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I'm trying to solve a question that asks:

If 5 pairs of fair dice are rolled, approximate the probability that the sum of the values obtained is between 30 and 40 inclusive.

My approach so far, was to create a random variable called $T$ whose range is from $10$ to $60$ and look for $$P\{|T-35| \le 5\} $$

I don't know what to do next. Can I proceed like this is a normal distribution or is five too small a sample to qualify for the Central Limit Theorem?

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You were asked to approximate the probability. If you use the normal distribution, you will certainly get an approximation of the probability. It is not immediately clear to me how good that approximation will be, but on the other hand, you weren't told how good an approximation you were expected to get. If I had to answer such a poorly specified question on a homework assignment, I would pick some number between 0 and 1 and present it as my approximation, and dare the instructor to establish that my answer was wrong. You may not be as combative as I am. – Gerry Myerson Nov 14 '12 at 2:44
See review problem 6.2 here: – Benjamin Dickman Nov 14 '12 at 4:57
Are the variance and mean of 10 dice the same as the variance and mean of 5 pairs of 2 dice? Does anything need to be added to the calculation? – CodyBugstein Nov 14 '12 at 15:43
TO ANSWER MY OWN QUESTION; Yes it is. I just tested it. – CodyBugstein Nov 14 '12 at 16:10
@B.D Take a look at the question again... how did $ \frac{30-10 \frac{7}{2}}{\sqrt{10} \sqrt{\frac{35}{12}}}$ become $.38$? Is that a miscalculation? – CodyBugstein Nov 15 '12 at 16:41
up vote 0 down vote accepted

I just made a spreadsheet to calculate it and get $\frac {41539796}{60466176}\approx 0.687$

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I got $.6456$. How do you get all those numbers? It should just be $\frac{40-35}{\sqrt{\frac{350}{12}}}$ and $\frac{30-35}{\sqrt{\frac{350}{12}}}$ – CodyBugstein Nov 15 '12 at 21:10
@Imray: this is the exact answer. I added up how many ways to get numbers from 30 to 40, and divided by 6^10. The disagreement shows the error in the approximation. – Ross Millikan Nov 15 '12 at 22:09
That must have taken you a long time! Are you sure you got every possible way? – CodyBugstein Nov 15 '12 at 22:19
@Imray: took a couple minutes. I made a column with 1 to 60 for the sum. The next ten columns were the number of ways to make that sum with 1 through 10 dice. In the column for 1 die, I put 1 in 1 through 6. Then in each successive column, a cell added the six cells above it and one column to the right (a little perturbation at the small end). Fill down/fill right, then sum the appropriate cells in the 10 dice column. – Ross Millikan Nov 15 '12 at 22:35

This can be determined exactly with the use of generating functions (also known as probability generating functions) and a computer algebra system (like GP/PARI).

The probability is the sum of the coefficients of the terms in the polynomial $$ \left(\frac{1}{6}x+\frac{1}{6}x^2+\frac{1}{6}x^3+\frac{1}{6}x^4+\frac{1}{6}x^5+\frac{1}{6}x^6\right)^{10} $$ with degree 30 through 40, which is the sum of the coefficients of the terms in the polynomial $$ (x+x^2+x^3+x^4+x^5+x^6)^{10} $$ with degree 30 through 40, divided by $6^{10}$.

Using GP, this is calculated instantly as


which gives the result $\frac{10384949}{15116544} \approx 0.68699227812918084$.

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My course hasn't covered generating functions, but can you explain how you got $(x + x^2 + .... )$ etc, where did that come from? – CodyBugstein Dec 3 '12 at 15:35
I added in an extra expression to make it clearer. The polynomial $P=\frac{1}{6}x+\frac{1}{6}x^2+...+\frac{1}{6}x^6$ is the probability generating function of a fair 6-sided die: the coefficient on $x^i$ is the probability of face $i$ appearing on a single roll. Then, the distribution of sums of $n$ such dice is given via the coefficients of $P^n$. You can read more about it on the Wikipedia page I linked to in my answer. – Matthew Conroy Dec 4 '12 at 5:28

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