View Full Version : Doubling doubles
Necromancy Black
21022009, 00:02
I keep reading this over and over and over...
Can someone please link or direct to where it says, in the current rule set that if you double a double (example banehead plus Flaming attack on a flamable model) if becomes 3 wounds, not 4.
People keep saying this is the case, and yet I'm yet to see a single rule saying so!
It's gone and it was very silly while it was here.
Please use real world maths.
Necromancy Black
21022009, 00:18
No it did use real maths, it was just ambiguous.
I know DnD had it (and I think some people have been confused by it) because they defined doubling as adding x to y where x = y. Thus when you got too doubles you got z = x + y + x, which is the same as z = 3y.
As far as I can tell this isn't the nromal definition of "double", which is multitive (is this even a real word?), not additive, so in warhammer and anywhere else two doubles is z = 2y*2 which is the same as z = 4y.
Basically I want to know if there is anything at all in the current rules to support the first one over the second, or are people still playing old rules?
There is no way that "double" or "doubling" can naturally mean any sort of addition.
All definitions I can find (and I've just looked up a few) use terms like "twice," "two times," "twofold," "factor of two" and so on, all indicative of multiplication, not addition.
havoc626
21022009, 00:55
The theory being the 2 doubles is a triple is that you double the BASE score twice, which is 2x+x. It wouldn't be 4x, because you are counting the first wound twice.
I might have been one of the ones to start this thought, but as I said in that thread, it was me getting confused with DnD. In Fantasy, a double doube equals 4, not 3. So Bane Head + Flaming attacks = 4 wounds per fail save(s).
Lord Dan
21022009, 01:18
There is no way that "double" or "doubling" can naturally mean any sort of addition.
Farmer brown has two apples. Farmer Joe brings him two apples as a gift. Farmer Brown has doubled the number of apples he previously owned.
No multiplication involved, and the result is double.
That's still multiplication rather than (or in addition to) addition, even though you have physically added the apples.
That's because it's impossible to "multiply" items in the real world, but note that you have still increased the number of apples by "a factor of two" or "twice as much" or "multiplied" them by two.
Occasionally adding will generate the same result as multiplying, but that is nothing more than coincidence, and it is by referring to division that you can determine if you have, indeed, doubled anything.
Note that if you added 1 apple (which has been suggested for the wounds) to result in 3 you would have doubled nothing.
Dokushin
21022009, 02:01
?
Multiplication is defined in terms of addition; it's simply adding a number to itself a given number of times. It is impossible to define multiplication without using addition.
The argument that doubling a number twice results in 3x is perfectly valid, if less intuitive than the 4x result. In game terms, 'doubling' gives you a bonus equal to the original. So you hit, and wound, and then two effects trigger: the first source gives you an extra wound, doubling your wound; and the second source gives you an extra wound, doubling your wound. End result, 3 wounds.
If we're done talking about apples or whatever, though, I myself would like to see where double double = 3 is cited. Anyone have a link or a page number or whatever?
Necromancy Black
21022009, 02:04
The thing we are dealing with words, not equations, in the rules.
As said before, the word itself "double" referes to multiplying something by 2. This can also be represented mathimatically by x + x but while it yeild the same results it's not exactly the same as 2x.
Thus when you have the sentence "double the double of the number of wounds" you'll get 2(2x) which is 4x. We would say that a double of a double is a quadruble, not a triple.
So by words alone it's multiplication, not addition. And seeing as no one has yet to give me any sources to say otherwise, this is how it works.
Dokushin
21022009, 02:07
Oh, absolutely, I'm with ya and that's how I play it *grin* I was just getting dizzy from all that mess up above with the wordmincing. I agree we need a cite to rule double double = 3; just trying to explain the logic behind it if that's the case, coming from a DnD background.
Hey, everyone! Necro and I are still waiting! Links! Please :D
What if the slann did d3 wounds (instead of flammable) with banehead? If the multipliers did stack, which do you apply first as it can make a huge difference. While both range from 26 wounds, the distribution varies (89% chance to cause 3+ wounds vs 66% for example).
