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In Out of the Abyss, there is something called the Maze Engine which, once activated, requires the DM to roll on a d100 table for random possible effects (pp. 186-188).

One of these entries is:

81—00: The engine emits a flash of violet-white light. All extraplanar creatures within 100 miles of the engine instantly return to their native planes of existence.

The Maze Engine is slowly sliding into magma below and will take 12 rounds to sink into the magma and be destroyed. One of these random effects happens per turn. So if I'm rolling on the d100 table with the hope of landing on the 81-00 option, I have 12 rolls before the Maze Engine is destroyed.

I suck as statistics, so can someone who understand statistics better than I tell me: what is the probability of rolling a 81-00 at least once on the d100 table over 12 attempts?

The reason I want this number is so that a modron (tridrone) NPC can tell the PCs exactly how likely it is to send them home with a precise-sounding decimal number (this is how I'm running the modron's motivation for helping us find the "Orderer" a.k.a. Maze Engine; it's so we can help them get back to Mechanus). And yes, as DM I can just fudge the dice roll so that it lands on this entry; this is more about wanting the exact probability for the purposes of the modron's in-game dialog.

The probability of not rolling 81–100 on a single attempt is 0.8 (80%), on 12 attempts the chance of never rolling it is 0.8^12 = 0.068719476736. So the chance of rolling 81–100 at least once is 1 − 0.068719476736 = 0.931280523264, or 93.1280523264% – as precise as it can get here.

The probability of an event is 1 minus the probability of the inverse of the event. Therefore, the probability that you do not roll an 81-100 is 1 minus the probability that you do roll an 81-100 in those turns.

Since the dice rolls are independent this is

p(X < 81)^12 = (0.80)^12

             ~= 0.069

And therefore the probability that the dice roll was an 81 or higher on any of the rolls is 1 - 0.069 which is approximately 93%, or exactly 93.1280523264% to 10 d.p. (derived from 227363409/244140625).

The other answers are accurate, but maybe you'd like a countdown on the probability (giving a savvy PC a chance to chime in with, "Never tell me the odds!").

On any particular round, the chance of rolling 81-100 is 20% (0.2). The chance that such a roll will come up in the time remaining = (1 - (1-0.2)^n)*100%, where n is the number of rounds remaining. For all twelve rounds, your percentages are: