Gamble model gamble_consecutive_numbers_in_lotto_tickets.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, ppl_common_utils.
import util.

main => go.

/*

  For a single Lotto ticket row in the pick 7 of 40 Lotto, the probabilities 
  of the number of consecutive numbers are:

  [num = 40,num_pick = 7]
  var : num consecutive
  Probabilities:
  1: 0.4341000000000000
  0: 0.2889000000000000
  2: 0.2227333333333333
  3: 0.0500000000000000
  4: 0.0041333333333333
  5: 0.0001333333333333
  mean = 1.04677

  var : p
  Probabilities:
  true: 0.7111000000000000
  false: 0.2889000000000000
  mean = [true = 0.7111,false = 0.2889]

  
  However, the example shows that for the 18 rows in the Lotto ticket,
  17 of these rows has at least one occurrence of consecutive numbers.

  Now, the model shows that the probability of at least one occurrence is quite 
  (surprisingly?) high: about 71%.  

  So what is the probability of 17 ticket rows of 18 has at least one occurrence
  of consecutive numbers. Here's a simple way, using binomial. First a simulation:

  Picat> binomial_dist_n(18,0.7111,10000).show_freq_prob
  Show freqs:
  5: 3 (0.0003000000000000)
  6: 15 (0.0015000000000000)
  7: 37 (0.0037000000000000)
  8: 114 (0.0114000000000000)
  9: 306 (0.0306000000000000)
  10: 712 (0.0712000000000000)
  11: 1311 (0.1311000000000000)
  12: 1868 (0.1868000000000000)
  13: 2052 (0.2052000000000000)
  14: 1761 (0.1761000000000000)
  15: 1132 (0.1132000000000000)
  16: 526 (0.0526000000000000)
  17: 137 (0.0137000000000000)
  18: 26 (0.0026000000000000)

  Given that we can rely on p=0.7111 in the experiment, the 
  exact PDF for exactly 17 such ticket rows is thus

  Picat> X=binomial_dist_pdf(18,0.7111,17)
  X = 0.015806713578139

  Here are the exact probabilities for 0..18 rows with at least one occurrence 
  of consecutive numbers (again, assuming that p=0.714 is correct):
  Picat> foreach(V=T in [V=binomial_dist_pdf(18,0.7111,V) : V in 0..18]) println(V=T) end
  0 = 1.96542e-10
  1 = 8.70786e-09
  2 = 1.82185e-07
  3 = 2.39164e-06
  4 = 2.20755e-05
  5 = 0.000152143
  6 = 0.000811383
  7 = 0.00342367
  8 = 0.0115872
  9 = 0.0316898
  10 = 0.0702012
  11 = 0.125668
  12 = 0.180437
  13 = 0.204982
  14 = 0.180194
  15 = 0.118275
  16 = 0.0545855
  17 = 0.0158067
  18 = 0.00216148

  The probability of (exact) 17 such row is small, about 1.6%, but 
  it's definity not very rare, though given a "surprise level" 0.01
  (i.e. how rare an event should in order to raise a surprise),
  it _is_ a little surprising.

  However, we would probably also be surprised if there were - say - 15 or 
  more such consecutive rows. And the probability of 15 or more 
  such rows is quite high: 19%!

  Picat> X=1-binomial_dist_cdf(18,0.7111,14)           
  X = 0.190828710530796

  The average number of ticket rows (of 18) with at least one occurrence is
  Picat> X=binomial_dist_mean(18,0.7111)    
  X = 12.799799999999999

  So, we should expect quite many occurrences of consecutive numbers in 
  a given Lotto ticket.

  * Other Lottos

  For the Swedish Lotto variant a draw is to pick 7 numbers from 1..35.
  The probability of at least one occurrence of consecutive numbers is 
  about 77%:

  [num = 35,num_pick = 7]
  var : num consecutive
  Probabilities:
  1: 0.4244000000000000
  2: 0.2632333333333333
  0: 0.2317333333333333
  3: 0.0713000000000000
  4: 0.0089000000000000
  5: 0.0004333333333333
  mean = 1.20253

  var : p
  Probabilities:
  true: 0.7682666666666667
  false: 0.2317333333333333
  mean = [true = 0.768267,false = 0.231733]


  For the UK Lotto with 6 numbers from 1..49, the probability is lower: about 49.6%

  [num = 49,num_pick = 6]
  var : num consecutive
  Probabilities:
  0: 0.5041000000000000
  1: 0.3931333333333333
  2: 0.0930666666666667
  3: 0.0094333333333333
  4: 0.0002666666666667
  mean = 0.608633

  var : p
  Probabilities:
  false: 0.5041000000000000
  true: 0.4959000000000000
  mean = [false = 0.5041,true = 0.4959]


*/
go ?=>
  member([Num,NumPick],[[40,7],
                        [35,7],
                        [49,6]]),
  println([num=Num,num_pick=NumPick]),
  reset_store,
  run_model(30_000,$model(Num,NumPick),[show_probs_trunc,mean]),
  nl,
  % show_store_lengths,
  fail,
  nl.
go => true.
  
  
model(Num,NumPick) =>
  Nums = 1..Num,

  Picks = draw_without_replacement(NumPick,Nums).sort,
  Diffs = differences(Picks),
  
  % Here we count any occurrence of two consecutive numbers as an occurrence.
  % This mean that if three numbers are consecutive, it's counted as 2 occurrences, etc.
  NumConsecutive = [1 : D in Diffs, D == 1].sum,

  % Probability of at least one consecutive occurrence
  P = check(NumConsecutive > 0),

  add("num consecutive",NumConsecutive),
  add("p",P).