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Does this AnyDice function accurately calculate the number of ogres you make unconcious with three 4th-level castings of Sleep?
Help Calculating Dice Odds for a Unique d6 Dice PoolHow can I execute code depending on a die value in AnyDice?How do I find the highest number rolled in a pool if dice, and the number of times it is rolled using AnyDice?Help Writing an AnyDice Function for a Weird Dice MechanicAnyDice function to use a die as high/low flip and criticalImplementing multiattack in AnyDiceHow do I model the fighter's Great Weapon Fighting fighting style in Anydice?How can I write and compare these formulas using Anydice?Exploding “D9” success-counting pools in anydice?AnyDice formula for rolling 2d20 and dropping farthest value from 10
$begingroup$
Say we have three ogres, and want to put them to sleep as a handful of level 7 casters with the sleep spell. We want to know how many of them, on average, will get put to sleep.
I tried to make an AnyDice function that searches a listing of all of the results of 11d8 for the HP totals that would put to sleep one of the ogres (using the average of 59 HP). Is this AnyDice function accurate to the goal?
output [count 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88 in 3d(11d8)]
If not, how would I properly create the function?
dnd-5e anydice
edited 17 hours ago
V2Blast
24.6k383155
asked 20 hours ago
David CoffronDavid Coffron
37.6k3129265
$endgroup$
add a comment |
$begingroup$
Say we have three ogres, and want to put them to sleep as a handful of level 7 casters with the sleep spell. We want to know how many of them, on average, will get put to sleep.
I tried to make an AnyDice function that searches a listing of all of the results of 11d8 for the HP totals that would put to sleep one of the ogres (using the average of 59 HP). Is this AnyDice function accurate to the goal?
output [count 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88 in 3d(11d8)]
If not, how would I properly create the function?
dnd-5e anydice
edited 17 hours ago
V2Blast
24.6k383155
asked 20 hours ago
David CoffronDavid Coffron
37.6k3129265
$endgroup$
add a comment |
$begingroup$
Say we have three ogres, and want to put them to sleep as a handful of level 7 casters with the sleep spell. We want to know how many of them, on average, will get put to sleep.
I tried to make an AnyDice function that searches a listing of all of the results of 11d8 for the HP totals that would put to sleep one of the ogres (using the average of 59 HP). Is this AnyDice function accurate to the goal?
output [count 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88 in 3d(11d8)]
If not, how would I properly create the function?
dnd-5e anydice
edited 17 hours ago
V2Blast
24.6k383155
asked 20 hours ago
David CoffronDavid Coffron
37.6k3129265
$endgroup$
Say we have three ogres, and want to put them to sleep as a handful of level 7 casters with the sleep spell. We want to know how many of them, on average, will get put to sleep.
I tried to make an AnyDice function that searches a listing of all of the results of 11d8 for the HP totals that would put to sleep one of the ogres (using the average of 59 HP). Is this AnyDice function accurate to the goal?
output [count 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88 in 3d(11d8)]
If not, how would I properly create the function?
dnd-5e anydice
dnd-5e anydice
edited 17 hours ago
V2Blast
24.6k383155
asked 20 hours ago
David CoffronDavid Coffron
37.6k3129265
edited 17 hours ago
V2Blast
24.6k383155
asked 20 hours ago
David CoffronDavid Coffron
37.6k3129265
edited 17 hours ago
V2Blast
24.6k383155
edited 17 hours ago
V2Blast
24.6k383155
edited 17 hours ago
V2Blast
24.6k383155
24.6k383155
asked 20 hours ago
David CoffronDavid Coffron
37.6k3129265
asked 20 hours ago
David CoffronDavid Coffron
37.6k3129265
asked 20 hours ago
David CoffronDavid Coffron
37.6k3129265
37.6k3129265
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The function is correct but can be made simpler
The function makes sense from a theoretical point of view (you are counting the number of times one of the listed numbers appears when doing 11d8 three times) and from practical corroboration from Xirema's answer.
However, it seems to be a particularly taxing function for anydice. Indeed, when changing it to 4 times instead of 3, anydice refused to provide results because it exceeded the maximum execution time.
A simpler and faster way to do the same function would be this anydice function:
output 3d(11d8 > 58)
Which is equivalent, also helps show that your initial thoughts were correct but is much faster and you can also use it for bigger values without anydice complaining.
