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How cumulative chance is calculated in Augury or similar spells?
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Part of the augury spell description says:
If you cast the spell two or more times before completing your next long rest, there is a cumulative 25 percent chance for each casting after the first that you get a random reading. The DM makes this roll in secret.
How should I roll this cumulative chance?
Using a single d4. A second casting of augury means that rolling 1 results in a random reading. A third casting means that rolling 1 or 2 results in a random reading, and so on.
Using more d4. A second casting of augury means that you roll 1d4, and if a 1 is rolled, it results in a random reading. A third casting means that you roll 2d4, and a 1 rolled on either die results in a random reading, and so on.
dnd-5e spells
edited 18 hours ago
V2Blast
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Vylix
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up vote
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Part of the augury spell description says:
If you cast the spell two or more times before completing your next long rest, there is a cumulative 25 percent chance for each casting after the first that you get a random reading. The DM makes this roll in secret.
How should I roll this cumulative chance?
Using a single d4. A second casting of augury means that rolling 1 results in a random reading. A third casting means that rolling 1 or 2 results in a random reading, and so on.
Using more d4. A second casting of augury means that you roll 1d4, and if a 1 is rolled, it results in a random reading. A third casting means that you roll 2d4, and a 1 rolled on either die results in a random reading, and so on.
dnd-5e spells
edited 18 hours ago
V2Blast
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asked 21 hours ago
Vylix
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2d4 can never result in 1. Do you mean taking the lowest of 2d4 (similar to disadvantage)?
– Ruse
20 hours ago
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up vote
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up vote
8
down vote
favorite
Part of the augury spell description says:
If you cast the spell two or more times before completing your next long rest, there is a cumulative 25 percent chance for each casting after the first that you get a random reading. The DM makes this roll in secret.
How should I roll this cumulative chance?
Using a single d4. A second casting of augury means that rolling 1 results in a random reading. A third casting means that rolling 1 or 2 results in a random reading, and so on.
Using more d4. A second casting of augury means that you roll 1d4, and if a 1 is rolled, it results in a random reading. A third casting means that you roll 2d4, and a 1 rolled on either die results in a random reading, and so on.
dnd-5e spells
edited 18 hours ago
V2Blast
14.1k23593
asked 21 hours ago
Vylix
5,85112187
Part of the augury spell description says:
If you cast the spell two or more times before completing your next long rest, there is a cumulative 25 percent chance for each casting after the first that you get a random reading. The DM makes this roll in secret.
How should I roll this cumulative chance?
Using a single d4. A second casting of augury means that rolling 1 results in a random reading. A third casting means that rolling 1 or 2 results in a random reading, and so on.
Using more d4. A second casting of augury means that you roll 1d4, and if a 1 is rolled, it results in a random reading. A third casting means that you roll 2d4, and a 1 rolled on either die results in a random reading, and so on.
dnd-5e spells
dnd-5e spells
edited 18 hours ago
V2Blast
14.1k23593
asked 21 hours ago
Vylix
5,85112187
edited 18 hours ago
V2Blast
14.1k23593
asked 21 hours ago
Vylix
5,85112187
edited 18 hours ago
V2Blast
14.1k23593
edited 18 hours ago
V2Blast
14.1k23593
edited 18 hours ago
V2Blast
14.1k23593
14.1k23593
asked 21 hours ago
Vylix
5,85112187
asked 21 hours ago
Vylix
5,85112187
asked 21 hours ago
Vylix
5,85112187
5,85112187
2d4 can never result in 1. Do you mean taking the lowest of 2d4 (similar to disadvantage)?
– Ruse
20 hours ago
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2d4 can never result in 1. Do you mean taking the lowest of 2d4 (similar to disadvantage)?
– Ruse
20 hours ago
2d4 can never result in 1. Do you mean taking the lowest of 2d4 (similar to disadvantage)?
– Ruse
20 hours ago
2d4 can never result in 1. Do you mean taking the lowest of 2d4 (similar to disadvantage)?
– Ruse
20 hours ago
add a comment |Â
6 Answers
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"Cumulative" in this context means Additive
Cumulative is a word formed from shortening "Accumulative", which means to add together. So when the text reads "a cumulative 25% chance of failure per casting", it's intended to mean "accumulate 25 percentage points worth of chance per casting".
Because of this, it is correct to treat the math as being (X / 4) chance of failure (random result), where X is the number of times prior to the current casting of Augury that you've cast Augury. This means that on the first casting, there's a 0% chance of failure, then a 25% chance, then 50%, then 75%, and on the fifth casting onwards, the result is always random.
Either roll a single d4, like you suggested, using rolls of (1,2,3,4) as the minimum required for success for each successive roll, or use d% when making this roll, using (1, 26, 51, 76) as the minimum required rolls for success.
If the rules intended you to roll multiple dice for a multiplicative calculation, like your second option suggests, the rules would expressly instruct you to do that.
answered 21 hours ago
Xirema
5,0551434
1
I think the answer would be even better if it explained why you think that cumulative means additive. Is it because of a generic definition of cumulative assumes addition? Or because the spell sais "cumulative ... chance" instead of "cumulative d4"? Or some other reason.
– Ruse
20 hours ago
2
@Ruse at which point, you might refer someone to a dictionary: "cumulative (adj.): increasing or increased in quantity, degree, or force by successive additions." When in doubt, the plain English version of the word is the one meant, unless the rules very specifically state otherwise.
