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Dice probability being modified
Clash Royale CLAN TAG#URR8PPP
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So for a roll20 D&D game I been working on I decided it'd be nice to have a macro that would calculate the probability of successfully hitting an enemy and showing it as a percentage.
To hit an enemy you must roll higher on your accuracy than their evasion, and both of these rolls are D20s. However, both accuracy and evasion are increased/decreased by various attributes and proficiencies.
From what I remember probability is the winning possibilities / total possibilities.
What I'm having problems with though is figuring out what the winning possibilities are when the accuracy/evasion bonuses/debuffs are applied.
I know that using triangular numbers you could calculate the chances of rolling a number equal to or higher than a 2nd dice roll. However, I do not know what math formula to use to calculate the modifiers.
For example: This is the possibilities for 2d6 with no modifiers
beginarrayc
&1&2&3&4&5&6 \ hline \
1&=&<&<&<&<&< \
2&>&=&<&<&<&< \
3&>&>&=&<&<&< \
4&>&>&>&=&<&< \
5&>&>&>&>&=&< \
6&>&>&>&>&>&= \
endarray
However, if you had a +1 added to your rolls then the possibilities for 2d6 would look like this:
beginarrayc
&1&2&3&4&5&6 \ hline \
1&<&<&<&<&<&< \
2&=&<&<&<&<&< \
3&>&=&<&<&<&< \
4&>&>&=&<&<&< \
5&>&>&>&=&<&< \
6&>&>&>&>&=&< \
endarray
So my problem is I can't figure out the formula used to calculate the increase each time. No mod 2d6 is 21 wins, but +1 would by 26 (+5), +2 would be 30 (+9) and +3 would be 33 (+12)
I apologise if this has been asked before in some way and I haven't found it. My internet has been pretty bad today so the amount of searching I done was limited.
probability sequences-and-series number-theory
asked Jul 26 at 20:25
Sargent179
31
add a comment |Â
up vote
0
down vote
favorite
So for a roll20 D&D game I been working on I decided it'd be nice to have a macro that would calculate the probability of successfully hitting an enemy and showing it as a percentage.
To hit an enemy you must roll higher on your accuracy than their evasion, and both of these rolls are D20s. However, both accuracy and evasion are increased/decreased by various attributes and proficiencies.
From what I remember probability is the winning possibilities / total possibilities.
What I'm having problems with though is figuring out what the winning possibilities are when the accuracy/evasion bonuses/debuffs are applied.
I know that using triangular numbers you could calculate the chances of rolling a number equal to or higher than a 2nd dice roll. However, I do not know what math formula to use to calculate the modifiers.
For example: This is the possibilities for 2d6 with no modifiers
beginarrayc
&1&2&3&4&5&6 \ hline \
1&=&<&<&<&<&< \
2&>&=&<&<&<&< \
3&>&>&=&<&<&< \
4&>&>&>&=&<&< \
5&>&>&>&>&=&< \
6&>&>&>&>&>&= \
endarray
However, if you had a +1 added to your rolls then the possibilities for 2d6 would look like this:
beginarrayc
&1&2&3&4&5&6 \ hline \
1&<&<&<&<&<&< \
2&=&<&<&<&<&< \
3&>&=&<&<&<&< \
4&>&>&=&<&<&< \
5&>&>&>&=&<&< \
6&>&>&>&>&=&< \
endarray
So my problem is I can't figure out the formula used to calculate the increase each time. No mod 2d6 is 21 wins, but +1 would by 26 (+5), +2 would be 30 (+9) and +3 would be 33 (+12)
I apologise if this has been asked before in some way and I haven't found it. My internet has been pretty bad today so the amount of searching I done was limited.
probability sequences-and-series number-theory
asked Jul 26 at 20:25
Sargent179
31
It appears that you are looking at two dice rolls. You are looking for the number of ways that one die roll can beat another die roll when one of the dice is modified. Is that correct?
