Tactica: Tomb Kings 8th Edition Book

[Jericho]

You can't do mathhammer like that for artillery dice since each Stalker cannot get 5 on an artillery dice. The average roll on a single artillery dice is a 6 and this is applied to each Stalker. You can only apply the above when rolling multiples of 6 artillery dice. Note however that in this case it doesn't matter, as 15 shots with 6's to wound yields 2.5 wounds, and the only way to get 2.5 wounds on a dice is to roll 3 wounds.

That really isn't how averages work

Mean/median/mode all mean very different things in Statistics.

Regarding Arcane items, the only one I plan on taking in lists is Enkhil's Kanopi. Dispel Scrolls are dumb. YMMV.

[Lilike]

Will I be taking 6? Yes. For prescisely the reasons listed. They annhilate heavy cavalry and monsters hard and aren't vulnulerable to support units like a unit of 3 is.

Couldn't agree with you more on the general effectiveness of stalkers. The gaze attack is spectacular and they can be an threat to any unit on the board since you can pop up just next to them. In addition their combat profile isn't nearly as bad as people make it out to be, they are no necropolis knights but they are strong enough to deal with most weaker units that will be directed their way. A unit of 6 strong will be magic in the shooting phase and will hit rather hard in CC. There are however problems with taking an single unit of 6, one is that you can lose a rather large amount of points if you have bad dice when they come up. The other is that their range is short and their bases big which can lead to not everyone in the unit being able to fire upon an target without getting really really close which may not always be possible.

[Citadel97501]

Couldn't agree with you more on the general effectiveness of stalkers. The gaze attack is spectacular and they can be an threat to any unit on the board since you can pop up just next to them. In addition their combat profile isn't nearly as bad as people make it out to be, they are no necropolis knights but they are strong enough to deal with most weaker units that will be directed their way. A unit of 6 strong will be magic in the shooting phase and will hit rather hard in CC. There are however problems with taking an single unit of 6, one is that you can lose a rather large amount of points if you have bad dice when they come up. The other is that their range is short and their bases big which can lead to not everyone in the unit being able to fire upon an target without getting really really close which may not always be possible.

The ETBS rule is why, I would rather have a couple units of 3, rather than 1 unit of 6. I always try to avoid randomness in my armies, its prevalent to much as is. Its one reason, I try to limit my exposure to chance as much as possible.

[TheKingInYellow]

I plan to try two units of three stalkers, they can have the first unit fire on a WM and the second can also fire if the WM is not destroyed, or find a new target if they are.

Assuming you have two WMs close enough together, anyhow.

[Spiney Norman]

You can't do mathhammer like that for artillery dice since each Stalker cannot get 5 on an artillery dice. The average roll on a single artillery dice is a 6 and this is applied to each Stalker. You can only apply the above when rolling multiples of 6 artillery dice. Note however that in this case it doesn't matter, as 15 shots with 6's to wound yields 2.5 wounds, and the only way to get 2.5 wounds on a dice is to roll 3 wounds.

Of course you can, thats like saying the average result for 1D6 is 4, therefore the average result for 3D6 is 12, its just plain wrong. The average score for one artillery dice is between 4 and 6 just as the average score for 1D6 is between 3 and 4 (conveniently written as 3.5, but then average scores don't have to be actually attainable, they are merely mathematically representative). The result "6" is no more the average result of an artillery dice than "4", but if I were to suggest that (4x3=) 12 was the average result for 3 artillery dice I'd be slammed for being overly negative (I'd also be just as wrong as suggesting it was 18).

The average for 2 artillery dice IS 10 and the average result for 1 artillery dice IS 5, because in a perfect statistical world if you rolled the artillery dice 100 times, added up all the results and divided the total by 100 you'd get 5.

Whats more "the only way to get 2.5 wounds on a dice is to roll 3 wounds" is also wrong, what that means is that on average half the time you will get 3 wounds, and half the time you will only get 2. All average are based on a projected series of events, NOT a single event.

[GreySeerZ]

Of course you can, thats like saying the average result for 1D6 is 4, therefore the average result for 3D6 is 12, its just plain wrong. The average score for one artillery dice is between 4 and 6 just as the average score for 1D6 is between 3 and 4 (conveniently written as 3.5, but then average scores don't have to be actually attainable, they are merely mathematically representative). The result "6" is no more the average result of an artillery dice than "4", but if I were to suggest that (4x3=) 12 was the average result for 3 artillery dice I'd be slammed for being overly negative (I'd also be just as wrong as suggesting it was 18).

