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Hey guys,

I have been using Method 216, http://mathhammer.blogspot.com/2009/11/everyone-deserves-second-chance.html?showComment=1414603112427#c3688465276386615972 .

How do you calculate -

Rending
Preferred Enemy
Melta Rule


If anyone has the formula, I would love to have it.

Thanks Dakka

Are you asking for a specific 'shortcut' method like that in the link or just how to do it in general?

I don't usually go for short cut methods because they just obscure the details and don't generalize to other situations easily. For instance, the 216 method linked is just multiplication in which you assume there are only 3 factions multiplied and the denominators are always 6. If you do things in more general ways these 'short cuts' become obvious after a few times.

216 is good for simple calculations, but does not work with complex rules in place.

Calculating special rules often revolves to big statistical calculations and are not simple formulas.

 BoomWolf wrote:
216 is good for simple calculations, but does not work with complex rules in place.

Calculating special rules often revolves to big statistical calculations and are not simple formulas.

I am sure someone who is smart enough could come up with an algorithm for just about any and every instance where a special rule may or may not apply. However, such knowledge would be highly suspect and secretly hoarded. So good luck finding it.

There ARE algorithms, but they are more complex and take a long time to do manually.

There are calculators across the interwebs, of varius levels of reliability that can cover the common rules (like melta, lance, rending, etc), but the really unique stuff might require manual work.


For example
http://www.heresy-online.net/combatcalculator/shooting.php
Is a pretty decent tool.

Just turn it into a probability instead of using all that 216 stuff.

Each side of a die has a 0.167 chance of being rolled (1/6)
Now imagine a space marine dreadnought is firing an assualt cannon. As he is hitting on a 3+, he has a 0.167*4 = 0.666 chance of hitting, as 4/6 sides of the die will cause a hit.

An assault cannon has 4 shots, so we can then multiply the number of shots it has by the chance that each shot will hit, that being 0.666*4= 2.67 to get the number of hits that it will score.

To work out the rending aspect of that, we then have to think about the probability of a rending hit occuring. This is acutally quite simple, as only one side of the die can cause this, and so the chance is 0.167 To find the chance of that dreadnought causing a rend, we therefore take the number of hits it will cause, and multiply it by the probability of each hit causing a rend, 2.67*0.167 = 0.445

I believe this is correct, although do feel free to correct me! You can use this general idea to work out any roll within warhammer

I don't quite understand the point of method 216... is it supposed to be a simplification?

The 216 just comes from the fact 3 rolls mean you are doing "something over 6" 3 times, so the denominator is 6*6*6 = 216.

So if you need a 3+ to hit, 4+ to wound, 3+ to save, your chance of killing is...

(4/6)*(3/6)*(2/6) = (4*3*2)/216

The way to figure out more complicated stats is just ignore the 216 simplification and do the full calculation.

So lets say you want to figure out a S4 rending attack against a 3+ save Space Marine.

Lets start by assuming we have already hit so I can drop that out of the calculation. You need to think of all possible outcomes that will result in success.

1. You wound, succeed in rending, the model gets no save.

2. You wound, but fail to rend, the model then fails their save.

Those are the 2 ways you can kill a model with a rending attack.

Sooo...

1. The probability of this happening is 1/6 (you roll a 6 on the "to wound" roll.

2. The probability of #2 happening is 2/6 (you roll either a 4 or a 5, thus wounding without rending) and then the model has a 3+ save so they have a 2/6 chance of failing their save (roll either a 1 or a 2).

Thus, the chance of killing a model given you've already hit is the sum of these 2 possible successful outcomes.

1/6 + (2/6)*(2/6) = 0.2777.

Now I ignored the chance to hit, if you say you have Bs4 and need a 3+ to hit, that's a 4/6 (chance of hitting, so you just multiply the entire result by 4/6.

4/6*(1/6 + (2/6)*(2/6)) = 4/6*0.2777 = 0.185185

+++++++++++++++++++++++++++++++
You can do the same thing with twinlinking. You need to think of all the possible ways of succeeding. So with twin linking that's...

