updated AI essay
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,slimbook,wins-slimbook,15.10.2024 16:32,file:///home/slimbook/.var/app/org.libreoffice.LibreOffice/config/libreoffice/4;
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# Analyzing the dataset of LoL's champions"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"https://www.kaggle.com/datasets/uskeche/leauge-of-legends-champions-dataset?resource=download"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 1,
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"metadata": {},
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"outputs": [],
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"source": [
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"import pandas as pd"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Read the data"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 2,
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"metadata": {},
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"outputs": [],
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"source": [
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"Dat = pd.read_csv(\"LoLChampions.csv\")\n",
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"Dat = Dat.drop(\"Id\", axis=1)"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"---"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Data quality and data preprocessing"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"The first thing that we should do is to check the quality of the data set."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 3,
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/html": [
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"<div>\n",
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"<style scoped>\n",
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" .dataframe tbody tr th:only-of-type {\n",
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" vertical-align: middle;\n",
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" }\n",
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"\n",
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" .dataframe tbody tr th {\n",
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" vertical-align: top;\n",
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" }\n",
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"\n",
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" .dataframe thead th {\n",
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" text-align: right;\n",
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" }\n",
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"</style>\n",
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"<table border=\"1\" class=\"dataframe\">\n",
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" <thead>\n",
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" <tr style=\"text-align: right;\">\n",
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" <th></th>\n",
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" <th>Name</th>\n",
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" <th>Class</th>\n",
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" <th>Style</th>\n",
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" <th>Difficulty</th>\n",
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" <th>DamageType</th>\n",
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" <th>Damage</th>\n",
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" <th>Sturdiness</th>\n",
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" <th>Crowd-Control</th>\n",
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" <th>Mobility</th>\n",
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" <th>Functionality</th>\n",
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" </tr>\n",
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" </thead>\n",
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" <tbody>\n",
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" <tr>\n",
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" <th>0</th>\n",
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" <td>Aatrox</td>\n",
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" <td>Warrior</td>\n",
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" <td>3</td>\n",
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" <td>2</td>\n",
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" <td>P</td>\n",
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" <td>3</td>\n",
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" <td>3</td>\n",
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" <td>2</td>\n",
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" <td>2</td>\n",
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" <td>2.0</td>\n",
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" </tr>\n",
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" <tr>\n",
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" <th>1</th>\n",
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" <td>Ahri</td>\n",
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" <td>Mage</td>\n",
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" <td>10</td>\n",
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" <td>2</td>\n",
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" <td>M</td>\n",
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" <td>3</td>\n",
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" <td>1</td>\n",
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" <td>2</td>\n",
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" <td>3</td>\n",
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" <td>1.0</td>\n",
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" </tr>\n",
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" <tr>\n",
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" <th>2</th>\n",
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" <td>Akali</td>\n",
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" <td>Assasin</td>\n",
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" <td>7</td>\n",
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" <td>2</td>\n",
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" <td>PM</td>\n",
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" <td>3</td>\n",
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" <td>1</td>\n",
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" <td>1</td>\n",
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" <td>3</td>\n",
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" <td>1.0</td>\n",
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" </tr>\n",
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" <tr>\n",
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" <th>3</th>\n",
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" <td>Alistar</td>\n",
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" <td>Tank-Support</td>\n",
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" <td>7</td>\n",
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" <td>1</td>\n",
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" <td>M</td>\n",
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" <td>1</td>\n",
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" <td>3</td>\n",
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" <td>3</td>\n",
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" <td>1</td>\n",
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" <td>2.0</td>\n",
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" </tr>\n",
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" <tr>\n",
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" <th>4</th>\n",
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" <td>Amumu</td>\n",
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" <td>Tank</td>\n",
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" <td>9</td>\n",
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" <td>1</td>\n",
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" <td>M</td>\n",
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" <td>2</td>\n",
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" <td>3</td>\n",
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" <td>3</td>\n",
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" <td>1</td>\n",
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" <td>1.0</td>\n",
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" </tr>\n",
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" </tbody>\n",