The average for 2 artillery dice IS 10 and the average result for 1 artillery dice IS 5, because in a perfect statistical world if you rolled the artillery dice 100 times, added up all the results and divided the total by 100 you'd get 5.

Whats more "the only way to get 2.5 wounds on a dice is to roll 3 wounds" is also wrong, what that means is that on average half the time you will get 3 wounds, and half the time you will only get 2. All average are based on a projected series of events, NOT a single event.

Yup, average does not necessarily mean that you CAN roll that number, but over the course of your dice slinging career the average of all of your rolls should be close to that number. You'll never roll a 3.5 but your number of 3 rolls should balance out with your number of 4 rolls just like your number of 2 rolls should balance out with your number of 5 rolls, likewise 1 and 6, since they all have the same probability (unless your dice aren't balanced). All of those pairings average to 3.5, just as the total does (1+2+3+4+5+6)/6 = 3.5.

[GodlessM]

That really isn't how averages work

Yes, but the normal mode of averages doesn't always work for dice due to everything being integer based and set within certain perameters. You can say something gets a 5 on an artillery dice, but that's impossible to do, however the closest result that allows you to achieve this is a 6 on the artillery dice.

People can say this is wrong all they want, but normal averages don't always work on dice plain and simple (you have to round to integers), however if they go roll a 2.5 then I will say they've proved me wrong.

[GreySeerZ]

Yes, but the normal mode of averages doesn't always work for dice due to everything being integer based and set within certain perameters. You can say something gets a 5 on an artillery dice, but that's impossible to do, however the closest result that allows you to achieve this is a 6 on the artillery dice.

People can say this is wrong all they want, but normal averages don't always work on dice plain and simple (you have to round to integers), however if they go roll a 2.5 then I will say they've proved me wrong.

well, if the dice landed on its edge..... Seriously though, at the same time you could argue why a 6, why not a 4? You can't just pick one, and say thats the closest, because they are equally close.

The true values of an artillery dice are 0, 2, 4, 6, 8, 10. It could be 4 OR 6, so 5 is a good way to average. With 3 you might roll 4 6 4 or 4 6 6 or 6 6 6 or 4 4 4. Add all those together, and divide by 12 and you get an average of 5 per model.

[EDMM]

GodlessM, your posts indicate a gross ignorance of statistics and probability.

I suggest you not post any further comment on this topic until you have availed yourself of a statistics class or textbook.

DISCLAIMER:
This post is intended as advice intended on helping another individual save face and ignore future shame. It is in no way intended as an insult, but should be taken as a face value, including the word "ignorance," which used only its technical sense and not with a derogatory intent.

[Nighthawke]

something odd came up at my store the other day, dont know if its been mentioned though, is it worth giving a tomb king the wizard hat to make him your hierophant, means you can do a chariot army etc without a special character or models no on them

[Siphon]

Is it worth it? No. Would it be fun and interesting and therefore something you should try, yes.

[GreySeerZ]

something odd came up at my store the other day, dont know if its been mentioned though, is it worth giving a tomb king the wizard hat to make him your hierophant, means you can do a chariot army etc without a special character or models no on them

Don't have my book, but I'm pretty sure your heirophant has to be a High Priest or Priest. Only special characters remove this requirement, such as Settra/Arkhan. I fear the only way to include a mobile hierophant is calvary. Though I may be wrong.

[Falkman]

The Hierophant has to use the Lore of Nehekhara (Arkhan is an exception), which rules out a King with Wizard hat.

[Nighthawke]

well it says the heirophant has to bea nehekharan undead wizard with the highest wizrd level, which if you give him the wizard hat he is.
and surely by it saying he must use the nehek lore that orverides his hats rules and makes him use it, since the only reason arkhan dosnt use it is his rules specifically say he uses the lore of death even if he is the hierophant?

[GreySeerZ]

The Hierophant has to use the Lore of Nehekhara (Arkhan is an exception), which rules out a King with Wizard hat.

Ah, yea, that was it, looks like Arkhan/Settra are the only way to go. Special characters are allowed in most tournament settings, and your allowed to remodel/rename them to be whatever you want, your just stuck with their rules, gear and points cost. I have no problem with them, and ours seem pretty balanced, but it is fun to build characters every once in a while.

That said, a unit of skeleton horse archers would not be that out of place in a chariot list.

well it says the heirophant has to bea nehekharan undead wizard with the highest wizrd level, which if you give him the wizard hat he is.