1. Rolling to hit the first time.

2. Failing to hit the first time, rerolling and succeeding to hit on the reroll.

For a Bs4 model, the chance of...

1. is 4/6 (because you need a 3, 4, 5 or 6 to hit)

2. is 2/6*4/6 (2/6 because you roll a 1 or 2 to miss, then 4/6 because you need a 3, 4, 5 or 6 to hit on the 2nd roll)

So then your chance of successfully hitting is the sum of these 2 results...

4/6 + 2/6*4/6 = 0.88888, or if you happen to be good at fractions you can figure it out in your head as 8/9.

So then your chance of hitting with Bs4 twin linked is 8/9, you then multiply this by your chance of wounding and chance of bypassing the save.

So I Bs4 model shooting with a twin linked S4 rending attack against a T4, 3+ save Space Marine is....

0.8888 (chance of hitting) multiplied by 0.27777 (chance of wounding and killing that we calculated earlier) is equal to 0.247 chance of killing with each shot.

++++++++++++++++++++++++++++++
Alright, so that's rending and twin linking.

Preferred enemy, you reroll 1's to hit. So the chances of successfully hitting (say a Bs3 model), you consider the possible ways of successfully hitting...

1. You roll a 4+, this has a 3/6 chance of happening.

2. You roll a 1 to hit (1/6 chance of happening) and then you roll a 4+ on your reroll (3/6 chance of happening)

So your chance of hitting becomes....

3/6 + (1/6*3/6)

The same applies to wounding as you also reroll 1's, so you just have to consider the 2 possible ways of success once again.

+++++++++++++++++++++++++++++
Melta

Ok, so melta is a bit different, because you're actually rolling an additional dice and then adding the results together. There is only 1 possible means of success....

1. The sum of the 2 results is greater than or equal to the armour value of the vehicle.

This then requires you to figure out the chance of rolling a 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 or 12 on 2D6. All the results that are high enough to give you a penetration or glance are then added together. This isn't difficult to do, I have a table where I've calculated these values, I don't really feel like going in how to calculate it, but I can tell you the chances are...

2 - 1/36
3 - 2/36
4 - 3/36
5 - 4/36
6 - 5/36
7 - 6/36
8 - 5/36
9 - 4/36
10 - 3/36
11 - 2/36
12 - 1/36

So... say you need to roll a 8+ to glance or penetrate, you would add 5/36 + 4/36 + 3/36 + 2/36 + 1/36.

You can also exploit the fact that 1 minus the chance of success = the chance of failure (because the chance of success + chance of failure has to equal 1, ie. you are 100% sure you will either succeed or fail, there is no other option). So say you only need a 4+ to penetrate, instead of adding up all the results which are successful, add up the chances of FAILURE, and do 1 minus that. So the chance of failure is 1/36 + 2/36, the chance of success is then 1 - (1/36 + 2/36).

You may not have understood any of that... but it's the most detailed description I'm willing to type out on an internet forum on how to calculate the things you asked If you don't understand it... most of this stuff should be covered in high school level mathematics/probability.

SYKOJAK wrote:

I am sure someone who is smart enough could come up with an algorithm for just about any and every instance where a special rule may or may not apply. However, such knowledge would be highly suspect and secretly hoarded. So good luck finding it.

Actually this algorithm would simply be a combination of basic combinatorial and probabilistic operations. It's far from being secret knowledge, just take books/articles on probability calculation and combinatorics and you'll be good to go. In fact applying this to 40k rules is rather elementary stuff.

The basic sequence for calculating shots --> casualties is:

Define n as the amount of shots fired. c is the number of casualties.

c = n * (p_hit) * (p_wound) * (p_notSaved)

p_hit can be calculated by taking the amount of positive results on a d6 for your case and divide them by 6 (the total number of possible results). So in case of BS 4 you will hit on 3,4,5,6 which is 4 out of 6 results. p_hit would be 4/6 = 0.67 = 66.67% in this example.

p_wound can be derived similarly. Let's take S3 vs T4 which means you wound on 5 and 6. So p_wound would be 2/6 = 0.33 = 33.33%.

p_notSaved can again be done the same way. Take a 5+ save, this will be 4/6 = 0.5 = 50%. Notice that you have to take the results that will be positive for the shooter to get the casualties, so the NOT saved results. In the case of a 5+ save you get wounds on a roll of 1,2,3 and 4, thus the 4/6 in the example.