– phyrfox
8 hours ago
@phyrfox You're correct, but the problem is the answer doesn't explain that. The answer needs to be backed by evidence, and right now the leading paragraph is just somebody's opinion/assertion. If this is being based on a combination of 1) dictionary definition and 2) plain language principles in 5e (which is missing) then that needs to be explained. Or, basically, the answer says it's "intended to mean" but doesn't justify that intent.
– Bloodcinder
1 hour ago
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Use the Single D4 Method
td;dr Roll a single d4.
The 2nd casting would have a 1 in 4 chance of random ruling. The 3rd casting 2 in 4. In this way the chance for a random reading accumulates by 25% with each additional casting.
Multiple Dice do not Accumulate by 25%
Rolling multiple dice is considering each chance to be an independent event. This leads to the subsequent castings adding diminishing probability of getting a random reading.
$beginarrayl
hline
textbfCasting per Long Rest & textbfChance of Random Reading & textbfCumulative % \
hline
text2nd casting of the day & text25% = 1 - 3/4 & text25% \
text3rd casting of the day & text43.75% = 1 - (3/4 * 3/4) & text18.75% \
text4th casting of the day & text57.8% = 1 - (3/4)^3 & text7.8% \
hline
endarray
$
The problem with rolling independent dice is each subsequent casting accumulates less chance of failure than the stated 25%.
edited 18 hours ago
V2Blast
14.1k23593
answered 21 hours ago
Grosscol
4,150734
The train of thought is that multiplicative accumulates by 25% of what's left, rather than 25% of the original value. "Armor" works like this in most video games. Each point of armor causes you to take 0.006% less damage relative to not having that point, which people perceive as "diminishing returns", when it's really not.
– Mooing Duck
20 hours ago
@MooingDuck I would expect a more specific wording if that were the case. My assumption was that the vernacular meaning of cumulative was intended.
– Grosscol
19 hours ago
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The "cumulative chances" add together (use Case 1).
There is no rule which explicitly clarifies whether your Case 1 or Case 2 is the correct way to adjudicate "cumulative chance." However, one of the design principles of D&D 5e is for mathematical calculations to be straightforward and simple to compute in your head.
Case 1 corresponds to simple arithmetic you can do in your head: the second time you cast the spell you have a 25% chance of failure, then 50% the next, then 75% the next, and so on (just by adding a flat 25% each time).
Case 2 corresponds to probability computations that most people can't do in their head: the second time you cast the spell you have a 25% chance of failure, then 43.75% the next time, then 57.8125% the next time, and so on (by using the principle of inclusion and exclusion in discrete probability).
Just because you can easily describe how to use d4's to do this math indirectly without doing any hard calculations doesn't mean that the text expects you to do that, otherwise there would be something in the text to explain how to do that. While there is a passage explaining how to use d% dice, there is no passage explaining how to use a d4 to simulate a 25% chance of failure, and without such guidance a person who doesn't see how to use d4's to solve the problem would be forced to do a lot of hard math using Case 2, whereas the rules already explain how to do Case 1.
It's also worth noting that all instances of this "cumulative chance" concept in the DMG and PHB use percentages that divide evenly into 100%, such as 5%, 10%, 20%, and 25%, which is helpful for Case 1 but would be unnecessary for Case 2 to the point of being a misleading contrivance.
So the clear intent in the text is to use your Case 1 for "cumulative chances."
answered 17 hours ago
Bloodcinder
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Your option 1 is correct: Roll a single d4.
The 2nd casting has a 25% chance of failing. This is the same as 1 in 4.
The 3rd casting has a 50% chance of failing. This is the same as 2 in 4.
The 4th casting has a 75% chance of failing. This is the same as 3 in 4.
You can always avoid the maths by using the percentage value and rolling a d100 (roll a d10 twice - the first roll is the 'tens', the second roll is the 'units'). It amounts to the same thing.
answered 21 hours ago
PJRZ
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-1
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- Roll a d100. If the result is equal to or lower than 25 * (number of times cast - 1), give a random reading.
This is effectively the same as option 1.
answered 21 hours ago
Derek Stucki
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You could also roll a d% (it's the one that goes 0, 10, 20, etc.) and a d10 at the same time, to get the same as a d100.
"If you cast the spell two or more times before completing your next long rest, there is a cumulative 25 percent chance for each casting after the first that you
get a random reading. The DM makes this roll in secret."
Cumulative: increasing or increased in quantity, degree, or force by successive additions
You could just take the long nap in between, but if you DO cast Augury more times before your next long rest, then it's up to the DM to choose how he/she wants to roll. I don't think it does fail, per say, but it does give a chance to give the player(s) what they don't want. For throwing outcomes, I think it'd be:
0%: First throw,
25%: Second throw,
50%: Third throw,
75%: Fourth throw,
100%: Fifth+ throw
Each percentage would be something that they did not expect (or maybe it just goes alright). I'd suggest having the d%, d10, and the d4 (if you are the DM) handy if the Cleric - since that seems to be the only class that can use it - is going to use the Augury spell.
I would treat the percentages like the Sorcerer's Wild Magic; if it is below the certain percentage (where d% and d10 come in), then the DM rolls the d4 to see which omen the player gets. If it is above, then they get what they wanted. They would have to vocally state what they wanted.
answered 19 hours ago
Dragonman0007
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
accepted
"Cumulative" in this context means Additive
Cumulative is a word formed from shortening "Accumulative", which means to add together. So when the text reads "a cumulative 25% chance of failure per casting", it's intended to mean "accumulate 25 percentage points worth of chance per casting".