I personally think it is only the original wound that is multiplied, such that you apply both effects simultaniously. Otherwise, one could argue infinite wounds for flammable+banehead. ie ...Slann wounds Treeman. It becomes 2 wounds due to flammable. Both of those wounds double due to banehead increasing it from 2 wounds to 4 wounds. Those 2 additional wound would be doubled since the treeman is flammable, making the total go from 4 wounds to 6 wounds. Again, those 2 newest additional wounds would now double from banehead (6 becomes 8 wounds), blah blah blah. Each additional wound added from an effect will trigger the other effect. This logic is rather stupid, but doubling doubles is not really that different (one effect is triggering the other effect).
Edit: I left out saying that applying both effects simultaniously still results in 4 wounds (2[x]+2[x]), but it prevents the endless logic loop that saying doubling doubles can create.
Necromancy Black
21022009, 12:31
What? Why would Banehead cause flammable wounds? By your logic you'd double from flammable, hen all those extra wounds would be doubled again as they're flammable and so on, meaning one wound from a flaming attack would kill any flammable model.
Yes the origanal wound is multiplied, but how much it multiplies by can changed. Basically you double a double meaning you get a quadruple. So you quadruple each wound meaning 1 wound = 4, 2 = 8, 3 = 12, etc. You don't aplly the doubling over an over, you aplly the doubling to each other ( ((*2)*2) = (*4) ) then aplly that to each wound once.
D3 is not a multiple. It's simply a random number, and whatever you get you times by the wound multiplier (so *2 or *4).
D3 is not a multiple. It's simply a random number, and whatever you get you times by the wound multiplier (so *2 or *4).
I didnt say d3 was a multiple. I only followed the logic you set up saying you double a double, hence apply one effect then apply the next effect onto the result. So which do you do first, d3 the wound, or double the wound? Both are done after all saves have been failed. If you apply the logic you presented, different statistical results can occur. If you do what I suggest, apply both at simultaniously, that doesnt happen.
What? Why would Banehead cause flammable wounds? By your logic you'd double from flammable, hen all those extra wounds would be doubled again as they're flammable and so on, meaning one wound from a flaming attack would kill any flammable model.
Umm... If banehead didnt cause flammable wounds (for a slann with flaming attacks), how are you doubling a double. Again, now with your new statement in the quote. ...A slann wounds a treeman. Double the wound cause of banehead. Now double the wound cause of flam...., oh wait, only the original wound is flamming, not the banehead one. Cant double the double, you can only double the original wound. In other words, you apply both effects to the original wound, and dont double a double.
What you are doing is 2[x]+2[x], which indeed does simplify to 4[x]. But be careful how you state that, cause lets say next army book introduces a "banehead" effect that triples the wounds. To say triple a double would be 6[x], which is wrong. The right answer would be 2[x]+3[x] or 5[x]. Gramatically it is wrong to state it as doubling a double and is setting it up for endless logic statments.
As far as I know, doubling stuff works on the original value.
You have 1 attack.
Double it because of reason x, means 1 extra attack.
Double it bacause of reason y, means 1 extra attack.
Each multiplyer gives 1 additional attack in this case.
1 original +1 for doubled (x) +1 for doubled (y) = 3
That's the idea behind 1 resulting in 3 instead of 4. Otherwise you add 1 to your original attacks.
Point is, you shouldn't double the result of the first doubling, as both modifiers apply at the same time.
At xragg: 2[x]+2[x] is not correct, as this means you started with 2 times an [x], but you started with only one, so where did the second come from? You start from the same [x] twice.
If you want to show the doublexdouble formula, it should be this: x*2*2 = 4*x
Result is the same, but the though behind it is not.
Now I don't know what the rules specifically say right now, but if they say: double your attacks, then it results in 3, as you don't start with 2 originals as many try to show with their sums, but with 1. It doesn't say, double your modified attacks/wounds.
Well...
