Forum adverts like this one are shown to any user who is not logged in. Join us by filling out a tiny 3 field form and you will get your own, free, dakka user account which gives a good range of benefits to you: No adverts like this in the forums anymore. Times and dates in your local timezone. Full tracking of what you have read so you can skip to your first unread post, easily see what has changed since you last logged in, and easily see what is new at a glance. Email notifications for threads you want to watch closely. Being a part of the oldest wargaming community on the net. If you are already a member then feel free to login now.

 skchsan wrote:

JakeSiren wrote:
Unless explicitly stated, such as in overwatch or the auto-fail, roll refers to the result after modifiers.

The excerpt "a roll of 1 always fails, irrespective of modifiers." makes no 'explicit statement' that the roll refers to unmodified dice roll. Therefore, as per your stance, the excerpt can only be referring to modified rolls of 1, at which point, a natural 1 can only auto-miss if there are no +hit modifiers present.

Uh, I think you need to reread what I said. The part where the rule says "irrespective of modifiers" is an explicit instruction to not apply modifiers - aka, an unmodified roll. I refered to it as an example. You have even quoted me saying this! (Bolded for you)

 skchsan wrote:

 Primark G wrote:
You would have to have WS0 to auto hit. The RB is very clear on this matter.

You still fail on roll of 1.

Exactly.

Wouldn't bypassing the "always fails on 1" make auras that allow you to reroll 1's be made redundant?

 skchsan wrote:
It APPEARS contradictory because it stands against your understaning of the RAW. The rulebook makes no distinction between what we colloquially refer to as "natural rolls" and "modified rolls." The assumption that "only natural rolls of 1 auto-fails" is a fallacious conclusion based on insufficient evidence.

It's not about modified or natural rolls (both are just "rolls"). The rule says to ignore modifiers. So we ignore modifiers. Unless you can give me a different take on "irrespective of modifiers".

Dandelion wrote:

 skchsan wrote:
It APPEARS contradictory because it stands against your understaning of the RAW. The rulebook makes no distinction between what we colloquially refer to as "natural rolls" and "modified rolls." The assumption that "only natural rolls of 1 auto-fails" is a fallacious conclusion based on insufficient evidence.

It's not about modified or natural rolls (both are just "rolls"). The rule says to ignore modifiers. So we ignore modifiers. Unless you can give me a different take on "irrespective of modifiers".

Correct. To really take it simply. If you roll 6 dice. And there is a -2 Modifier and a BS+1 on the model.

Your initial roll is 1, 2, 3, 3, 5, 6
Once you apply modifier, physically pick up the dice and take down by 2. 1 being the minimum. So you pick up the dice and take away 2 so you have now rolled. 1, 1, 1, 1, 3, 6. End result four misses and 2 hits. Even if you have a BS +1 as 1 is always a miss.

 SeanDavid1991 wrote:

Dandelion wrote:

 skchsan wrote:
It APPEARS contradictory because it stands against your understaning of the RAW. The rulebook makes no distinction between what we colloquially refer to as "natural rolls" and "modified rolls." The assumption that "only natural rolls of 1 auto-fails" is a fallacious conclusion based on insufficient evidence.

It's not about modified or natural rolls (both are just "rolls"). The rule says to ignore modifiers. So we ignore modifiers. Unless you can give me a different take on "irrespective of modifiers".

Correct. To really take it simply. If you roll 6 dice. And there is a -2 Modifier and a BS+1 on the model.

Your initial roll is 1, 2, 3, 3, 5, 6
Once you apply modifier, physically pick up the dice and take down by 2. 1 being the minimum. So you pick up the dice and take away 2 so you have now rolled. 1, 1, 1, 1, 3, 6. End result four misses and 2 hits. Even if you have a BS +1 as 1 is always a miss.

I kinda feel like you missed Dandelion's point.

Out of the rolls, irrespective of modifiers, which one's are ones? Those auto fail, everything else then has modifiers applied and checked against the shooters BS as to if they hit or not.

