Probability of one dice being twice the value of the other

Okay, so I know this is probably a pretty easy question but I can't figure out how to do it without writing out all the possibilities and counting them up.

"Given that two fair dice are rolled, what is the probability of one of the die being twice that of the other?"

So obviously the answer is 6/36. When you write down all the combinations where this will be true, there are 6 of them.

But I don't think the question intended me to simply enumerate the outcomes that met the condition so, I have been trying to figure out some more (for want of a better word) maths-y way of getting the answer. Any ideas?

Cheers, Davy.

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When you have found and very simple solution why try to find another? I understand wanting to simplify a complicated argument but what can be simplified here? –  Ittay Weiss Feb 4 '13 at 17:43
Advancing in math does not mean forgetting the easy ways to solve problems! –  Hurkyl Feb 4 '13 at 17:44
I guess as Henning showed, expanding the solution to dice with more than 6 sides is useful, but I didn't state that is the question which I really should have. But thanks for the encouragement. As always, I intensely wish I'd gone to all my Maths lectures now! –  Davy Kavanagh Feb 5 '13 at 9:25

Yes, getting past a simple enumeration of cases is indeed interesting -- for example, in order to find out how the probability varies with the number of sides on the dice. Here's one approach:

Suppose we have a red die and a green die. The event that one of them shows twice what the other shows is the union of

• $A$: The red die shows twice the number of what the green die shows.
• $B$: The green die shows twice the number of what the red die shows.

If sides are numbered starting with 1, these two events are clearly mutually exclusive, so $P(A\cup B)$ is just $P(A)+P(B)$. And these probabilities are clearly the same (by symmetry), so we can restrict our attention $A$ if we remember to double the probability at the end.

Now suppose we throw the red die first and then the green one. If the red number is odd, we're out of $A$ immediately, but if it is even there is exactly one of the $n$ numbers on the green die that will land us in $A$.

So $P(A)$ must be $\frac12\cdot\frac 1n$ for $n$-sided dice (I've assumed $n$ is even such that exactly half of the numbers on a die are even).

Therefore the probability of either $A$ or $B$ happening is $2\cdot\frac12\cdot\frac1n=\frac1n$, for dice with an even number of sides $n$.

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Thanks for the abstraction to more than 6 sides. Clearly, writing down all the cases of A for two 20-sided dice would not be a clever use of time. Cheers! –  Davy Kavanagh Feb 5 '13 at 9:22