CODEWITHSUNDEEP
pythonpython-3.x
Tim shaw
Tim shaw

Reputation: 37

Python dice rolling simulation

I'm having trouble with a code where I need to roll a six-sided die 1000 times and then return a list of how many times each number on the die was rolled.

The code runs just fine and I can get a list at the end, but my list keeps having 0 in place of four so it appears that my function is not keeping tabs on the number 4 being rolled or it's not being rolled at all.

I'm kind of stumped and I thought maybe someone here could help. Any and all help is appreciated.

Here's my code.

def rollDie(number):
    one = 0
    two = 0
    three = 0
    four = 0
    five = 0
    six = 0
    for i in range(0, number):
        roll=int(random.randint(1,6))
        if roll == 1:
            one = one+1
        elif roll == 2:
            two = two+1
        elif roll == 3:
            three = three+1
        elif roll == 4:
            four == four+1
        elif roll == 5:
            five = five+1
        elif roll == 6:
            six = six+1
    return [one,two,three,four,five,six]

Upvotes: 1

Views: 48327

Answers (4)

dedrick
dedrick

Reputation: 1

import random

def dice():

    print random.randint(1,6)

dice()

Upvotes: -3

steveha
steveha

Reputation: 76695

I can't improve on Martijn Pieters's answer. :-) But this problem can be more conveniently solved using a list.

import random

def rollDie(number):
    # create a list with 7 values; we will only use the top six
    rolls = [0, 0, 0, 0, 0, 0, 0]
    for i in range(0, number):
        roll=int(random.randint(1,6))
        rolls[roll] += 1
    return rolls

if __name__ == "__main__":
    result = rollDie(1000)
    print(result[1:])  # print only the indices from 1 to 6

And, this is a little bit tricky, but here is a better way to create a list of 7 entries all set to zero:

rolls = [0] * 7

Why count the zeros yourself? It's easier to just make Python do the work for you. :-)

EDIT: The list is length 7 because we want to use indices 1 through 6. There is also a position 0 in the list, but we don't use it.

Another way to do it is to map the dice rolls onto indices. It's a pretty simple mapping: just subtract 1. So, a die roll of 1 would go into index 0 of the list, a die roll of 2 would go into index 1, and so on. Now we will use every position in the list.

Here's that version:

import random

def rollDie(number):
    rolls = [0] * 6
    for i in range(0, number):
        roll=int(random.randint(1,6))
        rolls[roll - 1] += 1
    return rolls

if __name__ == "__main__":
    result = rollDie(1000)
    print(result)

Upvotes: 0

Martijn Pieters
Martijn Pieters

Reputation: 1121346

You have a small typo; you are testing for equality, not assigning:

four == four+1

should be:

four = four+1

However, you already have a number between 1 and 6, why not make that into an index into the results list? That way you don't have to use so many if statements. Keep your data out of your variable names:

def rollDie(number):
    counts = [0] * 6
    for i in range(number):
        roll = random.randint(1,6)
        counts[roll - 1] += 1
    return counts

Upvotes: 8

user278064
user278064

Reputation: 10170

You should do random.randint(1, 7), otherwise you will never get a 6.

...
roll = random.randint(1, 7)

Upvotes: -1

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