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stranger
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OP
stranger
Joined: Oct 2020
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Which weapon would be better? To me they are the same, wondering if I’m missing something.
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apprentice
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apprentice
Joined: Oct 2020
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I haven't personally noticed a difference.
I will say this, I think their rolls are wonky, in just about every aspect of the game. I don't believe they are being straight up with the math, I believe they have hidden code instead of just simply dice rolls that make the probability of lower rolls coming up more often. That seems to be how it is for me, and I've seen many others post the same. If that holds true for weapon damage rolls, then I'd rather get a roll of 3 twice, than a roll of 3 once.
But, who really knows. The rolls on just about everything simply doesn't seem on the up and up.
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journeyman
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journeyman
Joined: Oct 2020
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Mathematically a 2d6 weapon is marginally better. .5 higher average damage.
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stranger
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stranger
Joined: Oct 2020
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Yep,
This is assuming that the rolls are uniformly distributed and not pseudo random or cryptographically random.
2d6 on average deals (6+1)/2 * 2 = 7 damage
1d12 on average deals (12+1)/2 = 6.5 damage
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apprentice
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apprentice
Joined: Oct 2020
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SecSea, I don't believe your equation is correct. The answer for that equation is correct, but I believe it's (6)2+1. I believe the Plus 1 is applied after the two rolls of the 6 sided die. It's in their own description, 2d6+1 literally means (2)x+1, where x is the random roll of the 6 sided die.
I could be wrong in that, but it would make sense from a standpoint of keeping it on par with a 1d12+1.
Last edited by Rhend; 13/10/20 05:56 AM.
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stranger
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stranger
Joined: Oct 2020
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Suppose that my equation is wrong.
We still assume that it's uniformly distributed.
Case 2d6:
Let min=2 and max=12.
Let range=max-min=10
Average of infinite rolls is min+(range/2)=2+5=7
My equation is right for 2d6
Case 1d12:
Let min=1, max =12
Range=max-min=11
Average of infinite rolls is min+(range/2)=1+5.5=6.5
My equation is right for 1d12
Let's take it further.
Case 2d6+1:
Min=3
Max=13
Range=10
min+(range/2)=3+5=8
Average of infinite rolls is 8 (1 more than 2d6)
Now let's assume that it's not uniformly distributed. Id like you to prove that by writing down all rolls for a d6 and d12 made without taking into consideration other modifiers affecting the result (such as proficiency, threatened, etc...). Find the frequency of these rolls and compute the average.
Last edited by SecSea; 13/10/20 06:55 AM.
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old hand
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old hand
Joined: Sep 2015
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2 dice are always better than one when the max is the same. Cause your minimum damage will always be higher. It's a rule of thumb in pnp.
In a tight fight, higher minimal damage can save your life.
Last edited by Nyanko; 13/10/20 07:09 AM.
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stranger
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stranger
Joined: Oct 2020
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Another way to look at it.
If the odds of rolling {2,3,4...12} are the same then the average is the sum of these numbers / 11.
The sum series from I=0 to 10 of (I+2) is computed.
To solve this, take out the 2. Because it's repeated 11 times, then 2*11+ sum from I=0 to 10 of I remains.
Sum series solution is ((10+1)*10)/2.
So 22+(10+1)*(10/2)= 22+55=77.
Find the average by dividing by 11. 77/11=7
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stranger
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stranger
Joined: Oct 2020
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You are right about expected value (average), however you also have to consider distribution.
1k12 means every value is equally possible to roll. You get 1 as often as 3, 8 or 12. Exactly 8,33%.
2k6 is normal distribution (Gauss or bell curve). The most possible option to roll is 7 - 16,7%. It's 2 times more probable when compared to rolling 1k12. And there is only 2,78% to roll 2 or 12. Nearly 3 times less possible.
Rolling 2k6 is "safer" option to roll. There is almost 45% chance to roll 6, 7 or 8.
Of course my math is only accurate if game engine is actually rolling two (or more) dices. If implementation of 2k6 is done by rolling one k6 (as double precision number) and just multiplying the value by 2 then... there is almost no difference than rolling 1k12. Only this +0,5 to expected value as SecSea mentioned, however you will hardly notice that.
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apprentice
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apprentice
Joined: Oct 2020
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I get your point SecSea, I didn't account for 2 always being the minimum after a roll with 1 being the minimum for a 1d12. I still believe the +1 with the 2d6 is applied after the the 2 die rolls are added.