At least that what I think GW intended, as doubling the result of an earlier doubling tends to result in crazy situations where things do so much damage that they could wipe out armies singlehandedly.
EDIT
Okay, I though up a nice diagram.
XX > XXxx


\/
XXxx
Look here, I double (the arrows depict the modification) capital XX (the original 2 attacks/wounds in this example) two times. The result is XX (originals) and xx (new ones). However, if you combine the two doubling results, then XX are both the exact same thing. So why add them together?
I say: if the description says: double the original stat/count, then you come to this, with 6 as result. If the description says: double whatever, you double normal mathwise, like 2x2x2 = 8.
SteelTitan
21022009, 15:23
i thought that the general consensus on multiplying multiplied wounds resulted in 3 wounds, not 4.
the most classical case of this was the 6th edition blade of realities + bane head which each doubles the caused wounds.
Has this never been addressed in a faq by GW?
Hobgoblyn
21022009, 15:39
I'm with those who say it is x3, because there just isn't a reason to be doubling a doubled result from two different sources. It also screws with game balance because you could doubledoubledouble a D3 when each of those doubles was only supposed to provide 1 or 2 wounds more and now is producing 8x.
If you double a base result then you get 2x, if you double the BASE result again you get 3x. So 1 wound would get first increase (1x2) + second increase (1x2) instead of 2x2.
Desert Rain
21022009, 15:49
Mathematically it should be 2(2x)=4x if you double a double.
If you have a double and a D3 it becomes D3(2x) which results in one of these 2x, 4x, 6x.
Mathematically it should be 2(2x)=4x if you double a double.
Who says you double a double?
The 3 people you just reacted to all spoke of doubling the base number, not doubling a double. I'm unaware of any item that says you double the already just doubled result from something else.
If you ask why not double a double, then I just say not to screw over an otherwise fun game by introducing doubtfull side effects to what is supposed to be an item with an interesting effect.
All we know from GW is that they intended it to be 3 wounds or whatever max, even if there is no current ruling for that.
As far as I know, doubling stuff works on the original value.
You have 1 attack.
Double it because of reason x, means 1 extra attack.
Double it bacause of reason y, means 1 extra attack.
Each multiplyer gives 1 additional attack in this case.
1 original +1 for doubled (x) +1 for doubled (y) = 3
That's the idea behind 1 resulting in 3 instead of 4. Otherwise you add 1 to your original attacks.
Point is, you shouldn't double the result of the first doubling, as both modifiers apply at the same time.
At xragg: 2[x]+2[x] is not correct, as this means you started with 2 times an [x], but you started with only one, so where did the second come from? You start from the same [x] twice.
If you want to show the doublexdouble formula, it should be this: x*2*2 = 4*x
Result is the same, but the though behind it is not.
Now I don't know what the rules specifically say right now, but if they say: double your attacks, then it results in 3, as you don't start with 2 originals as many try to show with their sums, but with 1. It doesn't say, double your modified attacks/wounds.
Well...
At least that what I think GW intended, as doubling the result of an earlier doubling tends to result in crazy situations where things do so much damage that they could wipe out armies singlehandedly.
EDIT
Okay, I though up a nice diagram.
XX > XXxx


\/
XXxx
Look here, I double (the arrows depict the modification) capital XX (the original 2 attacks/wounds in this example) two times. The result is XX (originals) and xx (new ones). However, if you combine the two doubling results, then XX are both the exact same thing. So why add them together?
I say: if the description says: double the original stat/count, then you come to this, with 6 as result. If the description says: double whatever, you double normal mathwise, like 2x2x2 = 8.
If you look at your "edit", do you realize you are doing (2[x]+2[x])? The horizontal "double" is the first 2[x] and the vertical "double" is the second 2[x]. 'x' in (2[x]+2[x]) represent number of original wounds done.
