# Thread: Dice and Normal Distribution

1. ## Dice and Normal Distribution

Hey, I was trying to model a system that would calculate the probabilities of moves in a game, but was stumped by one thing so I thought I'd turn it into an exam type question:

If Julie has 4, 8 sided dice; and Hasish has 3, 8 sided dice, what is the probability that when they both roll their dice, the sum of Julie's dice is greater?

If I roll $\displaystyle x$, 8 sided dice; what is the probability that the sum of $\displaystyle y$ 8 sided dice is greater?

^^I'm hoping there isn't a ridiculously simple answer to it, but there may well be as I'm knackered at the moment.

Also I was wondering, does anyone have any real life applications of the normal (or Gaussian) distribution, and what type of things can be modelled with it?

Thanks!

2. Hi Nyoxsis,

The easiest way I know to answer your question about the dice is to use a normal approximation! The answer is that the probability that Julie’s sum is greater is about 0.745.

First let’s consider the result of rolling one 8-sided die. Let’s say the result is U. U has a “discrete uniform distribution” (see Uniform distribution (discrete) - Wikipedia, the free encyclopedia) so its mean is (1+8)/2 = 4.5 and its variance is (8^2 – 1) / 12 = 5.25.

Julie’s roll, say J, is the sum of 4 of the U random variables, so the mean of J is 4 * 4.5 = 18 and its variance is 4 * 5.25 = 21.

Hasish’s roll, say H, is the sum of 3 of the U random variables, so the mean of H is 3 * 4.5 = 13.5 and its variance is 3 * 5.25 = 15.75.

Define Y = J-H. We are interested in Pr(Y > 0). The exact distribution of Y is hard to work with, so let’s try a normal approximation. This seems reasonable because Y is the sum of several independent random variables (the U’s or their negatives). We know the mean of Y is 18 – 13.5 = 4.5 and its variance is 21 + 15.75 = 36.75, so its standard deviation is sqrt(36.75) = 6.06. Define Z = (Y – 4.5) / 6.06; then Z has mean 0 and standard deviation 1, so if we assume Z has a normal distribution (it doesn’t really, but it’s close) then we can look up the cumulative distribution in a table. Here we use a little trick to improve the accuracy of the approximation: we want to know Pr(Y>0), which is the same as Pr(Z > -4.5/6.06). But the distribution of Y is discrete and the distribution of Z is continuous, so let’s consider Pr(Y > 0.5) instead. This trick is so common that it has a name, the “correction for continuity”. Pr(Y > 0.5) = Pr(Z > -4.0/6.06 = -0.660), and looking up the cumulative normal distribution in a table we find this probability is 0.745. That’s the answer to your problem, approximately.

You might ask how accurate is the approximation. I used more advanced methods involving generating functions to find the exact distribution of Y and found that Pr(Y>0) = 1,554,908 / 8^7, which is about 0.7414. This would have been very tedious and time-consuming to find except that I used a computer algebra system to do the hard work. So in this case the normal approximation is pretty good, with a relative error of only about 0.5%.

jw

3. That's ingenious, thanks!

But I don't understand why you added the variance instead of subtracting it, Im sure it makes perfect logical sense, but I'm tired atm.

Yeah, thanks again, brilliant work- I'll get started on something for n and p dice. =)

4. Originally Posted by Nyoxis
That's ingenious, thanks!

But I don't understand why you added the variance instead of subtracting it, Im sure it makes perfect logical sense, but I'm tired atm.

Yeah, thanks again, brilliant work- I'll get started on something for n and p dice. =)
Theorem: Var(aX + bY) = a^2 Var(X) + b^2 Var(Y) + 2ab Cov(X, Y).

If X and Y are independent, Cov(X, Y) = 0 and Var(aX + bY) = a^2 Var(X) + b^2 Var(Y).

In particular, if X and Y are independent, a = 1 and b = -1: Var(X - Y) = Var(X) + Var(Y).

By the way, it makes perfect sense that you DON'T subtract variances - otherwise you could have a distribution with zero variance. Would that make sense .....?

5. Many thanks mr.f, it makes a little more sense now, although we haven't looked into covariance yet, if anyone was interested in how you would apply this work then take a look here: Unlock the Mines of Moria.

6. Originally Posted by Nyoxis
Many thanks mr.f, it makes a little more sense now, although we haven't looked into covariance yet, if anyone was interested in how you would apply this work then take a look here: Unlock the Mines of Moria.
Aha I might have guessed!

By the way, regarding the distribution of sums of random variables,

http://www.dartmouth.edu/~chance/tea...k/Chapter7.pdf

is of background interest (it even has an example of the sum of two uniform distributions).