– InterstellarProbe
Jul 26 at 20:31
@InterstellarProbe For the D&D style combat system that i've been making the attacker would roll accuracy and then the defender would roll evasion. Depending on a character's stats they may get bonuses to their accuracy or evasion. So Both dice could be modified or have no modifiers at all. My goal is to make a macro that would calculate what the probability is of you (the attacker) rolling higher than or equal to the defender with whatever bonuses both players may have.
– Sargent179
Jul 26 at 22:20
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
So for a roll20 D&D game I been working on I decided it'd be nice to have a macro that would calculate the probability of successfully hitting an enemy and showing it as a percentage.
To hit an enemy you must roll higher on your accuracy than their evasion, and both of these rolls are D20s. However, both accuracy and evasion are increased/decreased by various attributes and proficiencies.
From what I remember probability is the winning possibilities / total possibilities.
What I'm having problems with though is figuring out what the winning possibilities are when the accuracy/evasion bonuses/debuffs are applied.
I know that using triangular numbers you could calculate the chances of rolling a number equal to or higher than a 2nd dice roll. However, I do not know what math formula to use to calculate the modifiers.
For example: This is the possibilities for 2d6 with no modifiers
beginarrayc
&1&2&3&4&5&6 \ hline \
1&=&<&<&<&<&< \
2&>&=&<&<&<&< \
3&>&>&=&<&<&< \
4&>&>&>&=&<&< \
5&>&>&>&>&=&< \
6&>&>&>&>&>&= \
endarray
However, if you had a +1 added to your rolls then the possibilities for 2d6 would look like this:
beginarrayc
&1&2&3&4&5&6 \ hline \
1&<&<&<&<&<&< \
2&=&<&<&<&<&< \
3&>&=&<&<&<&< \
4&>&>&=&<&<&< \
5&>&>&>&=&<&< \
6&>&>&>&>&=&< \
endarray
So my problem is I can't figure out the formula used to calculate the increase each time. No mod 2d6 is 21 wins, but +1 would by 26 (+5), +2 would be 30 (+9) and +3 would be 33 (+12)
I apologise if this has been asked before in some way and I haven't found it. My internet has been pretty bad today so the amount of searching I done was limited.
probability sequences-and-series number-theory
asked Jul 26 at 20:25
Sargent179
31
So for a roll20 D&D game I been working on I decided it'd be nice to have a macro that would calculate the probability of successfully hitting an enemy and showing it as a percentage.
To hit an enemy you must roll higher on your accuracy than their evasion, and both of these rolls are D20s. However, both accuracy and evasion are increased/decreased by various attributes and proficiencies.
From what I remember probability is the winning possibilities / total possibilities.
What I'm having problems with though is figuring out what the winning possibilities are when the accuracy/evasion bonuses/debuffs are applied.
I know that using triangular numbers you could calculate the chances of rolling a number equal to or higher than a 2nd dice roll. However, I do not know what math formula to use to calculate the modifiers.
For example: This is the possibilities for 2d6 with no modifiers
beginarrayc
&1&2&3&4&5&6 \ hline \
1&=&<&<&<&<&< \
2&>&=&<&<&<&< \
3&>&>&=&<&<&< \
4&>&>&>&=&<&< \
5&>&>&>&>&=&< \
6&>&>&>&>&>&= \
endarray
However, if you had a +1 added to your rolls then the possibilities for 2d6 would look like this:
beginarrayc
&1&2&3&4&5&6 \ hline \
1&<&<&<&<&<&< \
2&=&<&<&<&<&< \
3&>&=&<&<&<&< \
4&>&>&=&<&<&< \
5&>&>&>&=&<&< \
6&>&>&>&>&=&< \
endarray
So my problem is I can't figure out the formula used to calculate the increase each time. No mod 2d6 is 21 wins, but +1 would by 26 (+5), +2 would be 30 (+9) and +3 would be 33 (+12)
I apologise if this has been asked before in some way and I haven't found it. My internet has been pretty bad today so the amount of searching I done was limited.
probability sequences-and-series number-theory
asked Jul 26 at 20:25
Sargent179
31
asked Jul 26 at 20:25
Sargent179
31
asked Jul 26 at 20:25
Sargent179
31
asked Jul 26 at 20:25
Sargent179
31
31
It appears that you are looking at two dice rolls. You are looking for the number of ways that one die roll can beat another die roll when one of the dice is modified. Is that correct?
