## Simple Monte Carlo Simulation in python

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What is the expectation of the number of times you need to roll a fair die until you have seen every number?

I have been asked to define a function that runs a monte carlo simulation that returns an estimate for the question above. My understanding of the solution is that I need to:

- Define a function that takes the number of trials as an input
- Generates a random integer between 1 and 6 (the number of sides on a die)
- Store that integer in a list if it is not already in the list
- Count the number of trials it takes to reach said condition, and return that count

I am fairly new to programming and Python in particular so I am struggling to identify why my syntax does not produce an output when the function is called and was hoping someone could help steer me in right direction

This is my code :

def roll(n=1000): trials = [] sides = 6 start = 1 for i in range (n): for x in range (sides): collection = [random.randint(1,sides)] while any([x not in collection]): collection.append(random.randint(1,6)) trials.append(len(collection)) return sum(trials)/ len(trials)

You are probably not *printing* whatever the funtion returns - that why it displays nothing.

Use `print(roll())`

instead of `roll()`

to print the result you get.

You have too many loops and your solution uses far too much memory space.

Consider being unlucky and having to roll 1.000.000.000.000 times to get the first 6 - you'll save 1.000.000.000.000 other numbers in a list .. thats a lot of memory.

You can use a `set`

to remember seen numbers and a counter to count how long it took to find all:

def roll(sides=6, n=1000): """Tests 'n' times to roll all numbers from 1 to 'sides' randomly. Returns average tries needed to see all numbers over 'n' tries.""" trials = [] # collects all sinly tried counters for _ in range(n): seen = set() # empty, will only ever store 6 elements at most tried = 0 # how long did it take to find all 6? while len(seen) < sides: # use sides here as well seen.add(random.randint(1,sides)) tried += 1 trials.append(tried) return sum(trials)/n print(roll())

Output (4 starts):

14.878 14.694 14.732 14.516

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Your `while`

condition does not express what you expect it to be. You probably wanted to use list comprehension

while any([x not in collection for x in [1, 2, 3, 4, 5, 6])

also, you don't want 3 layers of loop, only two: one `for`

each trial and the other `while`

trial is not complete. A working example, close to your original post would be

import random def roll(n=1000): trials = [] sides = 6 start = 1 possible_sides = [1, 2, 3, 4, 5, 6] for i in range (n): collection = [random.randint(1,sides)] while any([side not in collection for side in possible_sides]): collection.append(random.randint(1,6)) trials.append(len(collection)) return sum(trials)/ len(trials)

whereas a more efficient solution uses `set`

to efficiently do the same thing as the previous solution achieves with `any([side not in collection for side in possible_sides])`

:

import random def roll(n=1000): trials = [] sides = 6 start = 1 possible_sides = set([1, 2, 3, 4, 5, 6]) for i in range (n): n_rolls = 0 sides_rolled = set() while not sides_rolled == possible_sides: sides_rolled.add(random.randint(1, sides)) n_rolls += 1 trials.append(n_rolls) return sum(trials)/ len(trials)

Or, even more efficiently, just check if `len(sides_rolled) < 6`

, as pointed out in an answer by Patrick Artner.

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You can simplify your code considerably by using a `set`

instead of a list and also changing your iteration logic:

import random def roll_till_all_sides_appeared(): sides_seen = set() n = 0 while len(sides_seen) < 6: side = random.randint(1, 6) sides_seen.add(side) # will only be added if it isn't present n += 1 return n def test(repetitions): max_n = float('-inf') min_n = float('inf') sum_n = 0 for _ in range(repetitions): n = roll_till_all_sides_appeared() max_n = max(max_n, n) min_n = min(min_n, n) sum_n += n print('max:', max_n) print('min:', min_n) print('avg:', sum_n / repetitions)

This code can be used like this:

>>> test(10) max: 32 min: 8 avg: 14.3 >>> test(100) max: 45 min: 6 avg: 14.13 >>> test(1000) max: 56 min: 6 avg: 14.749 >>> test(10000) max: 62 min: 6 avg: 14.6422

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##### Comments

- Your original post is not properly indented: The
`while`

should have at least one line indented after it. - What do you mean by "does not produce an output when the function is called " ? How are you invoking/calling the function?
- @qu yes, but if you use it you need to use it throughout - and it is not needed for the task as formulated:
**YAGNI** - @Q you introduce a variable that you do not need. If you wanted to have it variable, I would use
`def roll(n=1000,sides=6)`

- this would give you the same readability but also make clear that this is used to calculate it for other sided dice as well ... and add docstring etc. - In the case of the code from the OP,
`sides`

is an explaining variable, ie it answers the question: "What is that number 6 doing?" (You had a bare 6 in a previous edit) `for _ in range(1000)`

should probably be`for _ in range(n)`

. Otherwise`n`

is not needed at all. You can also use it instead of`len(trials)`

, because if that is not`n`

something went seriously wrong...- @Graipher Thanks .. some blunders there after the last edit
- **while not sides_rolled == possible_sides: sides_rolled.add(random.randint(1, sides)) ** the "not" accomplishes the same thing as **while sides_rolled != possible_sides, then adds the result to the set and counts the number of rolls? Also, the .add is a specifc syntax for the set type that accomplishes the same thing as append?
- Thank you for this solution. Very helpful
- Yes, I could've written
`sides_rolled != possible_sides`

and it would perhaps be considered better. Honestly I was typing out rather quick and missed some possible idiosyncrasies. In general, as pointed out in other comments, checking the length of`sides_rolled`

is sufficient, and set equality adds unnecessary overhead.