If you roll a fair six sided die twice, what's the probability that you get the same number both times? So I know that rolling a fair six-sided die twice would mean the total possible outcomes would be 36, and rolling the same number twice would be 2/36 or 1/18, but I feel like that's wrong. What am I doing that isn't right?
 A: The first roll doesn't matter. It only sets a target. The chances of hitting a target in one roll are one in SIX, not one in three. You could have someone call out a number for the target. That doesn't matter.
A: There are six possible numbers that can be rolled twice.
For a specific number, say $1$, the probability of rolling it twice is equal to $$ \underbrace{\dfrac 16}_{1\text{st roll}}\;\times \;\underbrace{\dfrac 16}_{2\text{nd roll}}=\dfrac 1{36}$$ by the rule of the product.
Since this can happen for $2, 3, 4, 5, 6,$ as well as for $1$, 
the probability of rolling the same number twice is $$6\cdot \dfrac 16\cdot \dfrac 16 = \dfrac 16$$
A: I think this diagram might help. It's lifted from another website, so just imagine that the 'white die' denotes your first roll, and the 'red die' denotes your second roll. 
You'll notice that there are exactly $6$ pairs of rolls (out of all $36$ possible outcomes) in which the first roll equals the second roll.

The theoretical reason for this is well-explained in amWhy's answer.
A: Hint: as you stated, there is a total of $36$ outcomes. The outcomes you're interested in are $(1,1),....$ , a total of _ outcomes out of $36$.