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"</table>\n",
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"</div>"
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],
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"text/plain": [
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" Name Class Style Difficulty DamageType Damage Sturdiness \\\n",
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"0 Aatrox Warrior 3 2 P 3 3 \n",
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"1 Ahri Mage 10 2 M 3 1 \n",
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"2 Akali Assasin 7 2 PM 3 1 \n",
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"3 Alistar Tank-Support 7 1 M 1 3 \n",
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"4 Amumu Tank 9 1 M 2 3 \n",
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"\n",
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" Crowd-Control Mobility Functionality \n",
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"0 2 2 2.0 \n",
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"1 2 3 1.0 \n",
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"2 1 3 1.0 \n",
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"3 3 1 2.0 \n",
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"4 3 1 1.0 "
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]
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},
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"execution_count": 3,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"Dat.head()"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 4,
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/html": [
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"<div>\n",
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"<style scoped>\n",
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"\n",
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" .dataframe tbody tr th {\n",
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" text-align: right;\n",
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"</style>\n",
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"<table border=\"1\" class=\"dataframe\">\n",
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" <thead>\n",
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" <tr style=\"text-align: right;\">\n",
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" <th></th>\n",
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" <th>Style</th>\n",
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" <th>Difficulty</th>\n",
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" <th>Damage</th>\n",
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" <th>Sturdiness</th>\n",
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" <th>Crowd-Control</th>\n",
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" <th>Mobility</th>\n",
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" <th>Functionality</th>\n",
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" </tr>\n",
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" </thead>\n",
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" <tbody>\n",
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" <tr>\n",
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" <th>count</th>\n",
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" <td>145.000000</td>\n",
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" <td>145.000000</td>\n",
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" <td>145.000000</td>\n",
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" <td>145.000000</td>\n",
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" <td>145.000000</td>\n",
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" <td>145.000000</td>\n",
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" <td>144.000000</td>\n",
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" </tr>\n",
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" <tr>\n",
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" <th>mean</th>\n",
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" <td>6.544828</td>\n",
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" <td>1.882759</td>\n",
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" <td>2.434483</td>\n",
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" <td>1.662069</td>\n",
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" <td>2.013793</td>\n",
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" <td>1.717241</td>\n",
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" <td>1.465278</td>\n",
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" </tr>\n",
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" <tr>\n",
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" <th>std</th>\n",
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" <td>2.806192</td>\n",
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" <td>0.702213</td>\n",
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" <td>0.715054</td>\n",
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" <td>0.774782</td>\n",
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" <td>0.726351</td>\n",
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" <td>0.742588</td>\n",
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" <td>0.668086</td>\n",
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" </tr>\n",
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" <tr>\n",
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" <th>min</th>\n",
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" <td>1.000000</td>\n",
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" <td>1.000000</td>\n",
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" <td>1.000000</td>\n",
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" <td>1.000000</td>\n",
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" <td>0.000000</td>\n",
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" <td>1.000000</td>\n",
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" <td>1.000000</td>\n",
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" </tr>\n",
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" <tr>\n",
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" <th>25%</th>\n",
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" <td>4.000000</td>\n",
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" <td>1.000000</td>\n",
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" <td>2.000000</td>\n",
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" <td>1.000000</td>\n",
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" <td>2.000000</td>\n",
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" <td>1.000000</td>\n",
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" <td>1.000000</td>\n",
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" </tr>\n",
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" <tr>\n",
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" <th>50%</th>\n",
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" <td>7.000000</td>\n",
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" <td>2.000000</td>\n",
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" <td>3.000000</td>\n",
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" <td>1.000000</td>\n",
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" <td>2.000000</td>\n",
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" <td>2.000000</td>\n",
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" <td>1.000000</td>\n",
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" </tr>\n",
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" <tr>\n",
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" <th>75%</th>\n",
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" <td>9.000000</td>\n",
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" <td>2.000000</td>\n",
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" <td>3.000000</td>\n",
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" <td>2.000000</td>\n",
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" <td>3.000000</td>\n",
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" <td>2.000000</td>\n",
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" <td>2.000000</td>\n",
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" </tr>\n",
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" <tr>\n",
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" <th>max</th>\n",
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" <td>10.000000</td>\n",
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" <td>3.000000</td>\n",