– InterstellarProbe
Jul 26 at 20:31
@InterstellarProbe For the D&D style combat system that i've been making the attacker would roll accuracy and then the defender would roll evasion. Depending on a character's stats they may get bonuses to their accuracy or evasion. So Both dice could be modified or have no modifiers at all. My goal is to make a macro that would calculate what the probability is of you (the attacker) rolling higher than or equal to the defender with whatever bonuses both players may have.
– Sargent179
Jul 26 at 22:20
add a comment |Â
It appears that you are looking at two dice rolls. You are looking for the number of ways that one die roll can beat another die roll when one of the dice is modified. Is that correct?
– InterstellarProbe
Jul 26 at 20:31
@InterstellarProbe For the D&D style combat system that i've been making the attacker would roll accuracy and then the defender would roll evasion. Depending on a character's stats they may get bonuses to their accuracy or evasion. So Both dice could be modified or have no modifiers at all. My goal is to make a macro that would calculate what the probability is of you (the attacker) rolling higher than or equal to the defender with whatever bonuses both players may have.
– Sargent179
Jul 26 at 22:20
It appears that you are looking at two dice rolls. You are looking for the number of ways that one die roll can beat another die roll when one of the dice is modified. Is that correct?
– InterstellarProbe
Jul 26 at 20:31
It appears that you are looking at two dice rolls. You are looking for the number of ways that one die roll can beat another die roll when one of the dice is modified. Is that correct?
– InterstellarProbe
Jul 26 at 20:31
@InterstellarProbe For the D&D style combat system that i've been making the attacker would roll accuracy and then the defender would roll evasion. Depending on a character's stats they may get bonuses to their accuracy or evasion. So Both dice could be modified or have no modifiers at all. My goal is to make a macro that would calculate what the probability is of you (the attacker) rolling higher than or equal to the defender with whatever bonuses both players may have.
– Sargent179
Jul 26 at 22:20
@InterstellarProbe For the D&D style combat system that i've been making the attacker would roll accuracy and then the defender would roll evasion. Depending on a character's stats they may get bonuses to their accuracy or evasion. So Both dice could be modified or have no modifiers at all. My goal is to make a macro that would calculate what the probability is of you (the attacker) rolling higher than or equal to the defender with whatever bonuses both players may have.
– Sargent179
Jul 26 at 22:20
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Assuming no bonus, the probability of a successful hit (assuming ties go in favor of the attacker) would be $frac120+frac12(1-frac120)=frac2140$, seen easily by noting the probability that the two roll exactly the same is $frac120$ and the remaining cases split evenly between favoring the attacker and favoring the defender.
For a net difference of $1$ in favor of the attacker (E.g. by having +1 to attack rolls and +0 to defense, or +5 to attack and +4 to defense, etc...) we would include an additional $19$ favorable scenarios for the attacker out of the $400$ possible outcomes of rolling two dice giving a probability of $frac229400$
A net difference of $2$ in favor of the attacker would give an additional $18$ more favorable scenarios beyond those mentioned in the previous paragraph giving a probability of $frac247400$, so on and so forth.
In general: for a net difference of $ngeq 0$ in favor of the attacker, the probability would be $dfrac-n^2+39n+420800$
This was found by recurrence relations, letting $f(0)=frac2140$ and $f(n)=f(n-1)+frac20-n400$ by generalizing the observations made above.
A similar technique can be applied in the reverse direction to get the probabilities for if the defender has the advantage, or you can just use the values already found for the attacker having the advantage and subtracting away from $1$. The probability of a successful attack with a net difference of $ngeq 1$ in favor of the defender would correspond to $1$ minus the corresponding probability of a successful attack with net difference of $n-1$ in favor of the attacker.
