Thread: 3 dice probability (already solved but still having a hard time understanding)

1. 3 dice probability (already solved but still having a hard time understanding)

with a throw of 3 dice, what is the probability of getting a 9 or an 11?

Here is the solution posted

I am having really great difficulty in understanding what all the table parameters mean. I also don't know how to get those values on the table. maybe you can guide me?
$\displaystyle S_{12}$ = sum of outcomes in dice 1 and dice 2
$\displaystyle P_{1\&2}$ = number of permutations in $\displaystyle S_{12}$
$\displaystyle O_3$ = outcome of dice 3 to make 9 or 11
$\displaystyle P_3$ = Number of permutations each of $\displaystyle O_3$
$\displaystyle P = (P_{1\&2})(P_{3})$ = multiplying both

I have a background knowledge on permutation which is nPr where n = total ways , r = number of ways the total can be permutated

2. Re: 3 dice probability (already solved but still having a hard time understanding)

Originally Posted by TechnicianEngineer
with a throw of 3 dice, what is the probability of getting a 9 or an 11?
Here is the solution posted

I am having really great difficulty in understanding what all the table parameters mean. I also don't know how to get those values on the table. maybe you can guide me?
$\displaystyle S_{12}$ = sum of outcomes in dice 1 and dice 2
$\displaystyle P_{1\&2}$ = number of permutations in $\displaystyle S_{12}$
$\displaystyle O_3$ = outcome of dice 3 to make 9 or 11
$\displaystyle P_3$ = Number of permutations each of $\displaystyle O_3$
$\displaystyle P = (P_{1\&2})(P_{3})$ = multiplying both
I have a background knowledge on permutation which is nPr where n = total ways , r = number of ways the total can be permutated
Look at this webpage. Scroll down to the expansion.

If you add the coefficients of $\displaystyle x^9~\&~x^{11}$ you get $\displaystyle 52$.

The coefficient of $\displaystyle x^k$ tells us the number of ways to get a sum of $\displaystyle k$.

3. Re: 3 dice probability (already solved but still having a hard time understanding)

The table simply sets out to count the number of successful outcomes, a final total of 9 or 11.
Adding together the number of pips on two thrown dice, the total can be anything from 2 (two ones) through to 12 (two sixes).
These are listed down the first column.
The second column lists the number of ways in which each total can be achieved, (I think that using the word permutation is a bit of overkill).
So, using a two dice total of 5 as an example, a total of 5 could be arrived at in 4 ways, 4-1, 3-2, 2-3, 1-4.
The third column then lists the 'useful' throws with the third die. For a total of 5 with the first two dice, that would be a 4 taking the total to 9 or a 6 taking the total to 11.
The fourth column then gives the number of these 'useful' throws, 2 in this case.
Finally, the number in the fifth column is the product of the numbers in the second and fourth columns. (With each of the four ways of throwing a two dice total of 5 there are two 'successful' throws with the third die.)
The total number of successful outcomes is then the sum of the numbers down the fifth column.
Incidently, there are two incorrect entries in the table, in the third column the second and third entries should be reversed.