If you were doing a doubling double, thing the diagram would look like this:
XX > XXxx > XXXXxxxx
This can create an argument of infinite wounds. Update the diagram to show why your doubling each time so it looks like this:
XX (abanehead)> XXxx (bflammable)> XXXXxxxx
a. XX is doubled to XXxx due to banehead doubling wounds.
b. XXxx is doubled to XXXXxxxx due to flammable, but this can only be true if the new "xx" wounds created from banehead are also flammable.
It might appear to be alright at this point, and just state that new wounds dont qualify for banehead, but that cant be true. Why? Because GW doesnt state that you double with banehead first and flammable second, it just states they both double wounds. So this formula to be ture, should work either way. So this should also be true:
XX (aflammable)> XXxx (bbanehead)> XXXXxxxx
a. XX is doubled to XXxx due to flammable doubling wounds.
b. XXxx is doubled to XXXXxxxx due to banehead, but this can only be true if the new "xx" wounds created from flammable are also effected by banehead.
The two logic diagrams have shown that new flammable wounds are baneheadable, and new banehead wounds are flammable. Without drawing this out any longer, this creates an endless cycle of infinite wounds.
At the moment, its just an argument of mathmatical grammar, and I am really not going to lose sleep if people dont understand what I am saying.
Lord Dan
21022009, 21:00
At the moment, its just an argument of mathmatical grammar, and I am really not going to lose sleep if people dont understand what I am saying.
I usually stop reading when I stop comprehending. In the case of mathematical grammar it doesn't take too long.
Multiplying them infinitely is not correct in any way, including "mathematical grammar."
You have two operators, both tell you to double the initial result.
It doesn't matter what order you apply them in, but you don't reapply the operator after they have been applied.
For the result to be 3 wounds, GW would have to state something like, "Every wound from a flamming attack onto a flammable model does an extra wound."
This would then result in [x]+x
In the case of banehead being rewritten in the same parameters, this would create: [x]+x+x. Again, applying the results at the same time to avoid the quanday of infinite wounds.
Necromancy Black
21022009, 22:23
Thankfully this is not the case, and I shall continue to whipe treemen ancients off the board with 2 wounds from a Str 4 flaming magic missile :D
SteelTitan
21022009, 23:22
I think it is super weird that GW hasnt released a FAQ on this :S
If you look at your "edit", do you realize you are doing (2[x]+2[x])? The horizontal "double" is the first 2[x] and the vertical "double" is the second 2[x]. 'x' in (2[x]+2[x]) represent number of original wounds done.
I don't think you get my point here. What I say is: you double the initial result for one item, and double the initial result again for another item. You don't double the final result of the first doubling again, but instead double the intial twice and the add the result. Because you work from the initial result 2x, you don't suddenly get 2 initals to add together.
You get two results: doubling 1 and doubling 2. However, in both cases the original attack is the same one, so there's nothing to add.
So like in my previous example, if the result would have become XXXXxxxx (with X's being the original attacks, and x's being the doubled ones) then you are trying to say that the model started out with 4 attacks in the first place. It didn't.
For example this:
I have an apple. I multiply by 2, so now I have two apples, 1 new and 1 original.
I take the original apple back (so 1 apple, this is important here!), and double that one again. I now have again 1 original and 1 new apple. Now I take the first new apple to total up the result: 3 apples.
Why not 4? Because I applied the second doubling effect to the one original apple only! Not to the total result of the first doubling.
This is completely different from just saying: 2x+2x=4x. It has nothing to do with it.
Maybe I wasn't clear on the following though: I don't say that the rules imply that this is the way (don't have the actual rules for these items here so can't look if they say whether to use the original stat only), but this is what GW tried to say in their old errata that we used to have.
So there is no official ruling on that it is this way we should treat the matter, but they have previously shown us that this is the way they want us to play it. Only double the original stat/value, not the total result of an earlier multiplyer.
Maybe you could read it like this: both effects come into play at the exact same moment. So they cannot multiply the extra attacks from the other modifier as that one doesn't exist yet. So both give one extra attack, and since there suddenly are not 2 initial attacks but 1, the final result becomes 1+1+1=3.
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