I just wanted to add a more generalized answer to this question that will give you a formula for computing your odd of success with advantage and disadvantage rather than looking up the value in a table. I am going to do my best to make this clear to anyone with any math background, so let me know in the comments if any of the steps don't make sense.

Advantage

With advantage when you need to roll at least \$n\$ to succeed on your check (i.e. check - mod = \$n\$), you succeed if any one of your two dice rolls a value of \$n\$ or greater. Conversely, you fail when both of your two dice roll a value of \$n-1\$ or less. Since these are the only two options, you succeed or you fail, the probability of one of these two things happening is \$1\$, so we can say:

$$ P(success) + P(failure) = 1 $$

Where \$P(x)\$ indicates the probability of event \$x\$ occurring. We can re arrange this to get:

$$ P(success) = 1 - P(failure) $$

So now we know that we can find the value we want using the probability of failure, which we previously defined as:

$$ P(failure) = P(\text{both dice }\leq n-1) $$ the probability that both dice roll a value of \$n-1\$ or less. For one dice, we know that there are \$n-1\$ ways that you can roll \$n-1\$ or less (e.g. if \$n-1 = 5\$ you could roll \$1, 2, 3, 4, \text{or }5\$ so there are \$5\$ possible ways to do it). There are \$20\$ total possible ways to roll the dice. so the probability of one dice rolling \$n-1\$ or less is the number of ways to roll \$n-1\$ divided by the total number of ways to roll the dice or:

$$ P(\text{one die } \leq n-1) = \frac{n-1}{20} $$

Since both dice are the same, their probability of rolling \$n-1\$ is the same, so we know the probabilities for both dice. The two dice rolls are independent of one another, meaning that the number you roll one one die doesn't effect the number you roll on the other one. In other words, if you roll a 5 on the first die, the odds of rolling a 7 on the other one don't change. When two events are independent, we can find the probability of both events happening by multiplying their probabilities. In other words:

$$ P(\text{both dice }\leq n-1) = P(\text{one die }\leq n-1) \times P(\text{one die }\leq n-1)\\ P(\text{both dice }\leq n-1) = \frac{n-1}{20} \times \frac{n-1}{20}\\ P(\text{both dice }\leq n-1) = \Big( \frac{n-1}{20}\Big)^2 $$

Substituting this into our original equation we get:

$$ P(success) = 1 - \Big( \frac{n-1}{20}\Big)^2 $$

Disadvantage

Now let define what it means to succeed with disadvantage in the same way we defined what it meant to succeed with advantage. For disadvantage where you need to roll at least \$n\$ to succeed, both dice must roll a value of \$n\$ or greater. In other words, if we need to roll at least an \$18\$ to succeed, both dice must roll either \$18, 19, \text{or } 20\$. The total number of ways to roll at least \$n\$ on a 20 sided die are:

$$ \{\text{# of ways to roll }\geq n\} = \{\text{total # of ways to roll}\} - \{\text{# of ways to roll }\leq n-1\}\\ \{\text{# of ways to roll }\geq n\} = 20 - (n-1) = 21 - n $$

We can create a probability from this by dividing by the total number of ways to roll the die giving us:

$$ P(\text{one die }\geq n) = \frac{21 - n}{20} $$

As before, the dice rolls are independent, so we can get the probabilities of both dice being greater than or equal to \$n\$ is:

$$ P(success) = \Big( \frac{21-n}{20}\Big)^2 $$

Generalizations

Since we worked through the math, we can also see how we can easily change this formula to get new probabilities. For example, if we make a house rule of "super advantage" where you roll 3 dice instead of 2, we simply multiply our \$P(failure)\$ by one more die \$\frac{n-1}{20}\$ changing the \$^2\$ to \$^3\$. We can therefore generalize the formula to be:

$$ P(success) = 1 - \Big( \frac{n-1}{20}\Big)^m $$

Where \$m\$ is the number of dice. Similarly, the probabilities for "super disadvantage" would be:

$$ P(success) = \Big( \frac{21-n}{20}\Big)^m $$

Going further, if we want to we could also sub out the \$20\$ in the denominator for another number if you wanted to look at the odds for other dice. For example, lets say you are a rogue, and you just rolled a sneak attack where all 5d6 were ones and twos, and you want to know what the odds were of this happening. In other words, we want to know \$P(\text{all dice }\leq 2)\$. This is the same as our "failure" condition for advantage except with 5 6-sided dice instead of 2 20-sided dice. So we can sub the 2 out for 5 and the 20 out for 6 and get:

$$ P(\text{all dice }\leq \text{max roll}) = \Big( \frac{\text{max roll}}{\text{# of sides}}\Big)^\text{# of dice}\\ P(\text{all dice }\leq 2) = \Big( \frac{2}{6}\Big)^5 = 0.004 $$

So only a 0.4% chance of this happening.