JakeSiren wrote:

 SeanDavid1991 wrote:

Dandelion wrote:

 skchsan wrote:
It APPEARS contradictory because it stands against your understaning of the RAW. The rulebook makes no distinction between what we colloquially refer to as "natural rolls" and "modified rolls." The assumption that "only natural rolls of 1 auto-fails" is a fallacious conclusion based on insufficient evidence.

It's not about modified or natural rolls (both are just "rolls"). The rule says to ignore modifiers. So we ignore modifiers. Unless you can give me a different take on "irrespective of modifiers".

Correct. To really take it simply. If you roll 6 dice. And there is a -2 Modifier and a BS+1 on the model.

Your initial roll is 1, 2, 3, 3, 5, 6
Once you apply modifier, physically pick up the dice and take down by 2. 1 being the minimum. So you pick up the dice and take away 2 so you have now rolled. 1, 1, 1, 1, 3, 6. End result four misses and 2 hits. Even if you have a BS +1 as 1 is always a miss.

I kinda feel like you missed Dandelion's point.

Out of the rolls, irrespective of modifiers, which one's are ones? Those auto fail, everything else then has modifiers applied and checked against the shooters BS as to if they hit or not.

The end result is the final roll that matters.

 SeanDavid1991 wrote:

JakeSiren wrote:

 SeanDavid1991 wrote:

Dandelion wrote:

 skchsan wrote:
It APPEARS contradictory because it stands against your understaning of the RAW. The rulebook makes no distinction between what we colloquially refer to as "natural rolls" and "modified rolls." The assumption that "only natural rolls of 1 auto-fails" is a fallacious conclusion based on insufficient evidence.

It's not about modified or natural rolls (both are just "rolls"). The rule says to ignore modifiers. So we ignore modifiers. Unless you can give me a different take on "irrespective of modifiers".

Correct. To really take it simply. If you roll 6 dice. And there is a -2 Modifier and a BS+1 on the model.

Your initial roll is 1, 2, 3, 3, 5, 6
Once you apply modifier, physically pick up the dice and take down by 2. 1 being the minimum. So you pick up the dice and take away 2 so you have now rolled. 1, 1, 1, 1, 3, 6. End result four misses and 2 hits. Even if you have a BS +1 as 1 is always a miss.

I kinda feel like you missed Dandelion's point.

Out of the rolls, irrespective of modifiers, which one's are ones? Those auto fail, everything else then has modifiers applied and checked against the shooters BS as to if they hit or not.

The end result is the final roll that matters.

Sorry, could you be more precise? Because what I think you are saying is that when considering the rule "A roll of 1 always fails, irrespective of any modifiers that may apply", that we consider the roll respective of the modifiers that do apply? Ie: In your example we consider the result of 1, 1, 1, 1, 3, 6.

JakeSiren wrote:

 SeanDavid1991 wrote:

JakeSiren wrote:

 SeanDavid1991 wrote:

Dandelion wrote:

 skchsan wrote:
It APPEARS contradictory because it stands against your understaning of the RAW. The rulebook makes no distinction between what we colloquially refer to as "natural rolls" and "modified rolls." The assumption that "only natural rolls of 1 auto-fails" is a fallacious conclusion based on insufficient evidence.

It's not about modified or natural rolls (both are just "rolls"). The rule says to ignore modifiers. So we ignore modifiers. Unless you can give me a different take on "irrespective of modifiers".

Correct. To really take it simply. If you roll 6 dice. And there is a -2 Modifier and a BS+1 on the model.

Your initial roll is 1, 2, 3, 3, 5, 6
Once you apply modifier, physically pick up the dice and take down by 2. 1 being the minimum. So you pick up the dice and take away 2 so you have now rolled. 1, 1, 1, 1, 3, 6. End result four misses and 2 hits. Even if you have a BS +1 as 1 is always a miss.