I still do not trust the rolls in this game. I've had way too many times of save scumming rolling against a 10 and hitting under that 10, 7, 8, 9 times in a row. Statistically, it should be highly improbably to have that many rolls under 10. 55% chance of rolling a 10 or higher on a single role, the more you keep rolling under, the lower the probability should be for the next attempt to also be under. It just doesn't add up.
Last edited by Rhend; 13/10/20 08:44 AM.
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apprentice
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apprentice
Joined: Oct 2020
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I get your point SecSea, I didn't account for 2 always being the minimum after a roll with 1 being the minimum for a 1d12.
I still do not trust the rolls in this game. I've had way too many times of save scumming rolling against a 10 and hitting under that 10, 7, 8, 9 times in a row. Statistically, it should be highly improbably to have that many rolls under 10. 55% chance of rolling a 10 or higher on a single role, the more you keep rolling under, the lower the probability should be for the next attempt to also be under. It just doesn't add up. You, my friend, have fallen victim to the gambler's fallacy. The probability for each individual roll to be under 10, regardless of the results of all previous rolls, is always the same. Chances are, you're just unlucky.
Last edited by fishworshipper; 13/10/20 08:45 AM.
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apprentice
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apprentice
Joined: Oct 2020
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I get your point SecSea, I didn't account for 2 always being the minimum after a roll with 1 being the minimum for a 1d12.
I still do not trust the rolls in this game. I've had way too many times of save scumming rolling against a 10 and hitting under that 10, 7, 8, 9 times in a row. Statistically, it should be highly improbably to have that many rolls under 10. 55% chance of rolling a 10 or higher on a single role, the more you keep rolling under, the lower the probability should be for the next attempt to also be under. It just doesn't add up. You, my friend, have fallen victim to the gambler's fallacy. The probability for each individual roll to be under 10, regardless of the results of all previous rolls, is always the same. Chances are, you're just unlucky. Not at all. In probability your chances of getting a continued roll that is the less desirable 2 times in a row would be lower than the base 45%, 3 times is even lower than that, and so on and so forth. I've not fallen a victim to a gamblers fallacy, unless the odds are actually stacked and not truly represented (a hidden formula). Look at it like this, in flipping a quarter you would expect a 50% chance either way (it's actually been shown multiple times to be 51% heads for some reason). Anyways, your chance of getting Tail's would be 50%. Now, it goes Heads. Since it went heads, the chance of it going heads again is actually 25%. It would continue to dip like that. The same applies to the die roll, as long as there isn't an unknown formula/variable that's effecting those rolls. Many people have noticed the same as me though, and I wouldn't have even mentioned it without there being a lot of discussion about it already.
Last edited by Rhend; 13/10/20 09:42 AM.
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stranger
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stranger
Joined: Oct 2020
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Essentially, 2d6 will give you a higher likelihood of rolling a mid-range number on any given roll, and 2d6 is more likely to give you somewhere between about 4-9 damage on a roll, whereas a 1d12 will give you an even chance to earn any number, and tends to be more unpredictable. You'll end up with more large damage hits, but also with more small damage hits. Basically if you want more consistent, reliable damage, use 2d6, but if you're willing to risk the higher chance of a lower roll for better chance at higher rolls, use the 1d12. Mostly a personal choice based on your own risk tolerance. Even better, having both available can allow you to use whichever suits the situation you're in.
For 2d6 For 1d12
Total of dice | Permutations | Probability | | Total of dice | Probability |
1 0 0% 1 8.33%
2 1 (1,1) 2.77% 2 8.33%
3 2 (1,2)(2,1) 5.55% 3 8.33%
4 3 (1,3)(3,1)(2,2) 8.33% 4 8.33%
5 4 (1,4)(4,1)(2,3)(3,2) 11.11% 5 8.33%
6 5 (1,5)(5,1)(4,2)(2,4)(3,3) 13.88% 6 8.33%
7 6 (1,6)(6,1)(5,2)(2,5)(3,4)(4,3) 16.66% 7 8.33%
8 5 (2,6)(6,2)(3,5)(5,3)(4,4) 13.88% 8 8.33%
9 4 (3,6)(6,3)(4,5)(5,4) 11.11% 9 8.33%
10 3 (4,6)(6,4)(5,5) 8.33% 10 8.33%
11 2 (6,5)(5,6) 5.55% 11 8.33%
12 1 (6,6) 2.77% 12 8.33%
I added that to show a breakdown, but in D&D you shouldn't really be fishing for a certain number, so the individual percentages are not really useful on their own, and I think it's better to look at the odds of getting within a range of damage. Broken down into four ranges:
2d6 1d12
Roll | Range Roll | Range
1-3: 8.33% 1-3: 25%
4-6: 33.33% 4-6: 25%
7-9: 41.66% 7-9: 25%
10-12: 16.66% 10-12: 25%
Statistically, over many many many rolls, the 2d6 will typically do more total damage(about 7%), but that doesn't matter much when a battle is hinging on a single roll and where the line between success and failure lies.