Naturally this is not as versatile as your specific listing of each possible result but since you only care about results bigger than the creature's HP in this case it may work better.
Alternatively, as proposed by Ilmari Karonen, you could also use, for some more versatility in the results, this function:
output 3d[count 59..88 in 11d8 + 0]
(The + 0 makes AnyDice sum the 11d8 roll before passing it to [count VALUES in SEQUENCE])
answered 19 hours ago
SdjzSdjz
13.3k463108
$endgroup$
2
$begingroup$
Reason #473 I should really learn how AnyDice actually works..... Good answer.
$endgroup$
– Xirema
13 hours ago
$begingroup$
@IlmariKaronen Alright, I appreciate the contribution
$endgroup$
– Sdjz
12 hours ago
add a comment |
$begingroup$
This function appears to be correct
The output of the function in Anydice is
beginarrayl
hline
textNumber & textProbability \ hline
0 & 68.19% \ hline
1 & 27.85% \ hline
2 & 3.79% \ hline
3 & 0.17% \ hline
endarray
Using an independent method, I was able to calculate [within an acceptable Margin of Error] identical odds:
beginarray
hline
textNumber & textTrials & textProbability \ hline
0 & 432224441369339628206761607168 (4.322 * 10^29) & 68.1930% \ hline
1 & 176489190284293025714047057920 (1.765 * 10^29) & 27.8415% \ hline
2 & 24021805421679657469725081600 (2.402 * 10^28) & 3.7900% \ hline
3 & 1089863038802389357817856000 (1.090 * 10^27) & 0.1720% \ hline
endarray
So because these two methods appear to have identical outcomes, I'm led to conclude that the function you've provided is correct.
answered 19 hours ago
XiremaXirema
21.4k263126
$endgroup$
2
$begingroup$
may I ask what your independent method was?
$endgroup$
– Carmeister
9 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The function is correct but can be made simpler
The function makes sense from a theoretical point of view (you are counting the number of times one of the listed numbers appears when doing 11d8 three times) and from practical corroboration from Xirema's answer.
However, it seems to be a particularly taxing function for anydice. Indeed, when changing it to 4 times instead of 3, anydice refused to provide results because it exceeded the maximum execution time.
A simpler and faster way to do the same function would be this anydice function:
output 3d(11d8 > 58)
Which is equivalent, also helps show that your initial thoughts were correct but is much faster and you can also use it for bigger values without anydice complaining.
Naturally this is not as versatile as your specific listing of each possible result but since you only care about results bigger than the creature's HP in this case it may work better.
Alternatively, as proposed by Ilmari Karonen, you could also use, for some more versatility in the results, this function:
output 3d[count 59..88 in 11d8 + 0]
(The + 0 makes AnyDice sum the 11d8 roll before passing it to [count VALUES in SEQUENCE])
answered 19 hours ago
SdjzSdjz
13.3k463108
$endgroup$
2
$begingroup$
Reason #473 I should really learn how AnyDice actually works..... Good answer.
$endgroup$
– Xirema
13 hours ago
$begingroup$
@IlmariKaronen Alright, I appreciate the contribution
$endgroup$
– Sdjz
12 hours ago
add a comment |
$begingroup$
The function is correct but can be made simpler
The function makes sense from a theoretical point of view (you are counting the number of times one of the listed numbers appears when doing 11d8 three times) and from practical corroboration from Xirema's answer.
However, it seems to be a particularly taxing function for anydice. Indeed, when changing it to 4 times instead of 3, anydice refused to provide results because it exceeded the maximum execution time.
A simpler and faster way to do the same function would be this anydice function:
output 3d(11d8 > 58)
Which is equivalent, also helps show that your initial thoughts were correct but is much faster and you can also use it for bigger values without anydice complaining.
Naturally this is not as versatile as your specific listing of each possible result but since you only care about results bigger than the creature's HP in this case it may work better.
Alternatively, as proposed by Ilmari Karonen, you could also use, for some more versatility in the results, this function:
output 3d[count 59..88 in 11d8 + 0]
(The + 0 makes AnyDice sum the 11d8 roll before passing it to [count VALUES in SEQUENCE])
answered 19 hours ago
SdjzSdjz
13.3k463108
$endgroup$
2
$begingroup$
Reason #473 I should really learn how AnyDice actually works..... Good answer.