Because of this, it is correct to treat the math as being (X / 4) chance of failure (random result), where X is the number of times prior to the current casting of Augury that you've cast Augury. This means that on the first casting, there's a 0% chance of failure, then a 25% chance, then 50%, then 75%, and on the fifth casting onwards, the result is always random.
Either roll a single d4, like you suggested, using rolls of (1,2,3,4) as the minimum required for success for each successive roll, or use d% when making this roll, using (1, 26, 51, 76) as the minimum required rolls for success.
If the rules intended you to roll multiple dice for a multiplicative calculation, like your second option suggests, the rules would expressly instruct you to do that.
answered 21 hours ago
Xirema
5,0551434
1
I think the answer would be even better if it explained why you think that cumulative means additive. Is it because of a generic definition of cumulative assumes addition? Or because the spell sais "cumulative ... chance" instead of "cumulative d4"? Or some other reason.
– Ruse
20 hours ago
2
@Ruse at which point, you might refer someone to a dictionary: "cumulative (adj.): increasing or increased in quantity, degree, or force by successive additions." When in doubt, the plain English version of the word is the one meant, unless the rules very specifically state otherwise.
– phyrfox
8 hours ago
@phyrfox You're correct, but the problem is the answer doesn't explain that. The answer needs to be backed by evidence, and right now the leading paragraph is just somebody's opinion/assertion. If this is being based on a combination of 1) dictionary definition and 2) plain language principles in 5e (which is missing) then that needs to be explained. Or, basically, the answer says it's "intended to mean" but doesn't justify that intent.
– Bloodcinder
1 hour ago
add a comment |Â
up vote
10
down vote
accepted
"Cumulative" in this context means Additive
Cumulative is a word formed from shortening "Accumulative", which means to add together. So when the text reads "a cumulative 25% chance of failure per casting", it's intended to mean "accumulate 25 percentage points worth of chance per casting".
Because of this, it is correct to treat the math as being (X / 4) chance of failure (random result), where X is the number of times prior to the current casting of Augury that you've cast Augury. This means that on the first casting, there's a 0% chance of failure, then a 25% chance, then 50%, then 75%, and on the fifth casting onwards, the result is always random.
Either roll a single d4, like you suggested, using rolls of (1,2,3,4) as the minimum required for success for each successive roll, or use d% when making this roll, using (1, 26, 51, 76) as the minimum required rolls for success.
If the rules intended you to roll multiple dice for a multiplicative calculation, like your second option suggests, the rules would expressly instruct you to do that.
answered 21 hours ago
Xirema
5,0551434
1
I think the answer would be even better if it explained why you think that cumulative means additive. Is it because of a generic definition of cumulative assumes addition? Or because the spell sais "cumulative ... chance" instead of "cumulative d4"? Or some other reason.
– Ruse
20 hours ago
2
@Ruse at which point, you might refer someone to a dictionary: "cumulative (adj.): increasing or increased in quantity, degree, or force by successive additions." When in doubt, the plain English version of the word is the one meant, unless the rules very specifically state otherwise.
– phyrfox
8 hours ago
@phyrfox You're correct, but the problem is the answer doesn't explain that. The answer needs to be backed by evidence, and right now the leading paragraph is just somebody's opinion/assertion. If this is being based on a combination of 1) dictionary definition and 2) plain language principles in 5e (which is missing) then that needs to be explained. Or, basically, the answer says it's "intended to mean" but doesn't justify that intent.
– Bloodcinder
1 hour ago
add a comment |Â
up vote
10
down vote
accepted
up vote
10
down vote
accepted
"Cumulative" in this context means Additive
Cumulative is a word formed from shortening "Accumulative", which means to add together. So when the text reads "a cumulative 25% chance of failure per casting", it's intended to mean "accumulate 25 percentage points worth of chance per casting".
Because of this, it is correct to treat the math as being (X / 4) chance of failure (random result), where X is the number of times prior to the current casting of Augury that you've cast Augury. This means that on the first casting, there's a 0% chance of failure, then a 25% chance, then 50%, then 75%, and on the fifth casting onwards, the result is always random.
Either roll a single d4, like you suggested, using rolls of (1,2,3,4) as the minimum required for success for each successive roll, or use d% when making this roll, using (1, 26, 51, 76) as the minimum required rolls for success.
If the rules intended you to roll multiple dice for a multiplicative calculation, like your second option suggests, the rules would expressly instruct you to do that.
answered 21 hours ago
Xirema
5,0551434
"Cumulative" in this context means Additive
Cumulative is a word formed from shortening "Accumulative", which means to add together. So when the text reads "a cumulative 25% chance of failure per casting", it's intended to mean "accumulate 25 percentage points worth of chance per casting".
Because of this, it is correct to treat the math as being (X / 4) chance of failure (random result), where X is the number of times prior to the current casting of Augury that you've cast Augury. This means that on the first casting, there's a 0% chance of failure, then a 25% chance, then 50%, then 75%, and on the fifth casting onwards, the result is always random.
Either roll a single d4, like you suggested, using rolls of (1,2,3,4) as the minimum required for success for each successive roll, or use d% when making this roll, using (1, 26, 51, 76) as the minimum required rolls for success.
