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I know that rolling with disadvantage gives you a negative bonus in between 1 and 2, that's easy to calculate in anydice.com but how can I estimate the increasing in difficulty that adds an extra roll?.
I mean, if the probability of getting a 7-9 in 2d6 is 58,33% and 10+ is 16,67%, what is the probability of hitting again a 7-9 or 10+ result after the first roll results in a 7-9. And after a 10+?
HighDiceRoller' in their answer provide the correct commands in Anydice.com: if you click on the At Least option for data visualization, you can get the probabilities of obtaining a result greater or equal than your target.
If you are interested in mathematical details, keep on reading below.
The two rolls are statistically independent, which means that they are not influenced one by the other.
Let's take the case of just one d6 with advantage: rolling 1 or 2 has a probability of 33.33%, in the second roll obtaining a result of 1 or 2 has the same probability, because the result of the former roll does not influence the latter.
The same reasoning is applied when you add extra dice to the roll. If one has to roll 2d6 with disadvantage and obtains a results greater or equal than 10, which has a probability to happen of 16,67%, then the probability of obtaining another result greater or equal than 10 is still 16,67%.
If you want to compute the probability of success in two subsequent rolls, it depends if you roll with advantage or disadvantage. For example, with advantage it is sufficient that at least one of the rolls is successful: $$ \begin{split} P(\text{success}) =& P(\text{1st succeeds and 2nd does not succeed}) +\\ &+P(\text{1st does not succeed and 2nd succeeds})\\ &+ P(\text{1st succeeds and 2nd succeeds})=\\ =& P(\text{1st succeeds})P(\text{2nd does not succeed})+\\ &+ P(\text{1st does not succeed})P(\text{2nd succeeds})+\\ &+ P(\text{1st succeeds})P(\text{2nd succeeds}) \end{split} $$
Hence, if you need a result of 10+ on 2d6 with advantage, it will read as $$ P(\text{success}) = 16,67\% \cdot 83,33\% + 83,33\%\cdot 16,67\% + 16,67\% \cdot16,67\% = 30.56\% $$
You can use also the binomial distribution, if you prefer, with parameters \$n=2\$ (the number of trials) and \$p=16.67\%\$ (the probability of success).
For taking the higher or lower of two rolls you can use
output [highest 1 of 2d(2d6)]
output [lowest 1 of 2d(2d6)]
or
output 1@2d(2d6)
output 2@2d(2d6)
where the @ operator selects ranked dice.