I kinda feel like you missed Dandelion's point.

Out of the rolls, irrespective of modifiers, which one's are ones? Those auto fail, everything else then has modifiers applied and checked against the shooters BS as to if they hit or not.

The end result is the final roll that matters.

Sorry, could you be more precise? Because what I think you are saying is that when considering the rule "A roll of 1 always fails, irrespective of any modifiers that may apply", that we consider the roll respective of the modifiers that do apply? Ie: In your example we consider the result of 1, 1, 1, 1, 3, 6.

It is a weird scenario I'm only explaining it how the guys in our gaming centre play it and interpret it. We all agree tis fairest. End result that matters, but any 1's before the modifiers are kind of like a black hole. if it's a 1 initially modifiers just don't get applied to that roll.

 SeanDavid1991 wrote:

Spoiler:

JakeSiren wrote:

 SeanDavid1991 wrote:

JakeSiren wrote:

 SeanDavid1991 wrote:

Dandelion wrote:

 skchsan wrote:
It APPEARS contradictory because it stands against your understaning of the RAW. The rulebook makes no distinction between what we colloquially refer to as "natural rolls" and "modified rolls." The assumption that "only natural rolls of 1 auto-fails" is a fallacious conclusion based on insufficient evidence.

It's not about modified or natural rolls (both are just "rolls"). The rule says to ignore modifiers. So we ignore modifiers. Unless you can give me a different take on "irrespective of modifiers".

Correct. To really take it simply. If you roll 6 dice. And there is a -2 Modifier and a BS+1 on the model.

Your initial roll is 1, 2, 3, 3, 5, 6
Once you apply modifier, physically pick up the dice and take down by 2. 1 being the minimum. So you pick up the dice and take away 2 so you have now rolled. 1, 1, 1, 1, 3, 6. End result four misses and 2 hits. Even if you have a BS +1 as 1 is always a miss.

I kinda feel like you missed Dandelion's point.

Out of the rolls, irrespective of modifiers, which one's are ones? Those auto fail, everything else then has modifiers applied and checked against the shooters BS as to if they hit or not.

The end result is the final roll that matters.

Sorry, could you be more precise? Because what I think you are saying is that when considering the rule "A roll of 1 always fails, irrespective of any modifiers that may apply", that we consider the roll respective of the modifiers that do apply? Ie: In your example we consider the result of 1, 1, 1, 1, 3, 6.

It is a weird scenario I'm only explaining it how the guys in our gaming centre play it and interpret it. We all agree tis fairest. End result that matters, but any 1's before the modifiers are kind of like a black hole. if it's a 1 initially modifiers just don't get applied to that roll.

Ah, I see. It tends to help in these discussions if you prefix your house ruling with HIWPI, otherwise people (like me) think you are trying to discuss rules as written.

 skchsan wrote:
The excerpt "a roll of 1 always fails, irrespective of modifiers." makes no 'explicit statement' that the roll refers to unmodified dice roll.

Yes, it does. It explicitly says "irrespective of any modifiers that may apply". When something tells you "x always fails, ignoring modifiers", it's explicitly saying "unmodified x always fails". You seem to want "any modifiers" to only mean "any positive modifiers", but it doesn't. You ignore all modifiers when checking to see if your roll fails because it was a 1.

Benn Roe wrote:

 skchsan wrote:
The excerpt "a roll of 1 always fails, irrespective of modifiers." makes no 'explicit statement' that the roll refers to unmodified dice roll.

Yes, it does. It explicitly says "irrespective of any modifiers that may apply". When something tells you "x always fails, ignoring modifiers", it's explicitly saying "unmodified x always fails". You seem to want "any modifiers" to only mean "any positive modifiers", but it doesn't. You ignore all modifiers when checking to see if your roll fails because it was a 1.

You're reading the sentence as "a roll of 1, irrespective of any modifers that may apply, always fails." instead of reading it as written.

