# Dice Probability

• Nov 27th 2006, 05:04 PM
feiyingx
Dice Probability
Hi. Is there a way to solve the following without actually listing out all the possibilities?

What is the probability of rolling a total sum of 8 using three dice?

Thank you! :)
• Nov 27th 2006, 06:44 PM
galactus
Yes, there is a way.

There are 3 dice and you want to find all the sums of 8.

$\left(\sum_{k=1}^{6}x^{k}\right)^{3}$

Expand this out and check the coefficient of $x^{8}$

That is the number of ways to sum to 8.

The coefficent is $21x^{8}$

Therefore, there are 21 ways.

There are $6^{3}=216$ possible rolls with 3 dice.

Probability is $\frac{21}{216}$
• Nov 27th 2006, 09:00 PM
feiyingx
Wow! Ingenious xD.
Now can u explain why that is? I don't see it.

Thanks again!
• Nov 28th 2006, 03:52 AM
galactus
A text on advanced counting methods and/or generating functions may prove useful if you are curious of how it's derived.
• Nov 28th 2006, 08:51 AM
feiyingx
Thank you. =]
• Dec 18th 2006, 06:42 PM
F.A.P
Quote:

Originally Posted by galactus
A text on advanced counting methods and/or generating functions may prove useful if you are curious of how it's derived.

This is a more direct approach for those not familiar with generating functions.
Let Z be the sum of three dice. Let Y be the sum of the first two dice.

$P(Z=8)=\sum_{k=2}^{12}P(Z=8|Y=k)P(Y=k)$

Now

$P(Z = 8|Y = k) = 1/6$, for k = 2, 3,...,7, and 0 elsewhere. Also it's easy to show that

$P(Y=k)=\frac{k-1}{6^{2}}$, for k = 2, 3,...,7 (we won't consider higher k's)

Hence we end up with

$P(Z=8)=\sum_{k=2}^{7}\frac{1}{6}\frac{k-1}{6^{2}}=\frac{21}{6^{3}}=\frac{21}{216}$