A simple example will show that this is not that complicated. Let's say you shoot with 10 marines with bolters at Chaos Space Marines. It's in rapid fire range so n = 20. We have BS4 so p_hit will be 4/6. S4 vs T4 so p_wound is 3/6. The wounds will be not saved at 1 and 2, so p_wound is 2/6. Alltogether:
c = 20 * (4/6) * (3/6) * (2/6) = 2.22
In this example you will have 2.22 casualties.

Now if you want to add e.g. Rending you only have to consider how this influences any of the above steps. Correct me if I am wrong, but Rending means only that to wound rolls of 6 always wound and give AP2. Basically you need to calculate the probability for everything that does not trigger rending and add the probability with triggered rending.

For simplicity let's take our example from above (20 bolter shots at CSM) and assume that some effect gave those shots rending. In this case we can simply split up the calculation for the shots that do not gain rending and those that get.

c_noRending = 20 * (4/6) * (2/6) * (2/6) = 1.48
c_Rending = 20 * (4/6) * (1/6) * 1 = 2.22

c = c_noRending + c_Rending = 3.7

Now what happened there? For the non rending part we have 20 shots of which 4/6th will hit, then for 2 results on the d6 (4 and 5) we will wound. We leave out the 6 since that will be covered in the second term. And since there are non rending in the last part we have 2/6 since due to the 3+ armor save it will only kill on 1 and 2.

For rending it is the same for hitting - 4/6th of 20 shots. But now we consider only the rending shots, those that have a 6 for to wound. Those shots will not allow an armor save so we multiply by one. You could just leave multiplying by 1 out, but I kept it there for clarity. If the target model has a cover save you would take it into account there.

I hope this gave some insight for people who want to do those calculations. It's really pretty straightforward once you dig a bit into it.

edit: damn, you posted while I was writing and calculating ;D And yes, that "method" is supposed to be a simplification of doing the three staged part where you always multiply fractions with a denominator of 6.

 Murenius wrote:
edit: damn, you posted while I was writing and calculating ;D And yes, that "method" is supposed to be a simplification of doing the three staged part where you always multiply fractions with a denominator of 6.

As I was typing mine I kind of realised that someone who doesn't understand probability at all wouldn't understand it and someone who already knows a bit of probability probably wouldn't be helped by it as it's all stuff you learn pretty early in the piece, but I posted it anyway on the off chance it helps someone

I was thinking exactly the same. But maybe our posts help someone to use it for his problem and then adjust it.

I will say that the Method 216 thing seems like an unnecessary abstraction that's more likely to confuse people who don't understand probability when actually explaining how the ACTUAL probability works is no more complicated

Instead of method 216 I would simply break it down in to 3 fundamental principles...

Firstly...
On a given die roll, the chance of success is the number of faces which result in success divided by the total number of faces (and a D6 has 6 faces total, obviously ).

So, the chance your opponent will fail his 3+ save is 2/6... 2 faces result in failure, 6 faces total.

Ok, so now we know the probability of success for a single die.

Secondly....
Now, if we are rolling several dice in sequence, that is, the 2nd roll only occurs if the first one succeeds, the 3rd one only occurs if the 2nd one succeeds, etc, you MULTIPLY the results together. This is the case in hit/wound/save, as you roll to hit, if that was successful you roll to wound, if that was successful you roll to save. In this case you multiply the results together.

So a 3+ to hit, 4+ to wound and a 3+ save becomes

4/6 chance to hit multiplied by 3/6 chance to wound multiplied by 2/6 chance of your opponent failing their save.