Placement of commas make all the difference.

It really doesn't in this case. You're reading it as two final thoughts, loosely connected, but it's a single sentence that can't be considered properly in pieces. "A roll of 1 always fails", by itself, definitely means what you want it to mean, but you're otherwise imagining an implication that placing "irrespective of any modifiers that may apply" after that phrase just doesn't carry, namely that the only relevant modifiers are those that further reduce your result to 1.

Benn Roe wrote:
It really doesn't in this case. You're reading it as two final thoughts, loosely connected, but it's a single sentence that can't be considered properly in pieces. "A roll of 1 always fails", by itself, definitely means what you want it to mean, but you're otherwise imagining an implication that placing "irrespective of any modifiers that may apply" after that phrase just doesn't carry, namely that the only relevant modifiers are those that further reduce your result to 1.

They are two separate phrases of a related topic.

A roll of 1 always fail (independent clause), irrespective of any modifiers that may apply (dependent clause).

This sentence is a combined form of two sentences:
A roll of 1 always fails. A roll of 1 always fail regardless of any modifiers that may apply.

By claiming modified rolls of 1 doesn't always fail is a direct contradiction to the first complete thought, as the rulebook makes no distinction between modified rolls and natural rolls when it uses the word 'roll'.

You are understanding the clause "irrespective of any modifiers that may apply" as a relative clause whose function is to modify or act as an adjective to the phrase "A roll of 1".

The correct interpretation here is to say 'the failure stands irrespective of any effect modifiers may have,' and not 'roll of 1, in other words, a dice roll showing 1 facing up, fails irrespective of any modifiers.'

This argument is a matter of reading comprehension, not a matter of ambiguity in the rule writing.

 skchsan wrote:
This argument is a matter of reading comprehension, not a matter of ambiguity in the rule writing.

On that, we agree. The problem is that independent clauses are "independent" only because they could stand on their own structurally, not because they're a finished final thought. If "a roll of 1 always fails" were true without qualification, it could and would have been the whole sentence. One of the primary functions of a dependent clause is to modify an adjacent clause, and that's exactly what's happening here. You're asserting that the dependent clause is modifying the independent one only in so far as it widens the definition of "rolling a 1", but that's an implication that just isn't there. It's clarifying what it means by "rolling a 1" in this context. That's why it says "any modifiers" rather than "modifiers that would raise it".

 skchsan wrote:

Benn Roe wrote:
It really doesn't in this case. You're reading it as two final thoughts, loosely connected, but it's a single sentence that can't be considered properly in pieces. "A roll of 1 always fails", by itself, definitely means what you want it to mean, but you're otherwise imagining an implication that placing "irrespective of any modifiers that may apply" after that phrase just doesn't carry, namely that the only relevant modifiers are those that further reduce your result to 1.

They are two separate phrases of a related topic.

A roll of 1 always fail (independent clause), irrespective of any modifiers that may apply (dependent clause).

This sentence is a combined form of two sentences:
A roll of 1 always fails. A roll of 1 always fail regardless of any modifiers that may apply.

By claiming modified rolls of 1 doesn't always fail is a direct contradiction to the first complete thought, as the rulebook makes no distinction between modified rolls and natural rolls when it uses the word 'roll'.

This argument is a matter of reading comprehension, not a matter of ambiguity in the rule writing.

I think you have it wrong. "Irrespective of any modifiers that may apply" is a qualifier on "a roll of 1 always fail"

I'd like to point out that Games Workshop did in fact create a new phrase to differentiate "rolls of x" and "unmodified rolls of x".

Look at the Codex T'au Tidewall Shieldline - it reflects shots on an "unmodified save roll of 6", meaning you can reflect melta shots into mortal wounds.

The real significance of this statement though is that it is different phrasing from "rolls of 1 always fail, irrespective of modifiers". I think that the difference between the two phrases leans me towards the interpretation that ALL 1's fail, natural and modified.