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" <td>3.000000</td>\n",
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" <td>3.000000</td>\n",
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" <td>3.000000</td>\n",
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" <td>3.000000</td>\n",
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" <td>3.000000</td>\n",
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" </tr>\n",
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" </tbody>\n",
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"</table>\n",
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"</div>"
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],
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"text/plain": [
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" Style Difficulty Damage Sturdiness Crowd-Control \\\n",
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"count 145.000000 145.000000 145.000000 145.000000 145.000000 \n",
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"mean 6.544828 1.882759 2.434483 1.662069 2.013793 \n",
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"std 2.806192 0.702213 0.715054 0.774782 0.726351 \n",
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"min 1.000000 1.000000 1.000000 1.000000 0.000000 \n",
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"25% 4.000000 1.000000 2.000000 1.000000 2.000000 \n",
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"50% 7.000000 2.000000 3.000000 1.000000 2.000000 \n",
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"75% 9.000000 2.000000 3.000000 2.000000 3.000000 \n",
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"max 10.000000 3.000000 3.000000 3.000000 3.000000 \n",
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"\n",
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" Mobility Functionality \n",
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"count 145.000000 144.000000 \n",
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"mean 1.717241 1.465278 \n",
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"std 0.742588 0.668086 \n",
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"min 1.000000 1.000000 \n",
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"25% 1.000000 1.000000 \n",
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"50% 2.000000 1.000000 \n",
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"75% 2.000000 2.000000 \n",
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"max 3.000000 3.000000 "
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]
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},
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"execution_count": 4,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"Dat.describe()"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"We can observe that there are missing values in the Functionality variable."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 5,
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"metadata": {},
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"outputs": [
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{
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"data": {
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" <th></th>\n",
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" <th>Name</th>\n",
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" <th>Class</th>\n",
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" <th>Style</th>\n",
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" <th>Difficulty</th>\n",
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" <th>DamageType</th>\n",
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" <th>Damage</th>\n",
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" <th>Sturdiness</th>\n",
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" <th>Crowd-Control</th>\n",
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" <th>Mobility</th>\n",
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" <th>Functionality</th>\n",
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" </tr>\n",
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" </thead>\n",
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" <tbody>\n",
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" <tr>\n",
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" <th>23</th>\n",
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" <td>Ekko</td>\n",
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" <td>Mage-Warrior</td>\n",
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" <td>8</td>\n",
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" <td>3</td>\n",
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" <td>M</td>\n",
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" <td>3</td>\n",
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" <td>2</td>\n",
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" <td>2</td>\n",
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" <td>3</td>\n",
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" <td>NaN</td>\n",
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" </tr>\n",
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" </tbody>\n",
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"</table>\n",
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"</div>"
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],
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"text/plain": [
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" Name Class Style Difficulty DamageType Damage Sturdiness \\\n",
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"23 Ekko Mage-Warrior 8 3 M 3 2 \n",
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"\n",
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" Crowd-Control Mobility Functionality \n",
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"23 2 3 NaN "
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]
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},
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"execution_count": 5,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"Dat[Dat[\"Functionality\"].isnull()]"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"There are other issues regarding the quality of the data. For example, the type of the variables, several of them should be categorical.\n",
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"\n",
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"For now, we are going to keep the data set as it is, since these issues won't compromise our analysis."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 6,
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"Name object\n",
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"Class object\n",
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"Style int64\n",
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"Difficulty int64\n",
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"DamageType object\n",
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"Damage int64\n",
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"Sturdiness int64\n",
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"Crowd-Control int64\n",
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"Mobility int64\n",
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"Functionality float64\n",
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"dtype: object"
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]
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},
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"execution_count": 6,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"Dat.dtypes"
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]
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},
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{
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"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"Imagine that we select a champion at random (all of them with the same probability of being selected)."
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"---"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## Classical probability\n",
|
||||
"\n",
|
||||
"By classical probability, if we want to estimate the probability of the event $A$, we can do it as: $$P(A)=\\frac{1}{n}\\sum_{i=1}^n 1(\\text{observation}_i\\text{ satisfies A}).$$\n",
|
||||
"\n",
|
||||
"Which is just the average of $1(\\text{observation}\\text{ satisfies A})$."
