# Thread: Question on rolling dice.

1. ## Question on rolling dice.

I see the chart/table on rolling 2 dice, the outcomes.

If when it asks for doubles, is it just 6 ways or 12. Because if you used a total of 6, you can have(4,2)(2,4)exc. So that means for each count of a sum of die, you double it correct? Like I said for doubles, if you have 6,6, it is reversed as well as 6,6. Which I know looks weird.

Any help would be great, to get the wording down correctly.
Jo

2. ## Re: Question on rolling dice.

I see the chart/table on rolling 2 dice, the outcomes.
If when it asks for doubles, is it just 6 ways or 12. Because if you used a total of 6, you can have(4,2)(2,4)exc. So that means for each count of a sum of die, you double it correct? Like I said for doubles, if you have 6,6, it is reversed as well as 6,6. Which I know looks weird.
There is but one way to roll two six's. But two ways to roll a 2 and a 4.

3. ## Re: Question on rolling dice.

Hello, Jo!

Consider this problem . . .

A red die and a blue die are rolled.
How many outcomes have a sum of 6?

We can list the solutions:

. . $\displaystyle \begin{array}{cccccc}{\color{red}1} & {\color{blue}5} && \text{red 1, blue 5} \\ {\color{red}2} & {\color{blue}4} && \text{red 2, blue 4} \\ {\color{red}3}&{\color{blue}3} && \text{red 3, blue 3} & \text{(both are 3's)} \\ {\color{red}4}&{\color{blue}2}&& \text{red 4, blue 2} \\ {\color{red}5}&{\color{blue}1} && \text{red 5, blue 1} \end{array}$

There are five solutions.

But you might claim that there are six solutions
. . by insisting that: .$\displaystyle {\color{blue}3}\;\;{\color{red}3} \quad \text{blue 3, red 3} \quad\text{(both are 3's)}$
. . is yet another solution.

Can you see that this is wrong?