4/6*3/6*2/6 = 0.111111

Thirdly...
The average number of expected successes is the number of times you repeat an certain event multiplied by the chance of success of a single event. This means if you have a certain number of attacks, multiply the number of attacks by the chance that a single attack would be successful... this gives you the average number of successes.

So if you have 20 attacks, using the same 3+ to hit, 4+ to wound, 3+ save that we already calculated... the average number of kills is...

20*0.11111 = 2.2 on average.

Forget about this "method 216" junk when learning the actual correct way of doing things is no more complicated

Sonyca wrote:
Just turn it into a probability instead of using all that 216 stuff.

Each side of a die has a 0.167 chance of being rolled (1/6)
Now imagine a space marine dreadnought is firing an assualt cannon. As he is hitting on a 3+, he has a 0.167*4 = 0.666 chance of hitting, as 4/6 sides of the die will cause a hit.
)

That doesnt work, as each roll is independent (does not affect any other roll)

If you fire an avenger bolt cannon, which has 7 shots, then your method suggests you would always hit, when there is a still a chance (At BS4, 1/3x1/3x1/3x1/3x1/3x1/3x1/3 = not much) that it would miss.

The 216 method doesnt even deal with expectation, just gives a rough probabilty of a single event occuring. Expectation (what are my chances of *at least* 2 hits from 4 shots) is a much better wya of working out your chances of achieving the outcome youre looking for.

nosferatu1001 wrote:

Sonyca wrote:
Just turn it into a probability instead of using all that 216 stuff.

Each side of a die has a 0.167 chance of being rolled (1/6)
Now imagine a space marine dreadnought is firing an assualt cannon. As he is hitting on a 3+, he has a 0.167*4 = 0.666 chance of hitting, as 4/6 sides of the die will cause a hit.
)

That doesnt work, as each roll is independent (does not affect any other roll)

If you fire an avenger bolt cannon, which has 7 shots, then your method suggests you would always hit, when there is a still a chance (At BS4, 1/3x1/3x1/3x1/3x1/3x1/3x1/3 = not much) that it would miss.

You misunderstood what Sonyca was saying. It's 0.167*4 because there's 4 faces which could result in a hit when you need a 3+ to hit, which is entirely correct to get the chance of a single shot hitting.

The next sentence tells you to multiply by the number of shots (which is also 4) to get the average number of hits.


Automatically Appended Next Post:

nosferatu1001 wrote:
The 216 method doesnt even deal with expectation, just gives a rough probabilty of a single event occuring. Expectation (what are my chances of *at least* 2 hits from 4 shots) is a much better wya of working out your chances of achieving the outcome youre looking for.

It does deal with expectation, it deals with the expected average number of successes.

Granted I think "216" is a silly way of teaching probability, but that's another matter.

It is in some circumstances more useful to talk about "what are my chances of at least 2 hits from 4 shots"... but that requires using binomial distributions and whatnot and most wargamers don't know enough probability to calculate it and it does take more explaining to get someone to understand it. There is this useful site that will do the calculation for you though...

http://stattrek.com/online-calculator/binomial.aspx

Oops, apologies then, I need more coffee....which I dont drink. Awkward.

It is usually more useful to talk about "I need at least 3 casualties" than getting the average, as that doesnt tell you how well distributed the average is, at all. Its useful on large rolls, where things re more likely to distribute properly, but pretty horrible on 3, 4, 5, 6 dice rolls,

This is all bringing back horrible memories of stats theory courses at uni however....!

I think when evaluating the usefulness of a unit, averages are fine.

In the context of list building...
The difference between talking about "what are the chances of getting at least 4 kills" and "the average is 6" only matters when you're comparing a unit with a small number of attacks with a high % vs a unit with a small number of attacks with a low % but they both give similar averages.

All you really need to know there is that a larger number of attacks with a small % per attack but the same average will have a wider spread, so if you only need 3 wounds but the average is 5, you're more less likely to succeed than if you had a smaller number of attacks with a higher %. But then the flip side is if you need 8 wounds but the average is 5, you're more likely to get it if you have more attacks with a low % than if you have less attacks with a high %.

