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.
 A: I don't think you need to worry about finding a "mathsy" way to do it. Simple case enumeration is fine when it is easy to follow, and in this case it's certainly easy to follow.
You might phrase it like this:

Since you want one dice to have twice the value of the other, the possible valid pairings are (1,2), (2,4) and (3,6). The smaller of the two values might be on the first dice (3 possibilities) or on the second (another 3 possibilities, for a total of 6). There are 36 possible outcomes, so the probability we want is 6/36, or 1/6 when simplified.

A: 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 $n$.