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"What is the prob. of select a class warrior?"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 7,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
"text/plain": [
|
||||
"0.2138"
|
||||
]
|
||||
},
|
||||
"execution_count": 7,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
}
|
||||
],
|
||||
"source": [
|
||||
"round((Dat[\"Class\"]==\"Warrior\").mean(), 4)"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"---"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## Conditional Probability\n",
|
||||
"\n",
|
||||
"For conditional probabilities, we have, at least, two different ways to calculate them.\n",
|
||||
"\n",
|
||||
"For instance, if we want to calculate the probability of $P(A|B)$, we can do it as:\n",
|
||||
"\n",
|
||||
"1. Filter for the observations where the event $B$ happened. Then, calculate the probability of $A$ considering only those observations.\n",
|
||||
"2. Calculate it using the definition of conditional probability, i.e. $$P(A|B)=\\frac{P(A,B)}{P(B)}$$"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"What is the prob. of selecting a class warrior given that it is of difficulty 1?\n",
|
||||
"\n",
|
||||
"Let's do it first using the filtering approach:"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 8,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
"text/plain": [
|
||||
"0.2444"
|
||||
]
|
||||
},
|
||||
"execution_count": 8,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
}
|
||||
],
|
||||
"source": [
|
||||
"# Filter the observtions to consider only the champions where difficulty=1\n",
|
||||
"DatDifficult1 = Dat[Dat[\"Difficulty\"]==1]\n",
|
||||
"\n",
|
||||
"# Calculate the probability of selecting a class warrior, considering only the observations previosly filtered\n",
|
||||
"round((DatDifficult1[\"Class\"]==\"Warrior\").mean(), 4)"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"Using the definition:\n",
|
||||
"\n",
|
||||
"1. We need to calculate the join probability of selecting a class warrior and a difficulty equal 1: $$P(\\text{warrior and difficulty=1})$$\n",
|
||||
"2. We need to calculate the marginal probability of selecting a champion of difficulty 1 $$P(\\text{difficulty=1})$$\n",
|
||||
"3. We need to divide the former probability by the later: $$P(\\text{warrior|difficulty=1})=\\frac{P(\\text{warrior and difficulty=1})}{P(\\text{difficulty=1})}$$"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 9,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
"text/plain": [
|
||||
"0.2444"
|
||||
]
|
||||
},
|
||||
"execution_count": 9,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
}
|
||||
],
|
||||
"source": [
|
||||
"PWarriorAndDifficult1 = ((Dat[\"Class\"]==\"Warrior\") & (Dat[\"Difficulty\"]==1)).mean()\n",
|
||||
"PDifficult1 = (Dat[\"Difficulty\"]==1).mean()\n",
|
||||
"\n",
|
||||
"PWarriorGivenDifficult1 = PWarriorAndDifficult1 / PDifficult1\n",
|
||||
"\n",
|
||||
"round(PWarriorGivenDifficult1, 4)"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"---"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## Multiplication law or chain rule\n",
|
||||
"\n",
|
||||
"Consider that we want to calculate the probability of the events $A$, $B$, and $C$ at the same time, i.e. $P(A\\cap B\\cap C)$.\n",
|
||||
"\n",
|
||||
"The multiplication law, says that such probability can be calculated as: $$P(A\\cap B\\cap C)=P(A|B,C)P(B|C)P(C)$$\n",
|
||||
"\n",
|
||||
"Note that, it doesn't matter the order in which we condition."
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"What is the prob. of selecting a class warrior, difficulty 1, and style 4?\n",
|
||||
"\n",
|
||||
"First, let's calculate it using the deffinition of classical probability"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 10,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
"text/plain": [
|
||||
"0.0207"
|
||||
]
|
||||
},
|
||||
"execution_count": 10,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
}
|
||||
],
|
||||
"source": [
|
||||
"round(((Dat[\"Class\"]==\"Warrior\") & (Dat[\"Difficulty\"]==1) & (Dat[\"Style\"]==4)).mean(), 4)"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"Now, let's calculate it using the multiplication law:"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 11,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"name": "stdout",
|
||||
"output_type": "stream",
|
||||
"text": [
|
||||
"0.5\n",
|
||||
"0.46153846153846156\n",
|
||||
"0.0896551724137931\n"
|
||||
]
|
||||
}
|
||||
],
|
||||
"source": [
|
||||
"# Probability of being warrior given that it is difficulty 1 and style 4\n",
|
||||
"print(((Dat[\"Class\"]==\"Warrior\") & (Dat[\"Difficulty\"]==1) & (Dat[\"Style\"]==4)).mean() / ((Dat[\"Difficulty\"]==1) & (Dat[\"Style\"]==4)).mean())\n",
|
||||
"\n",
|
||||
"# Probability of difficulty 1 given style 4\n",
|
||||
"print(((Dat[\"Difficulty\"]==1) & (Dat[\"Style\"]==4)).mean() / (Dat[\"Style\"]==4).mean())\n",
|
||||
"\n",
|
||||
"# Probability of style 4\n",
|
||||