Another way to look at this is that everywhere else in the game that uses the term "roll of 1" seems to be using it to refer to the final modified roll. They don't have separate terminology for "unmodified rolls". A roll is a roll, including its modifiers. So, the independent clause on its own means "modified rolls of 1 always fail". The dependent clause then tells us, however, that no modifiers apply here ("irrespective of any modifiers"), which leaves us with "unmodified rolls of 1 always fail". There is no reading that leaves you with both modified and unmodified rolls of 1 always failing. Any assertion to the contrary is nonsense.

EDIT: I stand corrected. They do have terminology for unmodified rolls. It's "unmodified rolls", as pointed out by the poster above me. That actually reinforces my point, even though the poster above me drew the opposite conclusion from that discovery.

The reason I draw the opposite conclusion is as follows:

"Rolls of 1 always fail,.."

I see this as modified rolls of 1 always fails.

"... irrespective of any modifiers."

This part of the phrase seems to increase the scope of the 1's that fail to include 1's before modifiers.

I understand how you can interpret it your way though, as they should rewrite the rule in the Spring FAQ to say "unmodified rolls and rolls of 1 always fail" if that is in fact what they intended, so that it conforms with their new and old terminology.

Pieceocake wrote:
I think that the difference between the two phrases leans me towards the interpretation that ALL 1's fail, natural and modified.

This doesn't follow, as pointed out in my post above. Everywhere else in the game where they don't use the term "unmodified", they mean "modified", but the qualifier tells us explicitly to ignore modifiers for this calculation. "A roll of 1 always fails, irrespective of any modifiers that might apply" is synonymous with "an unmodified roll of 1 always fails". The only way that might not be true is if "a roll of 1 always fails" already meant "an unmodified roll of 1 always fails", and we know it doesn't. The use of "unmodified" is likely a recent change in terminology to make things clearer.

Pieceocake wrote:
The reason I draw the opposite conclusion is as follows:

"Rolls of 1 always fail,.."

I see this as modified rolls of 1 always fails.

"... irrespective of any modifiers."

This part of the phrase seems to increase the scope of the 1's that fail to include 1's before modifiers.

Well put.

 skchsan wrote:

Pieceocake wrote:
The reason I draw the opposite conclusion is as follows:

"Rolls of 1 always fail,.."

I see this as modified rolls of 1 always fails.

"... irrespective of any modifiers."

This part of the phrase seems to increase the scope of the 1's that fail to include 1's before modifiers.

Well put.

Except that it ignores the fact that "irrespective of any modifiers that may apply" doesn't mean "irrespective of any modifiers that would increase it". I understand what you guys are saying, but it's a misreading.

Except that it ignores the fact that "irrespective of any modifiers that may apply" doesn't mean "irrespective of any modifiers that would increase it". I understand what you guys are saying, but it's a misreading.

"irrespective of any modifiers that may apply" does mean "irrespective of any modifiers that would increase it".

"irrespective of any modifiers that may apply" also means irrespective of any modifiers that would decrease it.

Hence my reading above that essentially boils down to rolls of 1 always fail, and unmodified rolls of 1 too.

Pieceocake wrote:
I'd like to point out that Games Workshop did in fact create a new phrase to differentiate "rolls of x" and "unmodified rolls of x".

Look at the Codex T'au Tidewall Shieldline - it reflects shots on an "unmodified save roll of 6", meaning you can reflect melta shots into mortal wounds.

The real significance of this statement though is that it is different phrasing from "rolls of 1 always fail, irrespective of modifiers". I think that the difference between the two phrases leans me towards the interpretation that ALL 1's fail, natural and modified.

Or that they realized after writing the main rules how cumbersome the phrasing was, and by the time the T'au codex was written they streamlined it to "unmodified"

"Irrespective of any modifiers that may apply" is not the same as "ignore modifiers when determining an outcome".