If the rules intended you to roll multiple dice for a multiplicative calculation, like your second option suggests, the rules would expressly instruct you to do that.
answered 21 hours ago
Xirema
5,0551434
edited 19 hours ago
answered 21 hours ago
Xirema
5,0551434
answered 21 hours ago
Xirema
5,0551434
answered 21 hours ago
Xirema
5,0551434
5,0551434
1
I think the answer would be even better if it explained why you think that cumulative means additive. Is it because of a generic definition of cumulative assumes addition? Or because the spell sais "cumulative ... chance" instead of "cumulative d4"? Or some other reason.
– Ruse
20 hours ago
2
@Ruse at which point, you might refer someone to a dictionary: "cumulative (adj.): increasing or increased in quantity, degree, or force by successive additions." When in doubt, the plain English version of the word is the one meant, unless the rules very specifically state otherwise.
– phyrfox
8 hours ago
@phyrfox You're correct, but the problem is the answer doesn't explain that. The answer needs to be backed by evidence, and right now the leading paragraph is just somebody's opinion/assertion. If this is being based on a combination of 1) dictionary definition and 2) plain language principles in 5e (which is missing) then that needs to be explained. Or, basically, the answer says it's "intended to mean" but doesn't justify that intent.
– Bloodcinder
1 hour ago
add a comment |Â
1
I think the answer would be even better if it explained why you think that cumulative means additive. Is it because of a generic definition of cumulative assumes addition? Or because the spell sais "cumulative ... chance" instead of "cumulative d4"? Or some other reason.
– Ruse
20 hours ago
2
@Ruse at which point, you might refer someone to a dictionary: "cumulative (adj.): increasing or increased in quantity, degree, or force by successive additions." When in doubt, the plain English version of the word is the one meant, unless the rules very specifically state otherwise.
– phyrfox
8 hours ago
@phyrfox You're correct, but the problem is the answer doesn't explain that. The answer needs to be backed by evidence, and right now the leading paragraph is just somebody's opinion/assertion. If this is being based on a combination of 1) dictionary definition and 2) plain language principles in 5e (which is missing) then that needs to be explained. Or, basically, the answer says it's "intended to mean" but doesn't justify that intent.
– Bloodcinder
1 hour ago
1
1
I think the answer would be even better if it explained why you think that cumulative means additive. Is it because of a generic definition of cumulative assumes addition? Or because the spell sais "cumulative ... chance" instead of "cumulative d4"? Or some other reason.
– Ruse
20 hours ago
I think the answer would be even better if it explained why you think that cumulative means additive. Is it because of a generic definition of cumulative assumes addition? Or because the spell sais "cumulative ... chance" instead of "cumulative d4"? Or some other reason.
– Ruse
20 hours ago
2
2
@Ruse at which point, you might refer someone to a dictionary: "cumulative (adj.): increasing or increased in quantity, degree, or force by successive additions." When in doubt, the plain English version of the word is the one meant, unless the rules very specifically state otherwise.
– phyrfox
8 hours ago
@Ruse at which point, you might refer someone to a dictionary: "cumulative (adj.): increasing or increased in quantity, degree, or force by successive additions." When in doubt, the plain English version of the word is the one meant, unless the rules very specifically state otherwise.
– phyrfox
8 hours ago
@phyrfox You're correct, but the problem is the answer doesn't explain that. The answer needs to be backed by evidence, and right now the leading paragraph is just somebody's opinion/assertion. If this is being based on a combination of 1) dictionary definition and 2) plain language principles in 5e (which is missing) then that needs to be explained. Or, basically, the answer says it's "intended to mean" but doesn't justify that intent.
– Bloodcinder
1 hour ago
@phyrfox You're correct, but the problem is the answer doesn't explain that. The answer needs to be backed by evidence, and right now the leading paragraph is just somebody's opinion/assertion. If this is being based on a combination of 1) dictionary definition and 2) plain language principles in 5e (which is missing) then that needs to be explained. Or, basically, the answer says it's "intended to mean" but doesn't justify that intent.
– Bloodcinder
1 hour ago
add a comment |Â
up vote
6
down vote
Use the Single D4 Method
td;dr Roll a single d4.
The 2nd casting would have a 1 in 4 chance of random ruling. The 3rd casting 2 in 4. In this way the chance for a random reading accumulates by 25% with each additional casting.
Multiple Dice do not Accumulate by 25%
Rolling multiple dice is considering each chance to be an independent event. This leads to the subsequent castings adding diminishing probability of getting a random reading.
$beginarrayl
hline
textbfCasting per Long Rest & textbfChance of Random Reading & textbfCumulative % \
hline
text2nd casting of the day & text25% = 1 - 3/4 & text25% \
text3rd casting of the day & text43.75% = 1 - (3/4 * 3/4) & text18.75% \
text4th casting of the day & text57.8% = 1 - (3/4)^3 & text7.8% \
hline
endarray
$
The problem with rolling independent dice is each subsequent casting accumulates less chance of failure than the stated 25%.
edited 18 hours ago
V2Blast
14.1k23593
answered 21 hours ago
Grosscol
4,150734
The train of thought is that multiplicative accumulates by 25% of what's left, rather than 25% of the original value. "Armor" works like this in most video games. Each point of armor causes you to take 0.006% less damage relative to not having that point, which people perceive as "diminishing returns", when it's really not.
