What is the probability of the Maze Engine from Out of the Abyss activating this effect?
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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.
dnd-5e published-adventures statistics out-of-the-abyss
add a comment |
up vote
7
down vote
favorite
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.
dnd-5e published-adventures statistics out-of-the-abyss
Is the tridrone planning to account for the chance that the other results of the maze engine could affect the chance of them lasting ling enough to get sent home?
– David Coffron
1 hour ago
@DavidCoffron No, the tridrone is assuming that they can survive the other various effects like the one that tries to disintegrate them or throws a green slaad at them, etc.
– NathanS
58 mins ago
add a comment |
up vote
7
down vote
favorite
up vote
7
down vote
favorite
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.
dnd-5e published-adventures statistics out-of-the-abyss
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.
dnd-5e published-adventures statistics out-of-the-abyss
dnd-5e published-adventures statistics out-of-the-abyss
edited 19 mins ago
Sdjz
10.4k34993
10.4k34993
asked 3 hours ago
NathanS
22.1k6102238
22.1k6102238
Is the tridrone planning to account for the chance that the other results of the maze engine could affect the chance of them lasting ling enough to get sent home?
– David Coffron
1 hour ago
@DavidCoffron No, the tridrone is assuming that they can survive the other various effects like the one that tries to disintegrate them or throws a green slaad at them, etc.
– NathanS
58 mins ago
add a comment |
Is the tridrone planning to account for the chance that the other results of the maze engine could affect the chance of them lasting ling enough to get sent home?
– David Coffron
1 hour ago
@DavidCoffron No, the tridrone is assuming that they can survive the other various effects like the one that tries to disintegrate them or throws a green slaad at them, etc.
– NathanS
58 mins ago
Is the tridrone planning to account for the chance that the other results of the maze engine could affect the chance of them lasting ling enough to get sent home?
– David Coffron
1 hour ago
Is the tridrone planning to account for the chance that the other results of the maze engine could affect the chance of them lasting ling enough to get sent home?
– David Coffron
1 hour ago
@DavidCoffron No, the tridrone is assuming that they can survive the other various effects like the one that tries to disintegrate them or throws a green slaad at them, etc.
– NathanS
58 mins ago
@DavidCoffron No, the tridrone is assuming that they can survive the other various effects like the one that tries to disintegrate them or throws a green slaad at them, etc.
– NathanS
58 mins ago
add a comment |
2 Answers
2
active
oldest
votes
up vote
13
down vote
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.
add a comment |
up vote
10
down vote
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
).
I didn't know that was possible. If it is possible then the distribution is going to be way more complicated.
– mklingen
1 hour ago
It surprises me that 1-(4/5)^12 results in an even denominator
– Yakk
1 hour ago
My math was wrong. Fixed the rational.
– mklingen
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
13
down vote
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.
add a comment |
up vote
13
down vote
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.
add a comment |
up vote
13
down vote
up vote
13
down vote
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 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.
edited 3 hours ago
answered 3 hours ago
black_fm
1,90211321
1,90211321
add a comment |
add a comment |
up vote
10
down vote
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
).
I didn't know that was possible. If it is possible then the distribution is going to be way more complicated.
– mklingen
1 hour ago
It surprises me that 1-(4/5)^12 results in an even denominator
– Yakk
1 hour ago
My math was wrong. Fixed the rational.
– mklingen
1 hour ago
add a comment |
up vote
10
down vote
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
).
I didn't know that was possible. If it is possible then the distribution is going to be way more complicated.
– mklingen
1 hour ago
It surprises me that 1-(4/5)^12 results in an even denominator
– Yakk
1 hour ago
My math was wrong. Fixed the rational.
– mklingen
1 hour ago
add a comment |
up vote
10
down vote
up vote
10
down vote
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 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
).
edited 1 hour ago
answered 3 hours ago
mklingen
2184
2184
I didn't know that was possible. If it is possible then the distribution is going to be way more complicated.
– mklingen
1 hour ago
It surprises me that 1-(4/5)^12 results in an even denominator
– Yakk
1 hour ago
My math was wrong. Fixed the rational.
– mklingen
1 hour ago
add a comment |
I didn't know that was possible. If it is possible then the distribution is going to be way more complicated.
– mklingen
1 hour ago
It surprises me that 1-(4/5)^12 results in an even denominator
– Yakk
1 hour ago
My math was wrong. Fixed the rational.
– mklingen
1 hour ago
I didn't know that was possible. If it is possible then the distribution is going to be way more complicated.
– mklingen
1 hour ago
I didn't know that was possible. If it is possible then the distribution is going to be way more complicated.
– mklingen
1 hour ago
It surprises me that 1-(4/5)^12 results in an even denominator
– Yakk
1 hour ago
It surprises me that 1-(4/5)^12 results in an even denominator
– Yakk
1 hour ago
My math was wrong. Fixed the rational.
– mklingen
1 hour ago
My math was wrong. Fixed the rational.
– mklingen
1 hour ago
add a comment |
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Is the tridrone planning to account for the chance that the other results of the maze engine could affect the chance of them lasting ling enough to get sent home?
– David Coffron
1 hour ago
@DavidCoffron No, the tridrone is assuming that they can survive the other various effects like the one that tries to disintegrate them or throws a green slaad at them, etc.
– NathanS
58 mins ago