$endgroup$
– Xirema
13 hours ago
$begingroup$
@IlmariKaronen Alright, I appreciate the contribution
$endgroup$
– Sdjz
12 hours ago
add a comment |
$begingroup$
The function is correct but can be made simpler
The function makes sense from a theoretical point of view (you are counting the number of times one of the listed numbers appears when doing 11d8 three times) and from practical corroboration from Xirema's answer.
However, it seems to be a particularly taxing function for anydice. Indeed, when changing it to 4 times instead of 3, anydice refused to provide results because it exceeded the maximum execution time.
A simpler and faster way to do the same function would be this anydice function:
output 3d(11d8 > 58)
Which is equivalent, also helps show that your initial thoughts were correct but is much faster and you can also use it for bigger values without anydice complaining.
Naturally this is not as versatile as your specific listing of each possible result but since you only care about results bigger than the creature's HP in this case it may work better.
Alternatively, as proposed by Ilmari Karonen, you could also use, for some more versatility in the results, this function:
output 3d[count 59..88 in 11d8 + 0]
(The + 0 makes AnyDice sum the 11d8 roll before passing it to [count VALUES in SEQUENCE])
answered 19 hours ago
SdjzSdjz
13.3k463108
$endgroup$
The function is correct but can be made simpler
The function makes sense from a theoretical point of view (you are counting the number of times one of the listed numbers appears when doing 11d8 three times) and from practical corroboration from Xirema's answer.
However, it seems to be a particularly taxing function for anydice. Indeed, when changing it to 4 times instead of 3, anydice refused to provide results because it exceeded the maximum execution time.
A simpler and faster way to do the same function would be this anydice function:
output 3d(11d8 > 58)
Which is equivalent, also helps show that your initial thoughts were correct but is much faster and you can also use it for bigger values without anydice complaining.
Naturally this is not as versatile as your specific listing of each possible result but since you only care about results bigger than the creature's HP in this case it may work better.
Alternatively, as proposed by Ilmari Karonen, you could also use, for some more versatility in the results, this function:
output 3d[count 59..88 in 11d8 + 0]
(The + 0 makes AnyDice sum the 11d8 roll before passing it to [count VALUES in SEQUENCE])
answered 19 hours ago
SdjzSdjz
13.3k463108
edited 12 hours ago
answered 19 hours ago
SdjzSdjz
13.3k463108
answered 19 hours ago
SdjzSdjz
13.3k463108
answered 19 hours ago
SdjzSdjz
13.3k463108
13.3k463108
2
$begingroup$
Reason #473 I should really learn how AnyDice actually works..... Good answer.
$endgroup$
– Xirema
13 hours ago
$begingroup$
@IlmariKaronen Alright, I appreciate the contribution
$endgroup$
– Sdjz
12 hours ago
add a comment |
2
$begingroup$
Reason #473 I should really learn how AnyDice actually works..... Good answer.
$endgroup$
– Xirema
13 hours ago
$begingroup$
@IlmariKaronen Alright, I appreciate the contribution
$endgroup$
– Sdjz
12 hours ago
2
2
$begingroup$
Reason #473 I should really learn how AnyDice actually works..... Good answer.
$endgroup$
– Xirema
13 hours ago
$begingroup$
Reason #473 I should really learn how AnyDice actually works..... Good answer.
$endgroup$
– Xirema
13 hours ago
$begingroup$
@IlmariKaronen Alright, I appreciate the contribution
$endgroup$
– Sdjz
12 hours ago
$begingroup$
@IlmariKaronen Alright, I appreciate the contribution
$endgroup$
– Sdjz
12 hours ago
add a comment |
$begingroup$
This function appears to be correct
The output of the function in Anydice is
beginarrayl
hline
textNumber & textProbability \ hline
0 & 68.19% \ hline
1 & 27.85% \ hline
2 & 3.79% \ hline
3 & 0.17% \ hline
endarray
Using an independent method, I was able to calculate [within an acceptable Margin of Error] identical odds:
beginarray
hline
textNumber & textTrials & textProbability \ hline
0 & 432224441369339628206761607168 (4.322 * 10^29) & 68.1930% \ hline
1 & 176489190284293025714047057920 (1.765 * 10^29) & 27.8415% \ hline
2 & 24021805421679657469725081600 (2.402 * 10^28) & 3.7900% \ hline
3 & 1089863038802389357817856000 (1.090 * 10^27) & 0.1720% \ hline
endarray
So because these two methods appear to have identical outcomes, I'm led to conclude that the function you've provided is correct.
answered 19 hours ago
XiremaXirema
21.4k263126
$endgroup$
2
$begingroup$
may I ask what your independent method was?