– Mooing Duck
20 hours ago
@MooingDuck I would expect a more specific wording if that were the case. My assumption was that the vernacular meaning of cumulative was intended.
– Grosscol
19 hours ago
add a comment |Â
up vote
6
down vote
Use the Single D4 Method
td;dr Roll a single d4.
The 2nd casting would have a 1 in 4 chance of random ruling. The 3rd casting 2 in 4. In this way the chance for a random reading accumulates by 25% with each additional casting.
Multiple Dice do not Accumulate by 25%
Rolling multiple dice is considering each chance to be an independent event. This leads to the subsequent castings adding diminishing probability of getting a random reading.
$beginarrayl
hline
textbfCasting per Long Rest & textbfChance of Random Reading & textbfCumulative % \
hline
text2nd casting of the day & text25% = 1 - 3/4 & text25% \
text3rd casting of the day & text43.75% = 1 - (3/4 * 3/4) & text18.75% \
text4th casting of the day & text57.8% = 1 - (3/4)^3 & text7.8% \
hline
endarray
$
The problem with rolling independent dice is each subsequent casting accumulates less chance of failure than the stated 25%.
edited 18 hours ago
V2Blast
14.1k23593
answered 21 hours ago
Grosscol
4,150734
The train of thought is that multiplicative accumulates by 25% of what's left, rather than 25% of the original value. "Armor" works like this in most video games. Each point of armor causes you to take 0.006% less damage relative to not having that point, which people perceive as "diminishing returns", when it's really not.
– Mooing Duck
20 hours ago
@MooingDuck I would expect a more specific wording if that were the case. My assumption was that the vernacular meaning of cumulative was intended.
– Grosscol
19 hours ago
add a comment |Â
up vote
6
down vote
up vote
6
down vote
Use the Single D4 Method
td;dr Roll a single d4.
The 2nd casting would have a 1 in 4 chance of random ruling. The 3rd casting 2 in 4. In this way the chance for a random reading accumulates by 25% with each additional casting.
Multiple Dice do not Accumulate by 25%
Rolling multiple dice is considering each chance to be an independent event. This leads to the subsequent castings adding diminishing probability of getting a random reading.
$beginarrayl
hline
textbfCasting per Long Rest & textbfChance of Random Reading & textbfCumulative % \
hline
text2nd casting of the day & text25% = 1 - 3/4 & text25% \
text3rd casting of the day & text43.75% = 1 - (3/4 * 3/4) & text18.75% \
text4th casting of the day & text57.8% = 1 - (3/4)^3 & text7.8% \
hline
endarray
$
The problem with rolling independent dice is each subsequent casting accumulates less chance of failure than the stated 25%.
edited 18 hours ago
V2Blast
14.1k23593
answered 21 hours ago
Grosscol
4,150734
Use the Single D4 Method
td;dr Roll a single d4.
The 2nd casting would have a 1 in 4 chance of random ruling. The 3rd casting 2 in 4. In this way the chance for a random reading accumulates by 25% with each additional casting.
Multiple Dice do not Accumulate by 25%
Rolling multiple dice is considering each chance to be an independent event. This leads to the subsequent castings adding diminishing probability of getting a random reading.
$beginarrayl
hline
textbfCasting per Long Rest & textbfChance of Random Reading & textbfCumulative % \
hline
text2nd casting of the day & text25% = 1 - 3/4 & text25% \
text3rd casting of the day & text43.75% = 1 - (3/4 * 3/4) & text18.75% \
text4th casting of the day & text57.8% = 1 - (3/4)^3 & text7.8% \
hline
endarray
$
The problem with rolling independent dice is each subsequent casting accumulates less chance of failure than the stated 25%.
edited 18 hours ago
V2Blast
14.1k23593
answered 21 hours ago
Grosscol
4,150734
edited 18 hours ago
V2Blast
14.1k23593
edited 18 hours ago
V2Blast
14.1k23593
edited 18 hours ago
V2Blast
14.1k23593
14.1k23593
answered 21 hours ago
Grosscol
4,150734
answered 21 hours ago
Grosscol
4,150734
answered 21 hours ago
Grosscol
4,150734
4,150734
The train of thought is that multiplicative accumulates by 25% of what's left, rather than 25% of the original value. "Armor" works like this in most video games. Each point of armor causes you to take 0.006% less damage relative to not having that point, which people perceive as "diminishing returns", when it's really not.
– Mooing Duck
20 hours ago
@MooingDuck I would expect a more specific wording if that were the case. My assumption was that the vernacular meaning of cumulative was intended.
– Grosscol
19 hours ago
add a comment |Â
The train of thought is that multiplicative accumulates by 25% of what's left, rather than 25% of the original value. "Armor" works like this in most video games. Each point of armor causes you to take 0.006% less damage relative to not having that point, which people perceive as "diminishing returns", when it's really not.
– Mooing Duck
20 hours ago
@MooingDuck I would expect a more specific wording if that were the case. My assumption was that the vernacular meaning of cumulative was intended.
– Grosscol
19 hours ago
The train of thought is that multiplicative accumulates by 25% of what's left, rather than 25% of the original value. "Armor" works like this in most video games. Each point of armor causes you to take 0.006% less damage relative to not having that point, which people perceive as "diminishing returns", when it's really not.