As Pieceocake correctly said, the sentence refers to both modified and unmodified rolls of 1. A 1 fails regardless of whether it's had a -1 modifier or not applied to it. The "regardless of any modifiers" bit is clearly to get around auto hitting shenanigans with stacking +1's to hit.

WS1+ does nothing, don't apply that combat drug to your Succubus.

Pieceocake wrote:

Except that it ignores the fact that "irrespective of any modifiers that may apply" doesn't mean "irrespective of any modifiers that would increase it". I understand what you guys are saying, but it's a misreading.

"irrespective of any modifiers that may apply" does mean "irrespective of any modifiers that would increase it".

"irrespective of any modifiers that may apply" also means irrespective of any modifiers that would decrease it.

Hence my reading above that essentially boils down to rolls of 1 always fail, and unmodified rolls of 1 too.

If it's irrespective of any modifiers that would increase it and any modifiers that would decrease it, how would it be applying to both modified and unmodified rolls?

 doctortom wrote:
If it's irrespective of any modifiers that would increase it and any modifiers that would decrease it, how would it be applying to both modified and unmodified rolls?

"A roll of 1 always fails, irrespective of any modifiers that may apply." - this means that a 1 fails if it is modified and if it is not modified. The roll of 1 always fails.

Many of you seem to be misinterpreting "irrespective". Irrespective in this instance is better thought of as "regardless". It doesn't matter if the roll is modified to become a 1 or not, a natural 1 fails, a modified 1 fails. 1 always fails. This has been consistent across every edition of the game.

 An Actual Englishman wrote:

 doctortom wrote:
If it's irrespective of any modifiers that would increase it and any modifiers that would decrease it, how would it be applying to both modified and unmodified rolls?

"A roll of 1 always fails, irrespective of any modifiers that may apply." - this means that a 1 fails if it is modified and if it is not modified. The roll of 1 always fails.

Many of you seem to be misinterpreting "irrespective". Irrespective in this instance is better thought of as "regardless". It doesn't matter if the roll is modified to become a 1 or not, a natural 1 fails, a modified 1 fails. 1 always fails. This has been consistent across every edition of the game.

Exactly this. But don’t think logic will stop this Crazy Train... ;-)

Pieceocake wrote:

Except that it ignores the fact that "irrespective of any modifiers that may apply" doesn't mean "irrespective of any modifiers that would increase it". I understand what you guys are saying, but it's a misreading.

"irrespective of any modifiers that may apply" does mean "irrespective of any modifiers that would increase it".

"irrespective of any modifiers that may apply" also means irrespective of any modifiers that would decrease it.

Hence my reading above that essentially boils down to rolls of 1 always fail, and unmodified rolls of 1 too.

-A roll of 1 always fails, irrespective of any modifiers that may apply.
"May apply" is a present conditional. The modifiers are already applied. As soon as you roll the dice they are modified. It all happens in one step. There's no "remove natural 1s" and then "remove modified 1s". Natural 1s don't exist, only modified 1s. So we can't make the distinction between the two when determining whether something is a 1 or not.

All 1s are modified 1s. We must now look at any modified 1s irrespective of the modifiers that apply to see if they fail.

So now we can read this 2 ways:
- You remove the modifiers to see if any dice are 1s.
- You ignore whether or not any 1s have been modified and fail them all.

For case 2, rolls cannot be modified below 1. So, a 1-1 becomes a 1. And a 2-1 becomes a 1. In this second case, both 2s and 1s auto-fail.
But, if we have +1, then the 1 becomes a 2. This 2 is not a modified 1 and therefore must pass. Again natural 1s don't exist, only modified 1s. With +1 you simply rolled a 2. There's no step before this.

This of course does not help anything since a +1 allows auto-hits, so it can't be this.

The modifiers are already applied. As soon as you roll the dice they are modified. It all happens in one step.

This is false. They now reference unmodified rolls, at least since the Tau Codex. Check the Tidewall Shieldline reflecting shots mechanic for evidence.