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---
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
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- /2017/03/uniformity-math-random/
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---
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.
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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 <code class='language-javascript'>Math.random()</code> between 1 and 1000 inclusive and calls the function 1,000,000 times. Storing it&#8217;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
```bash
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grep -oE '[0-9]+' Random.csv &gt; 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&#8217;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
<u>Counted Data Condition:</u> The data can be converted from quantatative to count data.
<u>Independence Assumption:</u> One random value does not affect another.
<u>Expected Cell Frequency Condition:</u> 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
```R
rm(list=ls())
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chrome = read.csv("~/Chrome_corrected.csv", header = F)
firefox = read.csv("~/Firefox_corrected.csv", header = F)
ie11 = read.csv("~/IE11_corrected.csv", header = F)
```
```R
hist(ie11$V1, main = "Distribution of Random Values for IE11", xlab = "Random Value")
```
![](https://brandonrozek.com/wp-content/uploads/2017/03/ie11hist.png)
```R
hist(firefox$V1, main = "Distribution of Random Values for Firefox", xlab = "Random Value")
```
![](https://brandonrozek.com/wp-content/uploads/2017/03/firefoxhist.png)
```R
hist(chrome$V1, main = "Distribution of Random Values for Chrome", xlab = "Random Value")
```
![](https://brandonrozek.com/wp-content/uploads/2017/03/chromehist.png)
## Chi-Square Test
Before we run our test, we need to convert the quantatative data to count data by using the plyr package
```R
#Transform to count data
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library(plyr)
chrome_count = count(chrome)
firefox_count = count(firefox)
ie11_count = count(ie11)
```
Run the tests
```R
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# Chi-Square Test for Goodness-of-Fit
chrome_test = chisq.test(chrome_count$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.
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##
## Chi-squared test for given probabilities
##
## data: chrome_count$freq
## X-squared = 101.67, df = 99, p-value = 0.4069
`firefox_test`
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##
## Chi-squared test for given probabilities
##
## data: firefox_count$freq
## X-squared = 105.15, df = 99, p-value = 0.3172
`ie11_test`
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##
## 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.