– Mooing Duck
20 hours ago
The train of thought is that multiplicative accumulates by 25% of what's left, rather than 25% of the original value. "Armor" works like this in most video games. Each point of armor causes you to take 0.006% less damage relative to not having that point, which people perceive as "diminishing returns", when it's really not.
– Mooing Duck
20 hours ago
@MooingDuck I would expect a more specific wording if that were the case. My assumption was that the vernacular meaning of cumulative was intended.
– Grosscol
19 hours ago
@MooingDuck I would expect a more specific wording if that were the case. My assumption was that the vernacular meaning of cumulative was intended.
– Grosscol
19 hours ago
add a comment |Â
up vote
4
down vote
The "cumulative chances" add together (use Case 1).
There is no rule which explicitly clarifies whether your Case 1 or Case 2 is the correct way to adjudicate "cumulative chance." However, one of the design principles of D&D 5e is for mathematical calculations to be straightforward and simple to compute in your head.
Case 1 corresponds to simple arithmetic you can do in your head: the second time you cast the spell you have a 25% chance of failure, then 50% the next, then 75% the next, and so on (just by adding a flat 25% each time).
Case 2 corresponds to probability computations that most people can't do in their head: the second time you cast the spell you have a 25% chance of failure, then 43.75% the next time, then 57.8125% the next time, and so on (by using the principle of inclusion and exclusion in discrete probability).
Just because you can easily describe how to use d4's to do this math indirectly without doing any hard calculations doesn't mean that the text expects you to do that, otherwise there would be something in the text to explain how to do that. While there is a passage explaining how to use d% dice, there is no passage explaining how to use a d4 to simulate a 25% chance of failure, and without such guidance a person who doesn't see how to use d4's to solve the problem would be forced to do a lot of hard math using Case 2, whereas the rules already explain how to do Case 1.
It's also worth noting that all instances of this "cumulative chance" concept in the DMG and PHB use percentages that divide evenly into 100%, such as 5%, 10%, 20%, and 25%, which is helpful for Case 1 but would be unnecessary for Case 2 to the point of being a misleading contrivance.
So the clear intent in the text is to use your Case 1 for "cumulative chances."
answered 17 hours ago
Bloodcinder
16.3k255110
add a comment |Â
up vote
4
down vote
The "cumulative chances" add together (use Case 1).
There is no rule which explicitly clarifies whether your Case 1 or Case 2 is the correct way to adjudicate "cumulative chance." However, one of the design principles of D&D 5e is for mathematical calculations to be straightforward and simple to compute in your head.
Case 1 corresponds to simple arithmetic you can do in your head: the second time you cast the spell you have a 25% chance of failure, then 50% the next, then 75% the next, and so on (just by adding a flat 25% each time).
Case 2 corresponds to probability computations that most people can't do in their head: the second time you cast the spell you have a 25% chance of failure, then 43.75% the next time, then 57.8125% the next time, and so on (by using the principle of inclusion and exclusion in discrete probability).
Just because you can easily describe how to use d4's to do this math indirectly without doing any hard calculations doesn't mean that the text expects you to do that, otherwise there would be something in the text to explain how to do that. While there is a passage explaining how to use d% dice, there is no passage explaining how to use a d4 to simulate a 25% chance of failure, and without such guidance a person who doesn't see how to use d4's to solve the problem would be forced to do a lot of hard math using Case 2, whereas the rules already explain how to do Case 1.
It's also worth noting that all instances of this "cumulative chance" concept in the DMG and PHB use percentages that divide evenly into 100%, such as 5%, 10%, 20%, and 25%, which is helpful for Case 1 but would be unnecessary for Case 2 to the point of being a misleading contrivance.
So the clear intent in the text is to use your Case 1 for "cumulative chances."
answered 17 hours ago
Bloodcinder
16.3k255110
add a comment |Â
up vote
4
down vote
up vote
4
down vote
The "cumulative chances" add together (use Case 1).
There is no rule which explicitly clarifies whether your Case 1 or Case 2 is the correct way to adjudicate "cumulative chance." However, one of the design principles of D&D 5e is for mathematical calculations to be straightforward and simple to compute in your head.
Case 1 corresponds to simple arithmetic you can do in your head: the second time you cast the spell you have a 25% chance of failure, then 50% the next, then 75% the next, and so on (just by adding a flat 25% each time).
Case 2 corresponds to probability computations that most people can't do in their head: the second time you cast the spell you have a 25% chance of failure, then 43.75% the next time, then 57.8125% the next time, and so on (by using the principle of inclusion and exclusion in discrete probability).
Just because you can easily describe how to use d4's to do this math indirectly without doing any hard calculations doesn't mean that the text expects you to do that, otherwise there would be something in the text to explain how to do that. While there is a passage explaining how to use d% dice, there is no passage explaining how to use a d4 to simulate a 25% chance of failure, and without such guidance a person who doesn't see how to use d4's to solve the problem would be forced to do a lot of hard math using Case 2, whereas the rules already explain how to do Case 1.
It's also worth noting that all instances of this "cumulative chance" concept in the DMG and PHB use percentages that divide evenly into 100%, such as 5%, 10%, 20%, and 25%, which is helpful for Case 1 but would be unnecessary for Case 2 to the point of being a misleading contrivance.
So the clear intent in the text is to use your Case 1 for "cumulative chances."
answered 17 hours ago
Bloodcinder
16.3k255110
The "cumulative chances" add together (use Case 1).
