When I studied combinatorics, it seemed pretty clear where order mattered and where it didn't. Probability questions don't give as many hints as to whether order matters and I'm looking for some advice on properly interpreting questions. Here's an example below that I answered incorrectly:
Two fair dice are rolled. Let $X$ be the absolute value of the difference between
the two numbers on the dice. Calculate the probability that $X < 3$.
And here is my approach:
Since it seems like we're rolling both dice at once here, the roll is one event. The sample space should be $6 \choose 2$$+ 6$ because we should have $6 \choose 2$ unordered pairs of all possible rolls, plus the six cases where the number on each die is equal to the other on a particular roll. Then I calculated by hand all the cases where $X < 3$ ($X = 0$ was $6$, $X = 1$ was $5$ and $X = 2$ was $4$), added them up, and divided by the sample space calculated above.
But this solution is wrong. Order does matter and the sample space is $6^2$. For every combination I calculated, there are two ways to order them, so I had to multiply the frequencies of my events by two (except for the event that the two dice rolled the same number). Where is my thinking going wrong?
Best Answer
When determining probability by counting atoms of the event, you want to be counting equally probable atoms.
So order matters because all ordered pairs are equally probable.
Unordered pairs are not equally probable. Each of the $\binom 62$ pairs of distinct faces are twice as probable as each of the $6$ pairs of same faces.
$$2\times \binom 62+6=2\times\dfrac{6\times 5}{2\times 1}+6\\~=6^2$$