Probability of rolling x or lower twice in a row over multiple dice rolls? If I'm rolling, for example, 6 twenty-sided dice, one at a time, and I "fail" if I roll a 10 or lower twice in a row, how do I calculate the chances of success/failure?
Example:
6d20, success on 11 or higher.
I roll the dice, results in order are (13, 7, 17, 18, 2, 11). This is a success because the two failures (7, 2) were not rolled one after the other.
If instead I roll (15, 4, 8) it's a failure and I stop rolling, because I rolled <11 twice in a row.
How do I calculate the chances of success/failure? Over the 6 rolls in the example, or any number of rolls?
EDIT: I am not very knowledgeable when it comes to math and was hoping for a relatively simple explanation. If it's possible.
 A: Note that the chances of rolling a 10 or lower are the same as rolling an 11 or higher. That means this is the same as flipping a coin, and losing if we flip two tails in a row. (Think of it as if you roll an 11 -20, its the same as a tails and otherwise heads)
We want to know the chances that we don't ever get two adjacent tails.
If our most recent flip is a tail that is not after another tails, then we call that scary because we are close to losing: one more tails and we lose.
If our most recent flip was heads, we call that easy, since it means we can't lose in a single flip.
And, if our most recent flip was a tails after another tails, we call that a lose.
Now, initially, there is a 50-50 chance (on your first flip) that you flip heads or tails, meaning we can either go to an easy state or a scary state.
From the easy state, there is a 50-50 chance of ending up at an easy state (by flipping a heads) or flipping to a scary state (by flipping tails).
From the scary state, there is a 50-50 chance of ending up at the easy state and the lose state.
Let $P(i)$ represent the probability of losing by flip $i,$ and $S(i)$ represent the probability that we end up at a scary state by flip $i,$ as well as $Z(i)$ representing being at an easy state by flip i.
Note that $P(i) = \frac{1}{2} S(i-1) + P(i-1)$ because you need to get the 50% chance of flipping a tails if you are at a scary step, and already losing in a previous step means you have already lost.
$Z(i) = \frac{1}{2} (Z(i-1) + P(i-1) + S(i-1)),$ since at any step, we have the probability of flipping a heads and reaching an easy state.
$S(i) = \frac{1}{2}P(i-1).$
$P(1) = 0, S(1) = \frac{1}{2}, Z(1) = \frac{1}{2}.$
We can then use these recurrences and initial values to obtain $P(6),$ which should be the probability that we lose before or by the 6th flip.
Sorry for the long winded answer!
