# Dice Probability

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• Feb 4th 2011, 06:17 PM
rtblue
Dice Probability
Hey guys, here is my question:

Greg rolls 3 six-sided dice. Find the probability that the sum of the 3 dice is less than 9.

I realize that one possible way to go about this, would be to list all the cases in which the sum is less than 9, then divide by the total number of cases, but this seems too time consuming. Does anyone know of a more efficient way to go about this problem?
• Feb 4th 2011, 06:30 PM
mr fantastic
Quote:

Originally Posted by rtblue
Hey guys, here is my question:

Greg rolls 3 six-sided dice. Find the probability that the sum of the 3 dice is less than 9.

I realize that one possible way to go about this, would be to list all the cases in which the sum is less than 9, then divide by the total number of cases, but this seems too time consuming. Does anyone know of a more efficient way to go about this problem?

Well, that's probably the most efifcient approach.

You could define the random variable V = W + X + Y, calculate it's pmf and then use this to calculate Pr(V < 9) but I doubt you would find this easier.
• Feb 4th 2011, 06:37 PM
rtblue
Thanks Mr. Fantastic, I better start counting..
• Feb 4th 2011, 07:38 PM
CaptainBlack
Quote:

Originally Posted by rtblue
Hey guys, here is my question:

Greg rolls 3 six-sided dice. Find the probability that the sum of the 3 dice is less than 9.

I realize that one possible way to go about this, would be to list all the cases in which the sum is less than 9, then divide by the total number of cases, but this seems too time consuming. Does anyone know of a more efficient way to go about this problem?

The mean is 10.5, and the distribution is symmetric about the mean. Therefore the probability of 9 or fewer is 0.5 minus the probability of exactly 10.

CB
• Feb 4th 2011, 08:16 PM
Wilmer
From 1 to 18:
0,0,1,3,6,10,15,21,25,27,27,25,21,15,10,6,3,1 ; total 256
• Feb 5th 2011, 03:58 AM
Plato
Here is a web-tool to help with this question.
Look at the coefficients of the x’s.
The coefficient of \$\displaystyle x^3\$ is 1. There is one way to roll 3.
The coefficient of \$\displaystyle x^8\$ is 21. There are twenty-one ways to roll 8.

If you change the 3 to 4 on the whole sum, you get the ways to roll four dice.