# What is the probability of the Maze Engine from Out of the Abyss activating this effect?

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.

• Comments are not for extended discussion; this conversation has been moved to chat. – mxyzplk Dec 13 '18 at 12:52

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

$$\begin{array}{} p(X < 81)^{12} &=& (0.80)^{12}\\ &\approx& 0.069 \end{array}$$ 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 $$\93.1280523264\%\$$ to 10 d.p. (derived from $$\\frac{227363409}{244140625}\$$).

• Rather than using code formatting, you might want to use MathJax instead. – V2Blast Dec 13 '18 at 3:04

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:

$$\ \begin{array}{|l|c|l|} \hline \textbf{Round number} & \textbf{Probability}\\ \hline \textbf{12} & \textbf{93.1280523264%}\\ \hline \textbf{11} & \textbf{91.4100654080%}\\ \hline \textbf{10} & \textbf{89.2625817600%}\\ \hline \textbf{9} & \textbf{86.5782272000%}\\ \hline \textbf{8} & \textbf{83.2227840000%}\\ \hline \textbf{7} & \textbf{79.0284800000%}\\ \hline \textbf{6} & \textbf{73.7856000000%}\\ \hline \textbf{5} & \textbf{67.2320000000%}\\ \hline \textbf{4} & \textbf{59.0400000000%}\\ \hline \textbf{3} & \textbf{48.8000000000%}\\ \hline \textbf{2} & \textbf{36.0000000000%}\\ \hline \textbf{1} & \textbf{20.0000000000%}\\ \hline \textbf{0} & \textbf{0.0000000000%}\\ \hline \end{array} \$$

If you want an insanely complex way to say it, you should look at $$(\frac{4}{5} + \frac{1}{5}x)^{12}$$ $$= {{x^{12}}\over{244140625}}+{{48\,x^{11}}\over{244140625}}+{{1056\,x ^{10}}\over{244140625}}+{{2816\,x^9}\over{48828125}}+{{25344\,x^8 }\over{48828125}}+{{811008\,x^7}\over{244140625}}+{{3784704\,x^6 }\over{244140625}}+{{12976128\,x^5}\over{244140625}}+{{6488064\,x^4 }\over{48828125}}+{{11534336\,x^3}\over{48828125}}+{{69206016\,x^2 }\over{244140625}}+{{50331648\,x}\over{244140625}}+{{16777216}\over{ 244140625}}$$ The coefficients of the "x^n" terms where n isn't 0 are the probabilities that the Modron gets sent back home that number of times.

If there is a 10% chance that something else happens that undoes history: $$(\frac{7}{10} + \frac{1}{10}y + \frac{1}{5}x)^{12}$$ which expands to an insane polynomial.

This actually leads to two seemingly counterfactual cases; the chance that, given time continues, the Modron ends up banished, and the chance that the Modron is banished given you start it.

One of them is $$1-\frac{7}{10}^{12}$$, aka 0.986158712799, the other is $$\frac{\frac{9}{10}^{12} - \frac{7}{10}^{12}}{\frac{9}{10}^{12}}$$ or $$1-\frac{7}{9}^{12}$$, aka 0.95099206912471369407699877802074.

So you could have the Modron talk about "The properties of the Maze Engine make the conventional definition of linear reality non functional; depending on your specific definition what time means and what existence is, my probability of being banished varies from 0.931280523264 to 0.986158712799. These cases are best explained by the characteristic polynomial in tenths of 7 plus delta plus two beta, all raised to the 12th power." (Here I'm using "delta" for the time travel case, and "beta" for the banishment case).

• That's a lotta numbers. – V2Blast Dec 13 '18 at 21:24