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journeyman
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journeyman
Joined: Oct 2020
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In addition to 1d12 being a riskier option, 2d6 also benefits more from Great Weapon Fighting (reroll any damage die that rolls a 1 or 2, and use the new roll instead) under the 5e ruleset. There has been another thread questioning whether this works as per 5e rules in BG3, but it's hard to know for sure.
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stranger
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stranger
Joined: Oct 2020
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I get your point SecSea, I didn't account for 2 always being the minimum after a roll with 1 being the minimum for a 1d12.
I still do not trust the rolls in this game. I've had way too many times of save scumming rolling against a 10 and hitting under that 10, 7, 8, 9 times in a row. Statistically, it should be highly improbably to have that many rolls under 10. 55% chance of rolling a 10 or higher on a single role, the more you keep rolling under, the lower the probability should be for the next attempt to also be under. It just doesn't add up. You, my friend, have fallen victim to the gambler's fallacy. The probability for each individual roll to be under 10, regardless of the results of all previous rolls, is always the same. Chances are, you're just unlucky. Not at all. In probability your chances of getting a continued roll that is the less desirable 2 times in a row would be lower than the base 45%, 3 times is even lower than that, and so on and so forth. I've not fallen a victim to a gamblers fallacy, unless the odds are actually stacked and not truly represented (a hidden formula). Look at it like this, in flipping a quarter you would expect a 50% chance either way (it's actually been shown multiple times to be 51% heads for some reason). Anyways, your chance of getting Tail's would be 50%. Now, it goes Heads. Since it went heads, the chance of it going heads again is actually 25%. It would continue to dip like that. The same applies to the die roll, as long as there isn't an unknown formula/variable that's effecting those rolls. Many people have noticed the same as me though, and I wouldn't have even mentioned it without there being a lot of discussion about it already. If you flip a coin and it lands on heads, the odds of the next coin flip being heads is still 50%. On any given flip, whether it's your first or your thousandth, the odds of getting heads is always going to be 50%. Just because you got a heads on the previous flip does not mean you have a lower chance of heads or a higher chance of tails on the next flip. You're confusing the odds of a particular flip(or roll of the dice) with the probability of a streak. Going back to the dice, while yes, the odds of getting a roll of 1 or 2 twice in a row is lower than getting a 1 or 2 on the first and a higher number on the second, simply getting a 1 or 2 on the first roll does not decrease the probably that you will get a 1 or 2 on the second roll. The odds of getting a 1-2 on the first roll of a d12 is 1 in 6, and the odds of getting a 1 or 2 on the second roll is 1 in 6. While it's less likely that you'll maintain a streak as it goes on, each roll has the exact same probability to have the same result as the one before it, no matter how many rolls you do. Back to the coins, look at it this way, using your example. You've flipped your coin and it went heads, and now there's one flip left, with the same possible outcomes and probabilities of the first flip. There's a 50% chance it continues the streak, and a 50% chance that it doesn't.
Last edited by Patient; 13/10/20 04:36 PM.
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member
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member
Joined: Apr 2014
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I'm not sure if people throwing around equations are doing dice math correctly. The average of a 6 sided die is 3.5. You don't factor in zero because it's not an option in the set. Numerically, 2d6 is always better than 1d12. I saw similar conclusions but the work shown on how you got there seemed slightly off.
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apprentice
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apprentice
Joined: Oct 2020
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Patient, that does not align with probability in mathematics, at all. Probability is defined as a favorable outcome divided by the total number of outcomes. The more flips you have, the more total outcomes you have. If you stick to one "favorable" choice, the lower (or higher) the chance of that choice hitting becomes per turn, depending on the number of times it hit prior.
To put it simply, if you are flipping a coin 100 times (all things considered fair, the way it's flipped, and taking the coin weight distribution out of the equation), and it hit heads the first two times already. If you were asked to guess the total number of times it will now hit heads for the remaining flips, if you answer, "Well, 50% of the time" you'd be wrong. The variable, the equation essentially changed based on the number of flips made, and the number of times it has hit heads. At that point the likelihood of it landing heads would be 48% over time, but right at that point, the likelihood of the coin actually hitting heads a third time is 1/6, or a 16.666% chance.