There is no rule which explicitly clarifies whether your Case 1 or Case 2 is the correct way to adjudicate "cumulative chance." However, one of the design principles of D&D 5e is for mathematical calculations to be straightforward and simple to compute in your head.
Case 1 corresponds to simple arithmetic you can do in your head: the second time you cast the spell you have a 25% chance of failure, then 50% the next, then 75% the next, and so on (just by adding a flat 25% each time).
Case 2 corresponds to probability computations that most people can't do in their head: the second time you cast the spell you have a 25% chance of failure, then 43.75% the next time, then 57.8125% the next time, and so on (by using the principle of inclusion and exclusion in discrete probability).
Just because you can easily describe how to use d4's to do this math indirectly without doing any hard calculations doesn't mean that the text expects you to do that, otherwise there would be something in the text to explain how to do that. While there is a passage explaining how to use d% dice, there is no passage explaining how to use a d4 to simulate a 25% chance of failure, and without such guidance a person who doesn't see how to use d4's to solve the problem would be forced to do a lot of hard math using Case 2, whereas the rules already explain how to do Case 1.
It's also worth noting that all instances of this "cumulative chance" concept in the DMG and PHB use percentages that divide evenly into 100%, such as 5%, 10%, 20%, and 25%, which is helpful for Case 1 but would be unnecessary for Case 2 to the point of being a misleading contrivance.
So the clear intent in the text is to use your Case 1 for "cumulative chances."
answered 17 hours ago
Bloodcinder
16.3k255110
edited 17 hours ago
answered 17 hours ago
Bloodcinder
16.3k255110
answered 17 hours ago
Bloodcinder
16.3k255110
answered 17 hours ago
Bloodcinder
16.3k255110
16.3k255110
add a comment |Â
add a comment |Â
up vote
1
down vote
Your option 1 is correct: Roll a single d4.
The 2nd casting has a 25% chance of failing. This is the same as 1 in 4.
The 3rd casting has a 50% chance of failing. This is the same as 2 in 4.
The 4th casting has a 75% chance of failing. This is the same as 3 in 4.
You can always avoid the maths by using the percentage value and rolling a d100 (roll a d10 twice - the first roll is the 'tens', the second roll is the 'units'). It amounts to the same thing.
answered 21 hours ago
PJRZ
5,0571331
add a comment |Â
up vote
1
down vote
Your option 1 is correct: Roll a single d4.
The 2nd casting has a 25% chance of failing. This is the same as 1 in 4.
The 3rd casting has a 50% chance of failing. This is the same as 2 in 4.
The 4th casting has a 75% chance of failing. This is the same as 3 in 4.
You can always avoid the maths by using the percentage value and rolling a d100 (roll a d10 twice - the first roll is the 'tens', the second roll is the 'units'). It amounts to the same thing.
answered 21 hours ago
PJRZ
5,0571331
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Your option 1 is correct: Roll a single d4.
The 2nd casting has a 25% chance of failing. This is the same as 1 in 4.
The 3rd casting has a 50% chance of failing. This is the same as 2 in 4.
The 4th casting has a 75% chance of failing. This is the same as 3 in 4.
You can always avoid the maths by using the percentage value and rolling a d100 (roll a d10 twice - the first roll is the 'tens', the second roll is the 'units'). It amounts to the same thing.
answered 21 hours ago
PJRZ
5,0571331
Your option 1 is correct: Roll a single d4.
The 2nd casting has a 25% chance of failing. This is the same as 1 in 4.
The 3rd casting has a 50% chance of failing. This is the same as 2 in 4.
The 4th casting has a 75% chance of failing. This is the same as 3 in 4.
You can always avoid the maths by using the percentage value and rolling a d100 (roll a d10 twice - the first roll is the 'tens', the second roll is the 'units'). It amounts to the same thing.
answered 21 hours ago
PJRZ
5,0571331
answered 21 hours ago
PJRZ
5,0571331
answered 21 hours ago
PJRZ
5,0571331
answered 21 hours ago
PJRZ
5,0571331
5,0571331
add a comment |Â
add a comment |Â
up vote
-1
down vote
- Roll a d100. If the result is equal to or lower than 25 * (number of times cast - 1), give a random reading.
This is effectively the same as option 1.
answered 21 hours ago
Derek Stucki
19.6k665104
add a comment |Â
up vote
-1
down vote
- Roll a d100. If the result is equal to or lower than 25 * (number of times cast - 1), give a random reading.
This is effectively the same as option 1.
answered 21 hours ago
Derek Stucki
19.6k665104
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
- Roll a d100. If the result is equal to or lower than 25 * (number of times cast - 1), give a random reading.
This is effectively the same as option 1.
answered 21 hours ago
Derek Stucki
19.6k665104
- Roll a d100. If the result is equal to or lower than 25 * (number of times cast - 1), give a random reading.
This is effectively the same as option 1.
answered 21 hours ago
Derek Stucki
19.6k665104
edited 21 hours ago
answered 21 hours ago
Derek Stucki
19.6k665104
answered 21 hours ago
Derek Stucki
19.6k665104
answered 21 hours ago
Derek Stucki
19.6k665104
19.6k665104
add a comment |Â
add a comment |Â
up vote
-1
down vote
You could also roll a d% (it's the one that goes 0, 10, 20, etc.) and a d10 at the same time, to get the same as a d100.