Hes an undead wizard, sure, but has to have the Nehekhara lore... the rules for the hat specifically state that the wizard gains a random rulebook lore.

and surely by it saying he must use the nehek lore that orverides his hats rules and makes him use it, since the only reason arkhan dosnt use it is his rules specifically say he uses the lore of death even if he is the hierophant?

Nope, the Nehekhara lore is a pre-requisite to become the heirophant, he can't become the heirophant and then backstep to pickup the lore. Thats like (and excuse my American-ness) being president also gives you the role of commander and chief of the military. You cannot become the commander and chief of the military directly, and even if you could, it does not automatically give you the presidency.

[Nighthawke]

ah right fair enough, thanks for explaining it still might do it incase i get something nice , just have to have a liche in there as well

[GreySeerZ]

GodlessM, your posts indicate a gross ignorance of statistics and probability.

I suggest you not post any further comment on this topic until you have availed yourself of a statistics class or textbook.

DISCLAIMER:
This post is intended as advice intended on helping another individual save face and ignore future shame. It is in no way intended as an insult, but should be taken as a face value, including the word "ignorance," which used only its technical sense and not with a derogatory intent.

I don't think the word ignorance is the derogatory one, but the use of "gross" is offensive. Thats like saying, your not not getting this small principle, you are REALLY not getting anything related to statistics (implying that it is either impossible or unlikely that he will = he is unintelligent). Also telling him not to post anymore until he educates himself on this topic makes you come off as an elitest jerk. It would have been much more helpful/informative to post quotations from a text, or offered a resource or private conversation to explain, but your post really comes off as "your REALLY not getting this, stop posting".

Judging Godless's previous posts, he is not totally inept at statistics, just misunderstanding this one principle (I know hes not basing it on statistics, but aside from taking the weight of dice/shape/friction into account and basing it on physics I am not really sure where he is going, though it may be abstract, haha). Also, your disclaimer did very little to address what made your post insulting, if anything it added to it by implying that he should be ashamed or that he is losing face, neither of which is true...FYI

[GodlessM]

GodlessM, your posts indicate a gross ignorance of statistics and probability.

I suggest you not post any further comment on this topic until you have availed yourself of a statistics class or textbook.

DISCLAIMER:
This post is intended as advice intended on helping another individual save face and ignore future shame. It is in no way intended as an insult, but should be taken as a face value, including the word "ignorance," which used only its technical sense and not with a derogatory intent.

EDMM, your posts indicate a gross ignorance of literacy.

I suggest you not post any further comment on this topic until you have read my posts properly, particularly the part when I specifically said I don't believe normal statistics and probability work accurately for dice and slightly different methods need to be used, because I do understand probability and don't think it works here (normally any math can be done across any sets of numbers; dice limit you to integer values).

DISCLAIMER:
This post is intended as advice intended on helping another individual save face and ignore future shame. It is in no way intended as an insult, but should be taken as a face value, including the word "ignorance," which used only its technical sense and not with a derogatory intent. It is to point out I'm playing devil's advocate on the use of normal methods specifically because I do understand them and don't feel they are accurate here (if only you knew what I used to do for a living).

[Methios]

People can say this is wrong all they want, but normal averages don't always work on dice plain and simple (you have to round to integers), however if they go roll a 2.5 then I will say they've proved me wrong.

They dont ALWAYS work, but on avarage they will.
Saying the avarage wont always work is silly...ofcourse they wont.
But its the best we can work with.

[Spiney Norman]

Yes, but the normal mode of averages doesn't always work for dice due to everything being integer based and set within certain perameters. You can say something gets a 5 on an artillery dice, but that's impossible to do, however the closest result that allows you to achieve this is a 6 on the artillery dice.

People can say this is wrong all they want, but normal averages don't always work on dice plain and simple (you have to round to integers), however if they go roll a 2.5 then I will say they've proved me wrong.

Whew, I think we need to chill here.

The problem is that the very fact that you chose "6" to represent the average result demonstrates your bias towards proving the unit to be more effective than it is. You could also have chosen "4", which is just as close to the actual average result as "6", but obviously it would portray the unit as less effective than it really is, rather than more effective, which was clearly your intention.

You certainly can't carry that modified average into a calculation that would involve multiple dice rolls, as it would be expected that the more dice you roll the closer you would get to an average result, therefore an average of 4 artillery dice would not be (4x6=) 24, nor would it be (4x4=) 16, but in fact (2x4)+(2x6)= 20, which is the actual average result (and is coincidentally actually attainable by 4 stalkers shooting).