$endgroup$
– Carmeister
9 hours ago
add a comment |
$begingroup$
This function appears to be correct
The output of the function in Anydice is
beginarrayl
hline
textNumber & textProbability \ hline
0 & 68.19% \ hline
1 & 27.85% \ hline
2 & 3.79% \ hline
3 & 0.17% \ hline
endarray
Using an independent method, I was able to calculate [within an acceptable Margin of Error] identical odds:
beginarray
hline
textNumber & textTrials & textProbability \ hline
0 & 432224441369339628206761607168 (4.322 * 10^29) & 68.1930% \ hline
1 & 176489190284293025714047057920 (1.765 * 10^29) & 27.8415% \ hline
2 & 24021805421679657469725081600 (2.402 * 10^28) & 3.7900% \ hline
3 & 1089863038802389357817856000 (1.090 * 10^27) & 0.1720% \ hline
endarray
So because these two methods appear to have identical outcomes, I'm led to conclude that the function you've provided is correct.
answered 19 hours ago
XiremaXirema
21.4k263126
$endgroup$
2
$begingroup$
may I ask what your independent method was?
$endgroup$
– Carmeister
9 hours ago
add a comment |
$begingroup$
This function appears to be correct
The output of the function in Anydice is
beginarrayl
hline
textNumber & textProbability \ hline
0 & 68.19% \ hline
1 & 27.85% \ hline
2 & 3.79% \ hline
3 & 0.17% \ hline
endarray
Using an independent method, I was able to calculate [within an acceptable Margin of Error] identical odds:
beginarray
hline
textNumber & textTrials & textProbability \ hline
0 & 432224441369339628206761607168 (4.322 * 10^29) & 68.1930% \ hline
1 & 176489190284293025714047057920 (1.765 * 10^29) & 27.8415% \ hline
2 & 24021805421679657469725081600 (2.402 * 10^28) & 3.7900% \ hline
3 & 1089863038802389357817856000 (1.090 * 10^27) & 0.1720% \ hline
endarray
So because these two methods appear to have identical outcomes, I'm led to conclude that the function you've provided is correct.
answered 19 hours ago
XiremaXirema
21.4k263126
$endgroup$
This function appears to be correct
The output of the function in Anydice is
beginarrayl
hline
textNumber & textProbability \ hline
0 & 68.19% \ hline
1 & 27.85% \ hline
2 & 3.79% \ hline
3 & 0.17% \ hline
endarray
Using an independent method, I was able to calculate [within an acceptable Margin of Error] identical odds:
beginarray
hline
textNumber & textTrials & textProbability \ hline
0 & 432224441369339628206761607168 (4.322 * 10^29) & 68.1930% \ hline
1 & 176489190284293025714047057920 (1.765 * 10^29) & 27.8415% \ hline
2 & 24021805421679657469725081600 (2.402 * 10^28) & 3.7900% \ hline
3 & 1089863038802389357817856000 (1.090 * 10^27) & 0.1720% \ hline
endarray
So because these two methods appear to have identical outcomes, I'm led to conclude that the function you've provided is correct.
answered 19 hours ago
XiremaXirema
21.4k263126
edited 19 hours ago
answered 19 hours ago
XiremaXirema
21.4k263126
answered 19 hours ago
XiremaXirema
21.4k263126
answered 19 hours ago
XiremaXirema
21.4k263126
21.4k263126
2
$begingroup$
may I ask what your independent method was?
$endgroup$
– Carmeister
9 hours ago
add a comment |
2
$begingroup$
may I ask what your independent method was?
$endgroup$
– Carmeister
9 hours ago
2
2
$begingroup$
may I ask what your independent method was?
$endgroup$
– Carmeister
9 hours ago
$begingroup$
may I ask what your independent method was?
$endgroup$
– Carmeister
9 hours ago
add a comment |
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