Dice Probability

**feiyingx** · November 27th 2006, 04:04 PM

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!

**galactus** · November 27th 2006, 05:44 PM

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 $\frac{21}{216}$

**feiyingx** · November 27th 2006, 08:00 PM

Wow! Ingenious xD.
Now can u explain why that is? I don't see it.

Thanks again!

**galactus** · November 28th 2006, 02:52 AM

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

**feiyingx** · November 28th 2006, 07:51 AM

Thank you. =]

**F.A.P** · December 18th 2006, 05:42 PM

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

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

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