# Thread: Dice Toss Problem

1. ## Dice Toss Problem

Hi everyone,

I have the solution to a problem, but I am nore sure how the answer was arrived at, or whether it is 100% correct. The problem involves tossing 4 dice, and finding the probability of the sum being 10.

The solution to this problem is ((9 choose 3)-4) / 6^4.

Any ideas as to how this solution was derived? Any insight is appreciated.

2. ## Re: Dice Toss Problem

Originally Posted by KelvinScale
The problem involves tossing 4 dice, and finding the probability of the sum being 10.
There several ways to get a ten.
$\displaystyle \begin{gathered} 1234 \hfill \\ 1126 \hfill \\ 1225 \hfill \\ 2224 \hfill \\ 1333 \hfill \\ 1144 \hfill \\ 1135 \hfill \\ 2233 \hfill \\ \end{gathered}$
Now you have to count all of those.
See if I missed any.

POST SCRIPT.
Here is a neat trick.
Look at this webpage.
Look for the coefficient of $\displaystyle x^{10}$.
That tells us how many ways the the four dice add to 10.

3. ## Re: Dice Toss Problem

Hello, KelvinScale!

I have the solution to a problem, but I am not sure how the answer was arrived at,
or whether it is 100% correct.
The problem involves tossing 4 dice, and finding the probability of the sum being 10.

The solution to this problem is: .$\displaystyle \frac{{9\choose3} - 4}{6^4}$

Any ideas as to how this solution was derived?
Any insight is appreciated.

I have a theory . . . see if it makes sense.

Place ten objects in a row, leaving a space between each pair.
. . $\displaystyle \circ\;\circ\;\circ\;\circ\;\circ\;\circ\;\circ$-$\displaystyle \circ\;\circ\;\circ$

Select 3 of the 9 spaces and insert "dividers".

So that: .$\displaystyle \circ\;\circ\;\circ\:|\:\circ\:\circ\:|\:\circ\:| \:\circ\;\circ\;\circ\;\circ$ . represents the roll [3,2,1,4]

There are: .$\displaystyle {9\choose3}$ ways to place the dividers.

But these include the division: .$\displaystyle \circ\:|\:\circ\:|\:\circ\:|\:\circ\;\circ$-$\displaystyle \circ\;\circ\;\circ\;\circ\;\circ$
. . which represents [1,1,1,7] which is not permitted.

There are four such divisions: [1,1,1,7], [1,1,7,1], [1,7,1,1], and [7,1,1,1].

Hence, there are: .$\displaystyle {9\choose3} - 4$ .ways to get a sum of ten on four dice.

And there are: .$\displaystyle 6^4$ possible outcomes.

Therefore: .$\displaystyle P(\text{sum of 10}) \:=\:\frac{{9\choose3} - 4}{6^4}$

4. ## Re: Dice Toss Problem

And Plato's approach is absolutely correct!

. . $\displaystyle \begin{array}{cc} \text{Combination} & \text{Permutations} \\ \hline \\[-4mm]1234 & 4! \:=\:24 \\ 1126 & \frac{4!}{2!}\:=\:12 \\ \\[-4mm] 1225 & \frac{4!}{2!} \:=\:12 \\ \\[-4mm] 2224 & \frac{4!}{3!} \:=\:\;4 \\ \\[-4mm] 1333 & \frac{4!}{3!} \:=\:\;4 \\ \\[-4mm] 1144 & \frac{4!}{2!2!} \,=\:6 \\ \\[-4mm] 1135 & \frac{4!}{2!} \:=\:12 \\ \\[-4mm] 2233 & \frac{4!}{2!2!} \,=\:\:6 \\ \\[-4mm] \hline & \qquad\quad 80\end{array}$

Therefore: .$\displaystyle P(\text{sum of 10}) \:=\:\frac{80}{6^4}$