"If you cast the spell two or more times before completing your next long rest, there is a cumulative 25 percent chance for each casting after the first that you
get a random reading. The DM makes this roll in secret."
Cumulative: increasing or increased in quantity, degree, or force by successive additions
You could just take the long nap in between, but if you DO cast Augury more times before your next long rest, then it's up to the DM to choose how he/she wants to roll. I don't think it does fail, per say, but it does give a chance to give the player(s) what they don't want. For throwing outcomes, I think it'd be:
0%: First throw,
25%: Second throw,
50%: Third throw,
75%: Fourth throw,
100%: Fifth+ throw
Each percentage would be something that they did not expect (or maybe it just goes alright). I'd suggest having the d%, d10, and the d4 (if you are the DM) handy if the Cleric - since that seems to be the only class that can use it - is going to use the Augury spell.
I would treat the percentages like the Sorcerer's Wild Magic; if it is below the certain percentage (where d% and d10 come in), then the DM rolls the d4 to see which omen the player gets. If it is above, then they get what they wanted. They would have to vocally state what they wanted.
answered 19 hours ago
Dragonman0007
195
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Dragonman0007 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |Â
up vote
-1
down vote
You could also roll a d% (it's the one that goes 0, 10, 20, etc.) and a d10 at the same time, to get the same as a d100.
"If you cast the spell two or more times before completing your next long rest, there is a cumulative 25 percent chance for each casting after the first that you
get a random reading. The DM makes this roll in secret."
Cumulative: increasing or increased in quantity, degree, or force by successive additions
You could just take the long nap in between, but if you DO cast Augury more times before your next long rest, then it's up to the DM to choose how he/she wants to roll. I don't think it does fail, per say, but it does give a chance to give the player(s) what they don't want. For throwing outcomes, I think it'd be:
0%: First throw,
25%: Second throw,
50%: Third throw,
75%: Fourth throw,
100%: Fifth+ throw
Each percentage would be something that they did not expect (or maybe it just goes alright). I'd suggest having the d%, d10, and the d4 (if you are the DM) handy if the Cleric - since that seems to be the only class that can use it - is going to use the Augury spell.
I would treat the percentages like the Sorcerer's Wild Magic; if it is below the certain percentage (where d% and d10 come in), then the DM rolls the d4 to see which omen the player gets. If it is above, then they get what they wanted. They would have to vocally state what they wanted.
answered 19 hours ago
Dragonman0007
195
New contributor
Dragonman0007 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
You could also roll a d% (it's the one that goes 0, 10, 20, etc.) and a d10 at the same time, to get the same as a d100.
"If you cast the spell two or more times before completing your next long rest, there is a cumulative 25 percent chance for each casting after the first that you
get a random reading. The DM makes this roll in secret."
Cumulative: increasing or increased in quantity, degree, or force by successive additions
You could just take the long nap in between, but if you DO cast Augury more times before your next long rest, then it's up to the DM to choose how he/she wants to roll. I don't think it does fail, per say, but it does give a chance to give the player(s) what they don't want. For throwing outcomes, I think it'd be:
0%: First throw,
25%: Second throw,
50%: Third throw,
75%: Fourth throw,
100%: Fifth+ throw
Each percentage would be something that they did not expect (or maybe it just goes alright). I'd suggest having the d%, d10, and the d4 (if you are the DM) handy if the Cleric - since that seems to be the only class that can use it - is going to use the Augury spell.
I would treat the percentages like the Sorcerer's Wild Magic; if it is below the certain percentage (where d% and d10 come in), then the DM rolls the d4 to see which omen the player gets. If it is above, then they get what they wanted. They would have to vocally state what they wanted.
answered 19 hours ago
Dragonman0007
195
New contributor
Dragonman0007 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You could also roll a d% (it's the one that goes 0, 10, 20, etc.) and a d10 at the same time, to get the same as a d100.
"If you cast the spell two or more times before completing your next long rest, there is a cumulative 25 percent chance for each casting after the first that you
get a random reading. The DM makes this roll in secret."
Cumulative: increasing or increased in quantity, degree, or force by successive additions
You could just take the long nap in between, but if you DO cast Augury more times before your next long rest, then it's up to the DM to choose how he/she wants to roll. I don't think it does fail, per say, but it does give a chance to give the player(s) what they don't want. For throwing outcomes, I think it'd be:
0%: First throw,
25%: Second throw,
50%: Third throw,
75%: Fourth throw,
100%: Fifth+ throw
Each percentage would be something that they did not expect (or maybe it just goes alright). I'd suggest having the d%, d10, and the d4 (if you are the DM) handy if the Cleric - since that seems to be the only class that can use it - is going to use the Augury spell.
I would treat the percentages like the Sorcerer's Wild Magic; if it is below the certain percentage (where d% and d10 come in), then the DM rolls the d4 to see which omen the player gets. If it is above, then they get what they wanted. They would have to vocally state what they wanted.
answered 19 hours ago
Dragonman0007
195
New contributor
Dragonman0007 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 19 hours ago
answered 19 hours ago
Dragonman0007
195
New contributor
Dragonman0007 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 19 hours ago
Dragonman0007
195
answered 19 hours ago
Dragonman0007
195
195
New contributor
Dragonman0007 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Dragonman0007 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Dragonman0007 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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2d4 can never result in 1. Do you mean taking the lowest of 2d4 (similar to disadvantage)?
– Ruse
20 hours ago