Python Chest Opening Odds

Question

I've been trying to figure out how to recreate odds in python for a chest opening game thingy, but the random.choice isn't really considering different odds. The goal I'm trying to achieve is set certain odds for different "rarities", like for example having 50% probability for getting an "Uncommon".

Code example:

import random

skins = {
    'Common':'Knife',
    'Uncommon':'Pistol',
    'Rare':'Submachine Gun',
    'Epic':'Automatic Rifle',
    'Legendary':'Semi-Automatic Sniper Rifle'
}


print("Welcome to N/A")
print("Input 'open' to open a chest.")
n = input("> ")
if n == "open":
    print(random.choice(skins))

Because Its a "random" choice, I couldn't assign odds to it. like say Uncommon would be a 50% chance. — Jayotomix, Sep 14 '21 at 22:12
What do you mean by "the random.choice isn't really considered odds"? Are you asking how to pick a [_weighted_](https://stackoverflow.com/questions/3679694/a-weighted-version-of-random-choice) sample of a collection? — Brian61354270, Sep 14 '21 at 22:12
Check out [random.choices](https://docs.python.org/3/library/random.html#random.choices). It exposes a `weights` parameter. — gmdev, Sep 14 '21 at 22:14
Okay, so the only thing I still need to know is how to assign weights to rarities instead of just a numbered item in the list? — Jayotomix, Sep 14 '21 at 22:21
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Does this answer your question? [A weighted version of random.choice](https://stackoverflow.com/questions/3679694/a-weighted-version-of-random-choice) — Peter O., Sep 15 '21 at 03:01

Ted Klein Bergman · Answer 1 · 2021-09-14T22:42:16.763

A simple solution is to pick a random number between 0 and 100 and check which percentile it gets within.

import random

skins = {
    'Common':'Knife',
    'Uncommon':'Pistol',
    'Rare':'Submachine Gun',
    'Epic':'Automatic Rifle',
    'Legendary':'Semi-Automatic Sniper Rifle'
}

chance = random.randint(0, 100)
if chance < 50:
    print(skins["Common"])
elif chance < 80:
    print(skins["Uncommon"])
elif chance < 95:
    print(skins["Rare"])
elif chance < 99:
    print(skins["Epic"])
else:
    print(skins["Legendary"])

The first if has 50% chance of executing, the second 30%, the third 15%, the fourth 4% and the else has 1 % chance of executing (assuming linear perfect distribution of randint).

score 0 · Accepted Answer · answered Sep 14 '21 at 22:28

My advice is to use Random choice() method since it allows you to make a list where you can specify probability for each element. https://www.w3schools.com/python/ref_random_choices.asp

from random import choices
skins = {
    'Common':'Knife',
    'Uncommon':'Pistol',
    'Rare':'Submachine Gun',
    'Epic':'Automatic Rifle',
    'Legendary':'Semi-Automatic Sniper Rifle'
}
weights = [0.4, 0.25, 0.2, 0.1, 0.05]
print("Welcome to N/A")
print("Input 'open' to open a chest.")
n = input("> ")
if n == "open":
    print(choices(list(skins.items()), weights))

Python Chest Opening Odds

2 Answers2