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Message: <blockquote data-quote="ichabod" data-source="post: 1415483" data-attributes="member: 1257">This is actually incorrect. The chi-squared test for goodness of fit is good for saying that a random set of numbers is random, but it's not so good at noticing that a non-random set of numbers is non-random. If you use excel to make a whole slew of random numbers, and then subtract the nth random number from the random number after it, you will expect a certain distribution. Say you do random numbers from 1 to 10. There are 100 possible pairs of numbers. 1% of them should difference to -9, as 1 and 10 is the only combination that will do so. 2% of them should difference to -8, as (1,9) and (2,10) are the only two combinations that will give you -8. And so on. If you check the distribution of the differenced values with the chi-squared test, it fails the test. This is especially significant given how much trouble the chi-squred test has detecting a bad distribution.What does this mean? This means that each roll is highly dependent on what the last roll was. Not too surprising giving that Excel uses an itterated function to make the random numbers, but it is a definite weakness of such random numbers.</blockquote>