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Number Crunching
Topic Started: 19th January 2012 - 09:45 PM (927 Views)
Hotdogg
Clanrat
Is there a link to all things number related in 8th ed?
something that shos how to calculate things like average results on 3d6, what to expect a hellpit to do in combat, what average str is the WLC going to do, etc?
thanks for any help and for not bashing me too bad!
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Flem
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A blood curling cough a day, keeps the assassins away

The avarage roll of each dice is 3,5 . So if you roll 3D6 your average roll would be 3 * 3,5 , so 10,5.
The 3,5 result lies between3,4 . On an artillery dice this translates to 6-8, aka the average result of your warplightning cannon.
On 3D6 attacks if you do 10,5 hits, with WS 3 you'll mostly hit on 4+. So half of them hit, being 5,25. Of those 5,25 with a St of 6 against normal troops (T 3-4) one out of 6 won't wound. So 5,25 - (5,25/6) = 4,375wounds before saves.

Keep in mind this only statistics, in a real battle you won't roll average on every roll some will be good and some bad. For example on a D6 you have a 1 in 6 chance to roll a 3. If you role one D6 6 times you might have 1, more or even no 3's.

Flem,
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Krogholt
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Flem
20th January 2012 - 03:55 AM
... On an artillery dice this translates to 6-8, aka the average result of your warplightning cannon...
Well, no.

If there had been a "12" on the artillery die, then Yes. But since 12 = misfire, the average number rolled on artillery dice is 6

(2 + 4 + 6 + 8 + 10) / 5 = 30/5 = 6
Edited by Krogholt, 20th January 2012 - 08:42 AM.
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Flem
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A blood curling cough a day, keeps the assassins away

Krogholt
20th January 2012 - 08:42 AM
Flem
20th January 2012 - 03:55 AM
... On an artillery dice this translates to 6-8, aka the average result of your warplightning cannon...
Well, no.

If there had been a "12" on the artillery die, then Yes. But since 12 = misfire, the average number rolled on artillery dice is 6

(2 + 4 + 6 + 8 + 10) / 5 = 30/5 = 6
No cause you have 6 options so the if we follow the statistics of calculating the median it would lay between 3-4, which are the 6-8 results ,provided you see every number as the double of it's D6 result with the 6 as a misfire, and not see the order as Misfire, 2 , 4 ,6, 8, 10 cause then it would be 4-6.

Flem,

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Krogholt
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@Flem:

Well, that does not change that the "average number rolled" (which is equal to the average strength) is a 6.
Edited by Krogholt, 20th January 2012 - 12:17 PM.
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snowblizz
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Is it even meaningful to talk about an "average" with only one dice? Or dice in general for that matter? Like the artillery dice. To me it carries the connotation that it is the most likely outcome, but with one artillery dice I'm not more likely to roll 6 than anything else.

Doesn't it make more sense to say you are more likely to get a "high" S (6+) on the shot rather than talk about an "average S"?
Saying the average of a d6 is 3,5 really has no meaning in wargaming, but saying average of 2d6 is 7 would because that's the most common outcome.

Statistics and number theory has always baffled me though.
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SkavenDan
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Repent your evil prediction ways and trust in the power of dice! For ultimately the dice god controls the destiny of entire empires!
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Krogholt
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@ snowblizz: Definitely! And experience has probably taught us all that statistics has absolutely no influence once you start rolling those dice... Then its only a matter of praying to the Dice Gods.

Regardless, I just wanted to avoid some poor soul misinterpreting the second post of this topic and expecting to roll 6-8 strength for his WLC/Doomwheel, which is quite handy for killing... Dragons, and other high T monsters...
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Warlock Matik
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(Z–>)90º – (E–N²W)90ºt = 1

Gentlerats, we have a selection of mathhammer articles collected in a sticky in General Skaven here.

The beauty of maths is that it tends to be absolute; as such nearly all of those articles are still very relevant ;)
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SkavenDan
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Yeah that is some what right BUT the people who read things like that end up shouting "I should have got at least 3 6s and it should have died ARRRRRGGGGGG" *throws dice over board*
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Warlock Matik
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(Z–>)90º – (E–N²W)90ºt = 1

Very true Dan, those people need to realise that this sort of stuff only tells you what's most likely to happen, not what will :P If Quantum Mechanics has taught me anything it's to not rely on what you expect to happen ;)
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Rein bringer of the cheese
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Well Quantum mechanics will very much follow the road of most likely to happen ;) The thing I do not (want to) understand is that particles interfere with themselves and only themselves !

Ok back to statistics:

You have to understand what statistics mean in general. What does mean or average or standard deviation mean for me and the dice?

I always use the 'law of the large number of dice' which should translate to something close to the average. (which is not true but seems to happen a lot :) )
I also calculate the average of something. Then I expected my result to lie between -15% and + 15% of the average or something like that. Not so exact I just do this: if the average is 8 then I expect 6.5-9.5 wounds which in whole numbers is 6-10

To reply to the artillery dice: Yes the average of the dice is 6+1/6 misfire. There is no other way of expressing this average. But if you want to wound a high T monster, lets say a sphinx with T8, then 1 result will be very good (10), one result 50/50 (8) and 2-6 only 1/6 th good. This translates for me as: 8 and 10 are ok so I got like 30% chance of doing it wounds.

The number I want to add is: If you throw 3 d6 and discard the lowest then your average result is 8.6
This also implies that if you throw 3 d6 and discard the highest your average roll will be 5.4
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SkavenDan
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@Rein bringer of the cheese I don't disagree on any point you don't get .something everything should be rounded of for math hammer as you can't have in between values. so:

The number I want to add is: If you throw 3 d6 and discard the lowest then your average result is 8.6 =9
This also implies that if you throw 3 d6 and discard the highest your average roll will be 5.4 =5
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CapAmr05
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If Quantum Leap has taught me anything it's to not rely on what you expect to happen


Oh boy....

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Fixed that image for you ;)
Edited by Warlock Matik, 20th January 2012 - 10:01 PM.
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IronShark
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SkavenDan
20th January 2012 - 01:16 PM
Yeah that is some what right BUT the people who read things like that end up shouting "I should have got at least 3 6s and it should have died ARRRRRGGGGGG" *throws dice over board*
Well, at least that's what people who don't understand probability will say.

Here's a very basic example. Lets say you have 3 Jezzails shooting at long range. They need a 5+ to hit. On average, that's one hit. That's the average, not the probability.

The Probability works out as follows:

Zero Hits: 0.296
One Hit: 0.444
Two Hits: 0.222
Three Hits: 0.037


So, while you can see that a result of one hit is the most probable result, there is actually less than a 50% chance of getting exactly one hit. Which means more often than not, the Jezzails won't get exactly one hit.


I'm a fan of doing the math, partly because just doing the math is kind of entertaining for me. But you have to understand what the math is saying, and often it's not saying anything definitive. For example, look at the above results? Can you take anything from that? All it's saying is that 3 Jezzails have a decent chance of rolling one hit, but the odds of getting zero hits or two hits are significant as well. Most Warhammers know the approximate odds even without doing the actual calculations.



For the original poster (or anyone else interested) I can do some math for you if you have some specific questions.
Edited by IronShark, 20th January 2012 - 10:08 PM.
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