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| id | title | date | author | aliases | permalink | medium_post | mf2_syndicate-to | mf2_cite | format | tags | |||||
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| 2095 | Uniformity of Math.random() | 2017-03-07T21:50:52+00:00 | Brandon Rozek |
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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.
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 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, Chrome, and IE11
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$V1, main = "Distribution of Random Values for IE11", xlab = "Random Value")
hist(firefox$V1, main = "Distribution of Random Values for Firefox", xlab = "Random Value")
hist(chrome$V1, main = "Distribution of Random Values for Chrome", xlab = "Random Value")
Chi-Square Test
Before we run our test, we need to convert the quantatative data to count data by using the plyr package
#Transform to count data
library(plyr)
chrome_count = count(chrome)
firefox_count = count(firefox)
ie11_count = count(ie11)
Run the tests
# Chi-Square Test for Goodness-of-Fit
chrome_test = chisq.test(chrome_count$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.