Once you understand that, you don't really need to bother about calculating the specific % chance of each event because you can just intuitively know which is better.

In the context of an actual game...
It's really not practical to calculate "what is my chances of getting X wounds" because it's too hard to calculate on the spot. If my opponent whipped out a calculator or a laptop to try and calculate such statistics in the middle of a game I'd be like "Seriously? No, can we just play the game". The maths required to calculate a binomial distribution is too heavy to do mid-game unless you're Rain Man.

Where calculating specific percentages like that is useful is when you have anti-tank or anti-monster weapons and you're usually better off comparing them with "what are my chances of X hull points/wounds" rather than "what is my average hull points/wounds". Because the number is usually small and it gives you an idea how luck can sway things and may make you re-evaluate whether you plan to take either 2 or 3 of a given weapon (for example).

You may find out, for example, that on average you might be able to kill a tank with X number of weapons, but the % chance of it happening is actually less than 50%, and taking a few more weapons would raise that to 70%. So that's useful... but trying to do that sort of calculation mid-game isn't practical.

AllSeeingSkink wrote:

It's really not practical to calculate "what is my chances of getting X wounds" because it's too hard to calculate on the spot. If my opponent whipped out a calculator or a laptop to try and calculate such statistics in the middle of a game I'd be like "Seriously? No, can we just play the game". The maths required to calculate a binomial distribution is too heavy to do mid-game unless you're Rain Man.

For many defined problems you can develop "shortcuts", ways of thinking about known problems that will enable you to do such calculations on the spot. Two sequential rolls of d6 with rerolls... I've calculated this so often I can rapidly spit out the odds in a given situation in the game. I'm often abused by other players who know about this and want to know it in the game

 Murenius wrote:

AllSeeingSkink wrote:

It's really not practical to calculate "what is my chances of getting X wounds" because it's too hard to calculate on the spot. If my opponent whipped out a calculator or a laptop to try and calculate such statistics in the middle of a game I'd be like "Seriously? No, can we just play the game". The maths required to calculate a binomial distribution is too heavy to do mid-game unless you're Rain Man.

For many defined problems you can develop "shortcuts", ways of thinking about known problems that will enable you to do such calculations on the spot. Two sequential rolls of d6 with rerolls... I've calculated this so often I can rapidly spit out the odds in a given situation in the game. I'm often abused by other players who know about this and want to know it in the game

I'm talking specifically about calculating % chance of things happening though. Averages aren't so hard and can often be estimated mid-game. The % chance of something happening is much harder.

For example, lets say you have 5 Bs4 autocannons firing at an AV12 vehicle with 3HP, the average number of HP would be 2.22 per round of shooting, that's easy enough to figure out. But % wise, what chance do you actually have of killing it?

Well, you have...
39% chance of killing it
29.7% chance of inflicting 2HP of damage
23.1% chance of inflicting 1HP of damage
8.1% chance of inflicting no damage at all

I'm not aware of any shortcuts to figure that out and I doubt most regular non-savant people could figure it out in a time that would be reasonable during a game.

I find calculating percentages quite instructive. Sure averages are fine for most places, but they don't really tell you what to expect. If I want to know how many HPs I can reasonably expect a unit to take off a turn, percentages tell me that. Averages do not. This is because you often cannot reasonably expect to get average or better. So you can figure things more like '80% of the time I will do X or more HPs to such a target' that is a figure you can reasonably expect to achieve. I obviously wouldn't stall a game to do this on the fly but after you do it a few times for different set ups you start to get a feel for it and develop a more accurate intuition. For instance, I know my forgefiend averages 4 S8 hits a turn, but because I've looked at the percentages I know he will do less than that roughly 1/3rd of the time. Developing that kind of knowledge is why percentages are useful, not for calculating in the midst of battle. For me its much more helpful than just very qualitatively saying I have more attacks with lower % therefore I should be better off.

Ok, now I get what you meant and have to agree. It's pretty easy to shape this as a Bernoulli chains (3,4,..10 hits with p=0.22), calculating it with a tool is easy, but not in the head.