Damage

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Odin
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Damage

Postby Odin » Sun Nov 12, 2000 3:43 pm

There is only one thing i havent fully understood, and thats the damage aspector.
when it says 3d5,6d6, 4d6, 1d8 etc. what does that meen?




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Maurice
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Postby Maurice » Sun Nov 12, 2000 4:02 pm

The 'd' - also known as 'D' - stands for die (or dice). So 3d6 means three rolls with a six-sided die, while 5D20 means five rolls with a twenty-sided die.

As it is, the PnP version of AD&D uses a number of weird looking dice. There's a four-sided die (pyramid like), a six-sided one, an eight-sided one, a ten-sided one, a twelve-sided one and a twenty-sided one.

The ten-sided one is usually rolled for percentile checks, although there appears to be also a 100-sided die for that purpose - but I never saw that one.

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Gluteus Maximus
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Postby Gluteus Maximus » Sun Nov 12, 2000 4:06 pm

Simple. The d stands for die or dice. 1d8 means one eight-sided die. So damage is anywhere from 1 to 8 in this case. 2d8 stands for two eight-sided dice. Damage in this case ranges anywhere from 2 to 16. The numbers generated are entirely random. So here's a little test for you. Which is better, a sword that does 1d8 worth of damage or one that does 2d4?

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Postby Odin » Sun Nov 12, 2000 4:11 pm

If I got it right:
1d8 can do 1-(1*8) damage.
2d4 can do 2-(2*4) damage.
In this case, its better with the 2d4,
because you dont risk only be dealing out 1 damage. Though 2 isnt much better Image
As for a 7d8:
min 7 and max 56 ?
randomly between?

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Postby Gluteus Maximus » Sun Nov 12, 2000 5:11 pm

Yup, you got it.

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Postby Nighthawk » Mon Nov 13, 2000 9:15 am

2d4 and 1d8 are more different than just the bottom end. 2d4 is more consistent...it will only do 8 damage 1/16 of the time since that would take rolls of 4 on both dice.

BTW, to figure out the expected damage for a weapon, add the expected damage for each of its dice plus any bonuses. The quick way to get the expected damage for a die is it's size, plus 1, divided by 2.

d4 = 2.5, d6 = 3.5, d8 = 4.5, etc.

Since 2d4 = 5 > d8 = 4.5 a bastard sword will on average do 1/2 point more damage than a long sword.

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Postby Odin » Mon Nov 13, 2000 2:57 pm

So what you are saying is:
if you have a 3d5 +5, the expected damage is (5+5)/2=5 pr dies =15 damage?
What happens if you roll 15 as if it was without the bonus. only 3d5? and then in addition you have the +5 bonus?
Would it then be as if it where 3d10?
And you could then get max 30?
Or am I on the wrong road here

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Postby Lucian » Mon Nov 13, 2000 3:07 pm

if you have a hammer 2d4+5 it will do 2 to 8 damage +5 the bonus are added after the basic damage have been calculated

say you roll 3 and 2 then it would be 5+5 damage you inflict...and from there you have to add strength bonuses..and critical hit modifiers..that is when you roll 20 on thaco roll...some skills will let you go critical on 19 and 20..a critical doubles the damage you inflict

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Postby Lucian » Mon Nov 13, 2000 3:34 pm

and the higher the extra bonus is the better..some monsters can only be hit if the weapon has x or greather enchancement..kangaxx the lich is a good example..i believe he only can be hit by +4 weapons.

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Postby Odin » Mon Nov 13, 2000 4:49 pm

Hmmmm, I cant say I have fully understood it. Ill have to check it up in the manual. Maybe Ill understand it then.
It is nothing wrong with your answers guys, but this is new to me, and it may take some time.


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Postby Maurice » Mon Nov 13, 2000 4:59 pm

Well, it's not that difficult, really.

If we take a look at the 2D4+5 thingy ... if you roll 2D4, the score will be between 2 and 8, right? Statistically speaking, you have the highest chance to get a score of 5 in the roll (4+1, 3+2, 2+3 and 1+4), less to get 4 (1+3, 2+2, 3+1) or 6 (4+2, 3+3, 2+4), and even less to get 3 or 7. 2 can only be had with 1+1, while 8 can only be had with 4+4.

