Too many copies? Poor comparison? Urn Probability Problem

Question

full code: https://gist.github.com/QuantVI/79a1c164f3017c6a7a2d860e55cf5d5b

TLDR: sum(a3) gives a number like 770, when it should be more like 270 - as in 270 of 1000 trials where the results of drawing 4 contained (at least) 2 blue and 1 green ball.

---

I've rewritten both my way of creating the sample output, and my way of comparing the results twice already. Python as a syntax `all(x in a for x n b)` which I used initially, then change to something more deliberate to see if there was a change. I still have 750+ `True` evaluations of each trial. This is why I reassessed how I was selecting without replacement.

I've tested the draw function on its own with different Hats and was sure it worked.

The expected probability when drawing 4balls, without replacement, from a hat containing (blue=3,red=2,green=6), and having the outcome contain (blue=2,green=1) or ['blue','blue','green'] is around 27.2%. In my 1000 trials, I get higher then 700, repeatedly.

Is the error in Hat.draw() or is it in experiment()?

Note: Certain things are commented out, because I am debugging. Thus use sum(a3) as experiment is commented out to return things other than the probability right now.

import copy
import random
# Consider using the modules imported above.

class Hat:
    def __init__(self, **kwargs):
        self.d = kwargs
        self.contents = [
            key for key, val in kwargs.items() for num in range(val)
        ]
    def draw(self, num: int) -> list:
        if num >= len(self.contents):
            return self.contents
        else:
            indices = random.sample(range(len(self.contents)), num)

            chosen = [self.contents[idx] for idx in indices]

            #new_contents = [ v for i, v in enumerate(self.contents) if i not in indices]

            new_contents = [pair[1] for pair in enumerate(self.contents) 
                            if pair[0] not in indices]
            self.contents = new_contents
            return chosen

    def __repr__(self): return str(self.contents)

def experiment(hat, expected_balls, num_balls_drawn, num_experiments):
  trials =[]
  for n in range(num_experiments):
    copyn = copy.deepcopy(hat)
    result = copyn.draw(num_balls_drawn)
    trials.append(result)

  #trials = [ copy.deepcopy(hat).draw(num_balls_drawn) for n in range(num_experiments) ]

  expected_contents =  [key for key, val in expected_balls.items() for num in range(val)]

  temp_eval = [[o for o in expected_contents if o in trial] for trial in trials]

  temp_compare = [ evaled == expected_contents for evaled in temp_eval]

  return expected_contents,temp_eval,temp_compare, trials

  #evaluations = [ all(x in trial for x in expected_contents) for trial in trials ]

  #if evaluations: prob = sum(evaluations)/len(evaluations)
  #else: prob = 0

  #return prob, expected_contents


#hat3 = Hat(red=5, orange=4, black=1, blue=0, pink=2, striped=9)
#hat4 = Hat(red=1, orange=2, black=3, blue=2)

hat1 = Hat(blue=3,red=2,green=6)

a1,a2,a3,a4 = experiment(hat=hat1, expected_balls={"blue":2,"green":1}, num_balls_drawn=4, num_experiments=1000)
        #actual = probability
        #expected = 0.272
        #self.assertAlmostEqual(actual, expected, delta = 0.01, msg = 'Expected experiment method to return a different probability.')


hat2 = Hat(yellow=5,red=1,green=3,blue=9,test=1)

b1,b2,b3,b4 = experiment(hat=hat2, expected_balls={"yellow":2,"blue":3,"test":1}, num_balls_drawn=20, num_experiments=100)
        #actual = probability
        #expected = 1.0
        #self.assertAlmostEqual(actual, expected, delta = 0.01, msg = 'Expected experiment method to return a different probability.')

Answer 1

The issue is temp_eval = [[o for o in expected_contents if o in trial] for trial in trials]. It will always ad both blue to the list even if only one blue exists in the results of one trial.

However, I couldn't fix the error in a straight-forward way. Instead, my fix created a much lower answer, something less than 0.1, when around 0.27 is (270 of 1000 trials) is what I need.

The roundabout solution was to convert lists like ['red', 'green', 'blue', 'green'] into dictionaries using list on collections.Counter of that list. Then do a key-wose comparison of the values, such as [y[key]<= x.get(key,0) for key in y.keys()]). In this comparison y is the expected_balls variable, and x is the list of the counter object. If x doesn't have one of the keys, we get 0. Zero will be less than the value of any key in expected_balls.

From here we use functols.reduce to turn the output into a single True or False value. Then we map that functionality (compare all keys and get one T/F value) across all trials.

def experiment(hat, expected_balls, num_balls_drawn, num_experiments):
    trials =[]
    trials = [ copy.deepcopy(hat).draw(num_balls_drawn) 
              for n in range(num_experiments) ]
    trials_kvpairs = [dict(collections.Counter(trial)) for trial in trials]
    
    def contains(contained:dict , container:dict):
        each = [container.get(key,0) >= contained[key]
                for key in contained.keys()]
        return reduce(lambda item0,item1: item0 and item1, each)
    
    trials_success = list(map(lambda t: contains(expected_balls,t), trials_kvpairs))
    
    # expected_contents =  [pair[0] for pair in expected_balls.items() for num in range(pair[1])]
    # temp_eval = [[o for o in trial if o in expected_contents] for trial in trials]
    # temp_compare = [ evaled == expected_contents for evaled in temp_eval]
    # if temp_compare: prob = sum(temp_compare)/len(trials)
    # else: prob = 0
    return 'prob', trials_kvpairs, trials_success

When run using the this experiment(hat=hat1, expected_balls={"blue":2,"green":1}, num_balls_drawn=4, num_experiments=1000) the sum of the third part of the output was 276.