answered Jul 26 at 20:48
JMoravitz
44.1k33481
In D&D, I believe if $n=0$ the rule is that on a tie, you have a roll-off. Each player (attacker and defender) both roll until there is a clear winner. So, I think for $n=0$, the probability is $dfrac12$. This would give the formula: $$f(n) = begincasesdfrac12, & n=0 \ dfrac-n^2+39n+420800, & n>0endcases$$
– InterstellarProbe
Jul 26 at 21:00
@interstellarprobe in dnd the defender doesn't roll at all. They simply have a defense score which the attacker tries to roll higher than. If you were to assume a reroll on a tie, then the formula needs quite a bit of reworking and what you write in the comments would not be correct as that still has ties in favor of attacker for n>0
– JMoravitz
Jul 26 at 21:33
If advantage should go in favor of defender for ties, just replace every occurrence of the word "attacker" and it's variants with "defender" and vice versa
– JMoravitz
Jul 26 at 21:36
So N is the result of (attack mod - defence mod) meaning if you have 1 extra attack than their defence it'd be (1^2+(39*1)+420) / 800 ? Still confused about the 39, 420 and 800.
– Sargent179
Jul 26 at 23:55
Yes (though you seem to have missed the negative sign on the first term). The numbers you point at happen to appear in the closed form of the solution and so are relevant in that sense, but they are not easy to spot without calculations. I can assure you though that they didn't get pulled out of thin air.
– JMoravitz
Jul 27 at 1:20
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Assuming no bonus, the probability of a successful hit (assuming ties go in favor of the attacker) would be $frac120+frac12(1-frac120)=frac2140$, seen easily by noting the probability that the two roll exactly the same is $frac120$ and the remaining cases split evenly between favoring the attacker and favoring the defender.
For a net difference of $1$ in favor of the attacker (E.g. by having +1 to attack rolls and +0 to defense, or +5 to attack and +4 to defense, etc...) we would include an additional $19$ favorable scenarios for the attacker out of the $400$ possible outcomes of rolling two dice giving a probability of $frac229400$
A net difference of $2$ in favor of the attacker would give an additional $18$ more favorable scenarios beyond those mentioned in the previous paragraph giving a probability of $frac247400$, so on and so forth.
In general: for a net difference of $ngeq 0$ in favor of the attacker, the probability would be $dfrac-n^2+39n+420800$
This was found by recurrence relations, letting $f(0)=frac2140$ and $f(n)=f(n-1)+frac20-n400$ by generalizing the observations made above.
A similar technique can be applied in the reverse direction to get the probabilities for if the defender has the advantage, or you can just use the values already found for the attacker having the advantage and subtracting away from $1$. The probability of a successful attack with a net difference of $ngeq 1$ in favor of the defender would correspond to $1$ minus the corresponding probability of a successful attack with net difference of $n-1$ in favor of the attacker.
answered Jul 26 at 20:48
JMoravitz
44.1k33481
In D&D, I believe if $n=0$ the rule is that on a tie, you have a roll-off. Each player (attacker and defender) both roll until there is a clear winner. So, I think for $n=0$, the probability is $dfrac12$. This would give the formula: $$f(n) = begincasesdfrac12, & n=0 \ dfrac-n^2+39n+420800, & n>0endcases$$
– InterstellarProbe
Jul 26 at 21:00
@interstellarprobe in dnd the defender doesn't roll at all. They simply have a defense score which the attacker tries to roll higher than. If you were to assume a reroll on a tie, then the formula needs quite a bit of reworking and what you write in the comments would not be correct as that still has ties in favor of attacker for n>0
– JMoravitz
Jul 26 at 21:33
If advantage should go in favor of defender for ties, just replace every occurrence of the word "attacker" and it's variants with "defender" and vice versa
– JMoravitz
Jul 26 at 21:36
So N is the result of (attack mod - defence mod) meaning if you have 1 extra attack than their defence it'd be (1^2+(39*1)+420) / 800 ? Still confused about the 39, 420 and 800.
– Sargent179
Jul 26 at 23:55
Yes (though you seem to have missed the negative sign on the first term). The numbers you point at happen to appear in the closed form of the solution and so are relevant in that sense, but they are not easy to spot without calculations. I can assure you though that they didn't get pulled out of thin air.