"print((Dat[\"Style\"]==4).mean())"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"Thus, the probability that we are looking for should be the product of these three probabilities"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 12,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
"text/plain": [
|
||||
"0.0207"
|
||||
]
|
||||
},
|
||||
"execution_count": 12,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
}
|
||||
],
|
||||
"source": [
|
||||
"round(0.5 * 0.46153846153846156 * 0.0896551724137931, 4)"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"---"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## Law of total probability\n",
|
||||
"\n",
|
||||
"Law of total probability stays that, if $B_1,B_2,B_3,\\ldots$ form a partition of $\\Omega$, i.e. $\\cup_{i}B_i=\\Omega$ and $B_i\\cap B_j=\\emptyset$ for all $i\\neq j$. Then, $$P(A)=\\sum_iP(A|B_i)P(B_i)=\\sum_i P(A,B_i)$$"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"What is the prob. of selecting a warrior?\n",
|
||||
"\n",
|
||||
"First, let's calculate this probability using classical probability"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 13,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
"text/plain": [
|
||||
"0.2138"
|
||||
]
|
||||
},
|
||||
"execution_count": 13,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
}
|
||||
],
|
||||
"source": [
|
||||
"round((Dat[\"Class\"]==\"Warrior\").mean(), 4)"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"Now, let's calculate it using law of total probability.\n",
|
||||
"\n",
|
||||
"Note that, every champion should be of difficulty 1, 2, or 3. So the events $B_i:$ the champion is of difficulty $i$, form a partition of $\\Omega$.\n",
|
||||
"\n",
|
||||
"Thus, the probability of selecting a warrior can be calculate it as: $$P(\\text{warrior})=P(\\text{warrior and difficulty 1}) + P(\\text{warrior and difficulty 2}) + P(\\text{warrior and difficulty 3})$$"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 14,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"name": "stdout",
|
||||
"output_type": "stream",
|
||||
"text": [
|
||||
"0.07586206896551724\n",
|
||||
"0.09655172413793103\n",
|
||||
"0.041379310344827586\n"
|
||||
]
|
||||
}
|
||||
],
|
||||
"source": [
|
||||
"print(((Dat[\"Class\"]==\"Warrior\") & (Dat[\"Difficulty\"]==1)).mean())\n",
|
||||
"print(((Dat[\"Class\"]==\"Warrior\") & (Dat[\"Difficulty\"]==2)).mean())\n",
|
||||
"print(((Dat[\"Class\"]==\"Warrior\") & (Dat[\"Difficulty\"]==3)).mean())"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 15,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
"text/plain": [
|
||||
"0.2138"
|
||||
]
|
||||
},
|
||||
"execution_count": 15,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
}
|
||||
],
|
||||
"source": [
|
||||
"round(0.07586206896551724 + 0.09655172413793103 + 0.041379310344827586, 4)"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"---"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## Bayes' theorem"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"What is the probability of selecting a warrior given that it is of difficulty 1?\n",
|
||||
"\n",
|
||||
"We already calculated that probability previously."
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 16,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
"text/plain": [
|
||||
"0.24444444444444444"
|
||||
]
|
||||
},
|
||||
"execution_count": 16,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
}
|
||||
],
|
||||
"source": [
|
||||
"PWarriorGivenDifficult1"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"We are going to calculate it again. But this time, we are going to use Bayes' theorem.\n",
|
||||
"\n",
|
||||
"Accordign to Bayes' theorem, the probability that we are looking for, can be calculated as:\n",
|
||||
"\n",
|
||||
"$$P(\\text{Warrior|Difficulty=1})=\\frac{P(\\text{Difficulty=1|Warrior})P(\\text{Warrior})}{P(\\text{Difficulty=1})}$$"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 17,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
"text/plain": [
|
||||
"0.2444"
|
||||
]
|
||||
},
|
||||
"execution_count": 17,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
}
|
||||
],
|
||||
"source": [
|
||||
"PDifficult1GivenWarrior = ((Dat[\"Class\"]==\"Warrior\") & (Dat[\"Difficulty\"]==1)).mean() / (Dat[\"Class\"]==\"Warrior\").mean()\n",
|
||||
"PWarrior = (Dat[\"Class\"]==\"Warrior\").mean()\n",
|
||||
"PDifficult1 = (Dat[\"Difficulty\"]==1).mean()\n",
|
||||
"\n",
|
||||
"round(PDifficult1GivenWarrior * PWarrior / PDifficult1, 4)"
|
||||
]
|
||||
}
|
||||
],
|
||||
"metadata": {
|
||||
"kernelspec": {
|
||||
"display_name": "Python 3",
|
||||
"language": "python",
|
||||
"name": "python3"
|
||||
},
|
||||
"language_info": {
|
||||
"codemirror_mode": {
|
||||
"name": "ipython",
|
||||
"version": 3
|
||||
},
|
||||
"file_extension": ".py",
|
||||
"mimetype": "text/x-python",
|
||||
"name": "python",
|
||||
"nbconvert_exporter": "python",
|
||||
"pygments_lexer": "ipython3",
|
||||
"version": "3.12.5"
|
||||
}
|
||||
},
|
||||
"nbformat": 4,
|
||||
"nbformat_minor": 2
|
||||
}
|
||||
Binary file not shown.
Loading…
Reference in New Issue