So while you roll on average a score of 5 on 2D4, you'll roll on average a score of 5/2 on 1D4. This average is easily calculated from the highest score and lowest score on the die. Usually the lowest value is 1, but that hasn't got to be so. The average is then simply the sum of the lowest value and the highest value, divided by 2. This all provided that the numbers follow up on one another in a lineair way (1, 2, 3, 4, 5, etc .. or 2, 4, 6, 8, ... well, you get the idea), but this is *very* common on dice, I can assure you! Image

Anyway, once you have this average purely based on the involved dice, you can add the '+X' bonus to it. So while 2D4 has an average of 5, 2D4+5 has an average of (5+5)=10. 2D8 has an average of 9, while 2D8+4 has an average of 13.

I hope you didn't loose me somewhere in the explanation Image

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Odin
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Postby Odin » Mon Nov 13, 2000 5:09 pm

Sorry guys, still dont get it fully. Maurice, i followed you through the first section, and the last section. When you say that the average of a 2d4 is 5, i see it in the ( ), but isnt there a way to see it faster, a shortcut?

And i could always look in the manual if when you get tired of answering my stupid questions..


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Postby Sourced » Tue Nov 14, 2000 2:16 am

I am not gonna say anything about that statistical explination, Maurice.

But I usualy just check the max and min score a weapon can do. So a 1D6+3 weapon would do 4-9 points of damage and a 2D4+1 would do 3-9 point of damage (the first being the better choice).

I'm _very_ glad to have (somehow) made the statistical part of my study and hope never to see this stuff angain, thank you very much. Image
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Maurice
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Postby Maurice » Tue Nov 14, 2000 3:57 am

Well, Odin, seems my friend Source kinda answered my question already.

Anyway, yes, there's a shortcut, like I already wrote in my message.

From any die you simply take the lowest value and the highest value, add them, and divide by two. Add to this a possible '+' and you have the average for that particular roll.

If you have more dice (2D6, 3D6, etc...) simply multiply the average of a single die by the amount of dice.

In other words:
1D6: min = 1, max = 6, so average = (1+6)/2 = 3.5.

1D8: min = 1, max = 8, so average = (1+8)/2 = 4.5.

1D4+3: min = 1, max = 4, so average is (1+4)/2 = 2.5. Add to this the +3, and you get 5.5 as an average for this roll.

An alternate way is to say the following: min = (1+3) = 4, max = (4+3) = 7, so the average is (4+7)/2 = 5.5. Note that now you don't include the +3, since you already used it to modify the lower and upper limit.

2D12: min = 2, max = 24, so the average is (2+24)/2 = 13.

Alternately, 1D12 has a min of 1, and a max of 12, so the average is (1+12)/2 = 6.5 - for two such dice the average is then 2x 6.5 = 13 (which is correct accoring to what we just saw).

3D20: min = 3, max = 60, so the average is 63/2 = 31.5. Again, this can be computed by taking the average of 1D20 (which is 10.5 as you can check for yourself) and then multiply it by 3.

Okay, this is where I will end the lesson for now. Questions anyone? Image

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Postby daemon » Tue Nov 14, 2000 4:13 am

yeah,

whats (X^2+(y/4))/(y^3-(x^2/3.3)) in its third quadratic form?

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Maurice
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Postby Maurice » Tue Nov 14, 2000 4:30 am

We were talking about statistics here, not calculus Image

Has been a few years ...

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Postby daemon » Tue Nov 14, 2000 4:45 am

yeah - me too Image
daemon

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Odin
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Postby Odin » Tue Nov 14, 2000 4:53 am

Thanks guys.
I got it..
I am gratefull for your time. Image

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Odin
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Postby Odin » Tue Nov 14, 2000 5:01 am

Oh, Daemon. The answer to your question is:
d/dy[(4x^2)/y]+58e^-x
Image

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Maurice
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Postby Maurice » Tue Nov 14, 2000 5:04 am

I hope you're not saying that to get us to shut up about it, while you're still wandering in the woods, so to speak ... Image