– JMoravitz
Jul 27 at 1:20
 |Â
show 1 more comment
up vote
1
down vote
accepted
Assuming no bonus, the probability of a successful hit (assuming ties go in favor of the attacker) would be $frac120+frac12(1-frac120)=frac2140$, seen easily by noting the probability that the two roll exactly the same is $frac120$ and the remaining cases split evenly between favoring the attacker and favoring the defender.
For a net difference of $1$ in favor of the attacker (E.g. by having +1 to attack rolls and +0 to defense, or +5 to attack and +4 to defense, etc...) we would include an additional $19$ favorable scenarios for the attacker out of the $400$ possible outcomes of rolling two dice giving a probability of $frac229400$
A net difference of $2$ in favor of the attacker would give an additional $18$ more favorable scenarios beyond those mentioned in the previous paragraph giving a probability of $frac247400$, so on and so forth.
In general: for a net difference of $ngeq 0$ in favor of the attacker, the probability would be $dfrac-n^2+39n+420800$
This was found by recurrence relations, letting $f(0)=frac2140$ and $f(n)=f(n-1)+frac20-n400$ by generalizing the observations made above.
A similar technique can be applied in the reverse direction to get the probabilities for if the defender has the advantage, or you can just use the values already found for the attacker having the advantage and subtracting away from $1$. The probability of a successful attack with a net difference of $ngeq 1$ in favor of the defender would correspond to $1$ minus the corresponding probability of a successful attack with net difference of $n-1$ in favor of the attacker.
answered Jul 26 at 20:48
JMoravitz
44.1k33481
In D&D, I believe if $n=0$ the rule is that on a tie, you have a roll-off. Each player (attacker and defender) both roll until there is a clear winner. So, I think for $n=0$, the probability is $dfrac12$. This would give the formula: $$f(n) = begincasesdfrac12, & n=0 \ dfrac-n^2+39n+420800, & n>0endcases$$
– InterstellarProbe
Jul 26 at 21:00
@interstellarprobe in dnd the defender doesn't roll at all. They simply have a defense score which the attacker tries to roll higher than. If you were to assume a reroll on a tie, then the formula needs quite a bit of reworking and what you write in the comments would not be correct as that still has ties in favor of attacker for n>0
– JMoravitz
Jul 26 at 21:33
If advantage should go in favor of defender for ties, just replace every occurrence of the word "attacker" and it's variants with "defender" and vice versa
– JMoravitz
Jul 26 at 21:36
So N is the result of (attack mod - defence mod) meaning if you have 1 extra attack than their defence it'd be (1^2+(39*1)+420) / 800 ? Still confused about the 39, 420 and 800.
– Sargent179
Jul 26 at 23:55
Yes (though you seem to have missed the negative sign on the first term). The numbers you point at happen to appear in the closed form of the solution and so are relevant in that sense, but they are not easy to spot without calculations. I can assure you though that they didn't get pulled out of thin air.
– JMoravitz
Jul 27 at 1:20
 |Â
show 1 more comment
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Assuming no bonus, the probability of a successful hit (assuming ties go in favor of the attacker) would be $frac120+frac12(1-frac120)=frac2140$, seen easily by noting the probability that the two roll exactly the same is $frac120$ and the remaining cases split evenly between favoring the attacker and favoring the defender.
For a net difference of $1$ in favor of the attacker (E.g. by having +1 to attack rolls and +0 to defense, or +5 to attack and +4 to defense, etc...) we would include an additional $19$ favorable scenarios for the attacker out of the $400$ possible outcomes of rolling two dice giving a probability of $frac229400$
A net difference of $2$ in favor of the attacker would give an additional $18$ more favorable scenarios beyond those mentioned in the previous paragraph giving a probability of $frac247400$, so on and so forth.
In general: for a net difference of $ngeq 0$ in favor of the attacker, the probability would be $dfrac-n^2+39n+420800$
This was found by recurrence relations, letting $f(0)=frac2140$ and $f(n)=f(n-1)+frac20-n400$ by generalizing the observations made above.
