--- id: 2095 title: Uniformity of Math.random() date: 2017-03-07T21:50:52+00:00 author: Brandon Rozek layout: post guid: https://brandonrozek.com/?p=2095 permalink: /2017/03/uniformity-math-random/ medium_post: - 'O:11:"Medium_Post":11:{s:16:"author_image_url";N;s:10:"author_url";N;s:11:"byline_name";N;s:12:"byline_email";N;s:10:"cross_link";N;s:2:"id";N;s:21:"follower_notification";N;s:7:"license";N;s:14:"publication_id";N;s:6:"status";N;s:3:"url";N;}' mf2_syndicate-to: - 'a:1:{i:0;s:4:"none";}' mf2_cite: - 'a:4:{s:9:"published";s:25:"0000-01-01T00:00:00+00:00";s:7:"updated";s:25:"0000-01-01T00:00:00+00:00";s:8:"category";a:1:{i:0;s:0:"";}s:6:"author";a:0:{}}' tumblr_post_id: - "158123669889" format: aside kind: - note --- There are many cases where websites use random number generators to influence some sort of page behavior. One test to ensure the quality of a random number generator is to see if after many cases, the numbers produced follow a uniform distribution. Today, I will compare Internet Explorer 11, Chrome, and Firefox on a Windows 7 machine and report my results. ## Hypothesis H0: The random numbers outputted follow the uniform distribution HA: The random numbers outputted do not follow the uniform distribution ## Gathering Data I wrote a small [website](http://share.zeropointshift.com/files/2017/03/random.html) and obtained my data by getting the CSV outputted when I use IE11, Firefox, and Chrome. The website works by producing a random number using Math.random() between 1 and 1000 inclusive and calls the function 1,000,000 times. Storing it’s results in a file This website produces a file with all the numbers separated by a comma. We want these commas to be replaced by newlines. To do so, we can run a simple command in the terminal

grep -oE '[0-9]+' Random.csv > Random_corrected.csv
Do this with all three files and make sure to keep track of which is which. Here are a copy of my files for [Firefox](https://brandonrozek.com/wp-content/uploads/2017/03/Firefox_corrected.csv), [Chrome](https://brandonrozek.com/wp-content/uploads/2017/03/Chrome_corrected-1.csv), and [IE11](https://brandonrozek.com/wp-content/uploads/2017/03/IE11_corrected.csv) ## Check Conditions Since we’re interested in if the random values occur uniformly, we need to perform a Chi-Square test for Goodness of Fit. With every test comes some assumptions Counted Data Condition: The data can be converted from quantatative to count data. Independence Assumption: One random value does not affect another. Expected Cell Frequency Condition: The expected counts are going to be 10000 Since all of the conditions are met, we can use the Chi-square test of Goodness of Fit ## Descriptive Statistics For the rest of the article, we will use R for analysis. Looking at the histograms for the three browsers below. The random numbers all appear to occur uniformly
rm(list=ls())
chrome = read.csv("~/Chrome_corrected.csv", header = F)
firefox = read.csv("~/Firefox_corrected.csv", header = F)
ie11 = read.csv("~/IE11_corrected.csv", header = F)

hist(ie11$V1, main = "Distribution of Random Values for IE11", xlab = "Random Value")
![](https://brandonrozek.com/wp-content/uploads/2017/03/ie11hist.png)
hist(firefox$V1, main = "Distribution of Random Values for Firefox", xlab = "Random Value")
![](https://brandonrozek.com/wp-content/uploads/2017/03/firefoxhist.png)
hist(chrome$V1, main = "Distribution of Random Values for Chrome", xlab = "Random Value")
![](https://brandonrozek.com/wp-content/uploads/2017/03/chromehist.png) ## Chi-Square Test Before we run our test, we need to convert the quantatative data to count data by using the plyr package
#Transform to count data
library(plyr)
chrome_count = count(chrome)
firefox_count = count(firefox)
ie11_count = count(ie11)
Run the tests

# Chi-Square Test for Goodness-of-Fit
chrome_test = chisq.test(chrome_count$freq)
firefox_test = chisq.test(firefox_count$freq)
ie11_test = chisq.test(ie11_count$freq)

# Test results
chrome_test
As you can see in the test results below, we fail to reject the null hypothesis at a 5% significance level because all of the p-values are above 0.05. ## ## Chi-squared test for given probabilities ## ## data: chrome_count$freq ## X-squared = 101.67, df = 99, p-value = 0.4069
firefox_test
## ## Chi-squared test for given probabilities ## ## data: firefox_count$freq ## X-squared = 105.15, df = 99, p-value = 0.3172
ie11_test
## ## Chi-squared test for given probabilities ## ## data: ie11_count$freq ## X-squared = 78.285, df = 99, p-value = 0.9384 ## Conclusion At a 5% significance level, we fail to obtain enough evidence to suggest that the distribution of random number is not uniform. This is a good thing since it shows us that our random number generators give all numbers an equal chance of being represented. We can use Math.random() with ease of mind.