/* 

  Poisson fishing problem in Picat.

  https://www.reddit.com/r/probabilitytheory/comments/1fgmhgw/poisson_fishing_problem/
  """
  Question goes like this: A fisherman catches fish according to a Poisson process with 
  rate 0.6 per hour. The fisherman will keep fishing for two hours. If he has caught at 
  least one fish, he quits. Otherwise, he continues until he catches at least one fish.

  (a) Find the probability that the total time he spends fishing is between two and five hours.

  Solution and my conflicting approach:

  First of all he'll fish for more than 2 hrs if he catches no fish in first two hrs and the 
  probability of that is P(k=0,t=2).

  1.After two hrs, the probability that he fish for 3 more hrs is that he gets 1 fish in the 
  interval of 3 hrs which is P(k=1,t=3). So total probability is P1 = P(k=0,t=2).P(k=1,t=3)

  2. After 2 hrs, the probability that waiting time is less than 3hrs is P(0<T<3) = 1-exp(0.63) 
  (from exponential pdf). This is equivalent to saying there is atleast one fish caught in 
  3hrs interval which is equal to 1-P(k=0,t=3) = 1-exp(0.63. So the total probability is 
  now P2 = P(k=0,t=2)[1 - P(k=0,t=3)]

  You can see the results ate different but approach seems to me is correct. Can you please 
  clarify the results. Thank you.

  P.S. P(k,t) means k arrival in t interval
  """

  Cf my Gamble model gamble_poisson_fishing_problem.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.

main => go.

/*

  This model: The probability (p) that the fishing is between 2 and 5 hours 
  - according to this model - is very close to  0.5 (perhaps exactly 0.5?).

var : len
Probabilities:
1: 0.4425000000000000
2: 0.2480000000000000
3: 0.1328000000000000
4: 0.0811000000000000
5: 0.0410000000000000
6: 0.0256000000000000
7: 0.0132000000000000
8: 0.0068000000000000
9: 0.0032000000000000
10: 0.0022000000000000
11: 0.0018000000000000
12: 0.0008000000000000
14: 0.0004000000000000
15: 0.0003000000000000
13: 0.0002000000000000
17: 0.0001000000000000
mean = 2.2613
Histogram (total 10000)
      1:  4425 ############################################################ (0.443 / 0.443)
      2:  2480 ##################################                           (0.248 / 0.691)
      3:  1328 ##################                                           (0.133 / 0.823)
      4:   811 ###########                                                  (0.081 / 0.904)
      5:   410 ######                                                       (0.041 / 0.945)
      6:   256 ###                                                          (0.026 / 0.971)
      7:   132 ##                                                           (0.013 / 0.984)
      8:    68 #                                                            (0.007 / 0.991)
      9:    32                                                              (0.003 / 0.994)
     10:    22                                                              (0.002 / 0.996)
     11:    18                                                              (0.002 / 0.998)
     12:     8                                                              (0.001 / 0.999)
     13:     2                                                              (0.000 / 0.999)
     14:     4                                                              (0.000 / 1.000)
     15:     3                                                              (0.000 / 1.000)
     17:     1                                                              (0.000 / 1.000)

var : total
Probabilities:
1: 0.7333000000000000
2: 0.2153000000000000
3: 0.0457000000000000
4: 0.0052000000000000
5: 0.0005000000000000
mean = 1.3243
Histogram (total 10000)
      1:  7333 ############################################################ (0.733 / 0.733)
      2:  2153 ##################                                           (0.215 / 0.949)
      3:   457 ####                                                         (0.046 / 0.994)
      4:    52                                                              (0.005 / 0.999)
      5:     5                                                              (0.001 / 1.000)

var : p
Probabilities:
true: 0.5029000000000000
false: 0.4971000000000000
mean = [true = 0.5029,false = 0.4971]
Histogram (total 10000)
false     :  4971 ###########################################################  (0.497 / 0.497)
true      :  5029 ############################################################ (0.503 / 1.000)

*/
go ?=>
  reset_store,
  run_model(10_000,$model,[show_probs,mean,show_histogram]),
  nl,
  % show_store_lengths,nl,
  % fail,
  nl.
go => true.
  
  
model() =>
  Lambda = 6/10,

  Fishing = [],
  while (Fishing.sum == 0)
    NumFish = poisson_dist(Lambda),
    Fishing := Fishing ++ [NumFish]
  end,

  Len = Fishing.len,
  Total = Fishing.sum,

  P = check( (Len >= 2, Len <= 5)),
  
  add("len",Len),
  add("total",Total),
  add("p",P).