A similar technique can be applied in the reverse direction to get the probabilities for if the defender has the advantage, or you can just use the values already found for the attacker having the advantage and subtracting away from $1$. The probability of a successful attack with a net difference of $ngeq 1$ in favor of the defender would correspond to $1$ minus the corresponding probability of a successful attack with net difference of $n-1$ in favor of the attacker.
answered Jul 26 at 20:48
JMoravitz
44.1k33481
Assuming no bonus, the probability of a successful hit (assuming ties go in favor of the attacker) would be $frac120+frac12(1-frac120)=frac2140$, seen easily by noting the probability that the two roll exactly the same is $frac120$ and the remaining cases split evenly between favoring the attacker and favoring the defender.
For a net difference of $1$ in favor of the attacker (E.g. by having +1 to attack rolls and +0 to defense, or +5 to attack and +4 to defense, etc...) we would include an additional $19$ favorable scenarios for the attacker out of the $400$ possible outcomes of rolling two dice giving a probability of $frac229400$
A net difference of $2$ in favor of the attacker would give an additional $18$ more favorable scenarios beyond those mentioned in the previous paragraph giving a probability of $frac247400$, so on and so forth.
In general: for a net difference of $ngeq 0$ in favor of the attacker, the probability would be $dfrac-n^2+39n+420800$
This was found by recurrence relations, letting $f(0)=frac2140$ and $f(n)=f(n-1)+frac20-n400$ by generalizing the observations made above.
A similar technique can be applied in the reverse direction to get the probabilities for if the defender has the advantage, or you can just use the values already found for the attacker having the advantage and subtracting away from $1$. The probability of a successful attack with a net difference of $ngeq 1$ in favor of the defender would correspond to $1$ minus the corresponding probability of a successful attack with net difference of $n-1$ in favor of the attacker.
answered Jul 26 at 20:48
JMoravitz
44.1k33481
answered Jul 26 at 20:48
JMoravitz
44.1k33481
answered Jul 26 at 20:48
JMoravitz
44.1k33481
answered Jul 26 at 20:48
JMoravitz
44.1k33481
44.1k33481
In D&D, I believe if $n=0$ the rule is that on a tie, you have a roll-off. Each player (attacker and defender) both roll until there is a clear winner. So, I think for $n=0$, the probability is $dfrac12$. This would give the formula: $$f(n) = begincasesdfrac12, & n=0 \ dfrac-n^2+39n+420800, & n>0endcases$$
– InterstellarProbe
Jul 26 at 21:00
@interstellarprobe in dnd the defender doesn't roll at all. They simply have a defense score which the attacker tries to roll higher than. If you were to assume a reroll on a tie, then the formula needs quite a bit of reworking and what you write in the comments would not be correct as that still has ties in favor of attacker for n>0
– JMoravitz
Jul 26 at 21:33
If advantage should go in favor of defender for ties, just replace every occurrence of the word "attacker" and it's variants with "defender" and vice versa
– JMoravitz
Jul 26 at 21:36
So N is the result of (attack mod - defence mod) meaning if you have 1 extra attack than their defence it'd be (1^2+(39*1)+420) / 800 ? Still confused about the 39, 420 and 800.
– Sargent179
Jul 26 at 23:55
Yes (though you seem to have missed the negative sign on the first term). The numbers you point at happen to appear in the closed form of the solution and so are relevant in that sense, but they are not easy to spot without calculations. I can assure you though that they didn't get pulled out of thin air.
– JMoravitz
Jul 27 at 1:20
 |Â
show 1 more comment
In D&D, I believe if $n=0$ the rule is that on a tie, you have a roll-off. Each player (attacker and defender) both roll until there is a clear winner. So, I think for $n=0$, the probability is $dfrac12$. This would give the formula: $$f(n) = begincasesdfrac12, & n=0 \ dfrac-n^2+39n+420800, & n>0endcases$$
– InterstellarProbe
Jul 26 at 21:00
@interstellarprobe in dnd the defender doesn't roll at all. They simply have a defense score which the attacker tries to roll higher than. If you were to assume a reroll on a tie, then the formula needs quite a bit of reworking and what you write in the comments would not be correct as that still has ties in favor of attacker for n>0
– JMoravitz
Jul 26 at 21:33
If advantage should go in favor of defender for ties, just replace every occurrence of the word "attacker" and it's variants with "defender" and vice versa
– JMoravitz
Jul 26 at 21:36
So N is the result of (attack mod - defence mod) meaning if you have 1 extra attack than their defence it'd be (1^2+(39*1)+420) / 800 ? Still confused about the 39, 420 and 800.
