Two scenarios:
1. Using one die, roll a 6 twice.
$\frac16\times\frac16=\frac1{36}$
Why are these two probabilities different? Because the events are independent, isn't rolling a pair the same as rolling a die twice?
In a sense, rolling two dice at once is the same as rolling 1 die twice at the same time? How does this "timing" issue affect the probability?
In the second case, your pair can be any one of $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)$ and you'd satisfy "obtaining a pair".
That gives you six possible pairs, each of which one has probability of occurring $\dfrac 1{36}$ gives us $$6\times \frac 1{36} = \frac 16$$
Now, if you want to know what the probability of rolling two dice simultaneously and obtaining two sixes (one prespecified pair of the six possible pairs), that would be $\dfrac 1{6\cdot 6} = \dfrac 1{36}$.
With this distinction made, yes, the probability of obtaining two sixes when rolling one die twice, and the probability of rolling two sixes simultaneously are equal.
Note that you are asking two different questions. In the first problem, you are asking the probability of rolling a particular number twice, while in the second problem, you are asking the probability of rolling one of the numbers twice.
The probability of rolling a $6$ twice is $$\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$$ but so is the probability of rolling a $1$ twice, a $2$ twice, a $3$ twice, a $4$ twice, or a $5$ twice. Hence, the probability of rolling the same number twice is $$6 \cdot \frac{1}{36} = \frac{1}{6}$$
I think the simplest answer for the second, pairing, scenario is:
Your first rolling does not have any possibility expected, which means you will have any numbers. However, your second roll should be pairing the first number and the possibility should be 1/6.
so to roll a pair, the odds is 1 x 1/6 = 1/6