If you'd like though, we can agree to disagree. Mathematic nerdgasm is fun and all, but other things are cool also, lol =)
Last edited by Rhend; 13/10/20 08:31 PM.
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stranger
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stranger
Joined: Oct 2020
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// Remaps all scores - where a score of SCORE_MOD is the average dmg of this character.
SCORE_MOD 100.0
BASE_NEARBY_SCORE 0.2
....
function GetModifier(value)
return math.floor((value - 10) / 2)
end
The logic used in BG3 testing environment to compute several values in a normalised fashion (damage might be among them). There's a trick though, after several manipulations, they take the floor of a value from a range (note score_mod is manipulated) because the probability of a point that is one unit less than another is higher for any point greater than or equal to the mean. I would guess then that since 2d6 offers 7 and 1d12 offers 6(.5 is not taken into account) then 2d6 is def better. There is...however the matter with this snippet of code:
//2. operator | will calculate chance of either left or right to be trigger. Formula is 1-(1-leftSuccessChance)*(1-rightSuccessChance)
...
function invert(n)
local absolute = math.abs(n)
if absolute <= 1 and absolute >= 0
then return (1 - absolute)
else return n
end
end
...
__bor = function(a, b) -- | operator
local r = (type(b) == "boolean") and boolToNum(b) or b
return invert(invert(a)*invert(r));
end,
Which makes me believe that if damage is a range from neg to 0 then taking the floor of -6.5 is -7, and invert is 7. Which means that 2d6 offers no advantage of 1d12. I'll look into this further
Last edited by SecSea; 13/10/20 08:29 PM.
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journeyman
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journeyman
Joined: Oct 2020
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Patient, that does not align with probability in mathematics, at all. Probability is defined as a favorable outcome divided by the total number of outcomes. The more flips you have, the more total outcomes you have. If you stick to one "favorable" choice, the lower (or higher) the chance of that choice hitting becomes per turn, depending on the number of times it hit prior.
To put it simply, if you are flipping a coin 100 times (all things considered fair, the way it's flipped, and taking the coin weight distribution out of the equation), and it hit heads the first two times already. If you were asked to guess the total number of times it will now hit heads for the remaining flips, if you answer, "Well, 50% of the time" you'd be wrong. The variable, the equation essentially changed based on the number of flips made, and the number of times it has hit heads. At that point the likelihood of it landing heads would be 48% over time, but right at that point, the likelihood of the coin actually hitting heads a third time is 1/6, or a 16.666% chance.
If you'd like though, we can agree to disagree. Mathematic nerdgasm is fun and all, but other things are cool also, lol =) Been a while since I've done statistics - but I think both of you are right: So you have two sperate views of the probably: what is the probability of getting a head on a coin and its always 50% if its a truly independent toss, but the other way of looking at it is what is the probability of getting heads on the third toss GIVEN that the first two were heads. In the second statement you're giving a predictive probability i.e. if you tossed 3 coins in row rather than just looking at the probability of the the coin toss in isolation. If that makes sense.
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apprentice
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apprentice
Joined: Oct 2020
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Patient, that does not align with probability in mathematics, at all. Probability is defined as a favorable outcome divided by the total number of outcomes. The more flips you have, the more total outcomes you have. If you stick to one "favorable" choice, the lower (or higher) the chance of that choice hitting becomes per turn, depending on the number of times it hit prior.
To put it simply, if you are flipping a coin 100 times (all things considered fair, the way it's flipped, and taking the coin weight distribution out of the equation), and it hit heads the first two times already. If you were asked to guess the total number of times it will now hit heads for the remaining flips, if you answer, "Well, 50% of the time" you'd be wrong. The variable, the equation essentially changed based on the number of flips made, and the number of times it has hit heads. At that point the likelihood of it landing heads would be 48% over time, but right at that point, the likelihood of the coin actually hitting heads a third time is 1/6, or a 16.666% chance.
If you'd like though, we can agree to disagree. Mathematic nerdgasm is fun and all, but other things are cool also, lol =) Been a while since I've done statistics - but I think both of you are right: So you have two sperate views of the probably: what is the probability of getting a head on a coin and its always 50% if its a truly independent toss, but the other way of looking at it is what is the probability of getting heads on the third toss GIVEN that the first two were heads. In the second statement you're giving a predictive probability i.e. if you tossed 3 coins in row rather than just looking at the probability of the the coin toss in isolation. If that makes sense. Exactly what I'm saying. The probability, the odds change with successive flips. The same happens with rolls of the die. In isolation, a single flip, all things fair, it will be 50/50. But, that goes out the window with successive flips.
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