– Sargent179
Jul 26 at 23:55
Yes (though you seem to have missed the negative sign on the first term). The numbers you point at happen to appear in the closed form of the solution and so are relevant in that sense, but they are not easy to spot without calculations. I can assure you though that they didn't get pulled out of thin air.
– JMoravitz
Jul 27 at 1:20
In D&D, I believe if $n=0$ the rule is that on a tie, you have a roll-off. Each player (attacker and defender) both roll until there is a clear winner. So, I think for $n=0$, the probability is $dfrac12$. This would give the formula: $$f(n) = begincasesdfrac12, & n=0 \ dfrac-n^2+39n+420800, & n>0endcases$$
– InterstellarProbe
Jul 26 at 21:00
In D&D, I believe if $n=0$ the rule is that on a tie, you have a roll-off. Each player (attacker and defender) both roll until there is a clear winner. So, I think for $n=0$, the probability is $dfrac12$. This would give the formula: $$f(n) = begincasesdfrac12, & n=0 \ dfrac-n^2+39n+420800, & n>0endcases$$
– InterstellarProbe
Jul 26 at 21:00
@interstellarprobe in dnd the defender doesn't roll at all. They simply have a defense score which the attacker tries to roll higher than. If you were to assume a reroll on a tie, then the formula needs quite a bit of reworking and what you write in the comments would not be correct as that still has ties in favor of attacker for n>0
– JMoravitz
Jul 26 at 21:33
@interstellarprobe in dnd the defender doesn't roll at all. They simply have a defense score which the attacker tries to roll higher than. If you were to assume a reroll on a tie, then the formula needs quite a bit of reworking and what you write in the comments would not be correct as that still has ties in favor of attacker for n>0
– JMoravitz
Jul 26 at 21:33
If advantage should go in favor of defender for ties, just replace every occurrence of the word "attacker" and it's variants with "defender" and vice versa
– JMoravitz
Jul 26 at 21:36
If advantage should go in favor of defender for ties, just replace every occurrence of the word "attacker" and it's variants with "defender" and vice versa
– JMoravitz
Jul 26 at 21:36
So N is the result of (attack mod - defence mod) meaning if you have 1 extra attack than their defence it'd be (1^2+(39*1)+420) / 800 ? Still confused about the 39, 420 and 800.
– Sargent179
Jul 26 at 23:55
So N is the result of (attack mod - defence mod) meaning if you have 1 extra attack than their defence it'd be (1^2+(39*1)+420) / 800 ? Still confused about the 39, 420 and 800.
– Sargent179
Jul 26 at 23:55
Yes (though you seem to have missed the negative sign on the first term). The numbers you point at happen to appear in the closed form of the solution and so are relevant in that sense, but they are not easy to spot without calculations. I can assure you though that they didn't get pulled out of thin air.
– JMoravitz
Jul 27 at 1:20
Yes (though you seem to have missed the negative sign on the first term). The numbers you point at happen to appear in the closed form of the solution and so are relevant in that sense, but they are not easy to spot without calculations. I can assure you though that they didn't get pulled out of thin air.
– JMoravitz
Jul 27 at 1:20
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It appears that you are looking at two dice rolls. You are looking for the number of ways that one die roll can beat another die roll when one of the dice is modified. Is that correct?
– InterstellarProbe
Jul 26 at 20:31
@InterstellarProbe For the D&D style combat system that i've been making the attacker would roll accuracy and then the defender would roll evasion. Depending on a character's stats they may get bonuses to their accuracy or evasion. So Both dice could be modified or have no modifiers at all. My goal is to make a macro that would calculate what the probability is of you (the attacker) rolling higher than or equal to the defender with whatever bonuses both players may have.
– Sargent179
Jul 26 at 22:20