The Box-Muller transformation of two uniform random variables actually gives a true normal distribution–provided you have a decent way of generating random numbers in the interval from 0 to 1. It’s a pretty elegant trick based on a polar transformation of a bivariate normal distribution: if X and Y are uniformly distributed random variables in [0,1], then

Z = sqrt(-2*ln(X))*cos(2*pi*Y)

is a standard normal random variable. The Mathworld article is here:
http://mathworld.wolfram.com/Box-MullerTransformation.html. And here is an online calculator that produces normally distributed random variables (with links to other random generators): http://www.had2know.com/academics/gaussian-normal-random-generator.html

I ran into your webpage and I have a question, I hope you can help.
I am trying to find an equation that describe the probability of the following event:

We have 10 cards (nmbered from 1 to 10) and these cards are shuffled and then leid out in random order. ie the first time we could get 3, 6, 5, 7, 8, 10, 9, 2, 1, 4 and and different order for the second time. Each card is then given a rank according to the order in which they appear (in the example above, card 3 is 1st).
If one does that a number of times (eg 100), one can see that the average rank of each card has a mean of 5.5 and the distribution of these means is normal. The shape of the normal distribution gets narrower and narrowed the more events are carried out. There must be a function that describes this normal distrbution based on the number of cards and the number of shuffling events. I hope you can help, and /or maybe direct us to a reference (book or article) as a source of information.

Imagine the density for the dice. Half of the probability is for rolls between 5 and 17, the other half between 18 and 30. So the CDF of the discrete dice distribution reaches 1/2 at 17, but the CDF of the continuous normal distribution reaches 1/2 at 17.5. Or think about the end. The CDF of the discrete distribution reaches 1 at 30, but the CDF of the normal is only 1 in the limit. So the discrete distribution is a little bit ahead of the continuous distribution.

I tried this with the K-S test, but unfortunately the fact that the dice generator only has 26 possible values throws the test off: samples from a continuous distribution are not supposed to have repeats.

Another point probably worth mentioning is that the values generated by this method cannot exceed 6 in absolute value, even in theory. But the theoretical probability of a Normal r.v. being outside of (- 6,6) is about 2 per thousand million. And having bounded values is not necessarily bad.

The variance of a uniform[0,1] random variable is 1/12. So adding 12 together makes the variance 1. That’s why his trick produces a standard random sample.