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(WARNING - this is a math question, be ready for formulas)

The method for calculating the chance of scoring a wound is well known

Chance of hitting * chance of wounding * chance of failing save

So IG lasgun vs Marine
1/2*1/3*1/3=1/18 (6%)

Marine bolter vs IG
2/3*2/3*1=4/9 (44%)

But when you get multiple shots it gets harder. It's tempting to just double it (2 lasguns=2/18 chance of killing a marine) but that's wrong. 3 Bolters do not have a 12/9 chance of killing a guardsman, you can always fail.

Is this method correct?

1-(the chance of failing to wound)^number of shots

or specifically:
1-(1-(Chance to hit, chance to wound, chance to fail save)^number of shots

So an IG rapid firing lasgun vs marine is:
1-(1-(1/2*1/3*1/3)^2=11%
Not quite double.

IG heavy bolter is
1-(1-(1/2*2/3*1/3)^3=30%

Calculating the odds of getting 2 or even 3 kills is well beyond me.

Anyway my question is whether or not the math is right. I could simplify the formula one set by looking at inverses:
1-(chance of missing*chance of failing to wound*chance of making save)^number of shots
But since I usually already know the chance of wounding it's easier to just use that and subtract if from 1.

Here's a helpful link: http://en.wikipedia.org/wiki/Binomial_distribution

It's a tricky one, but Binomial Distribution is the right way forward:

Let n = the number of shots / attacks fired
Let w = the number of wounds you would like
Let p = the probability of scoring a wound (as calculated in your formula)

The formula is then nCw * p^w * (1-p)^(n-w), where nCw is the expression for combinations (n!/(w! x (n-w)!)) It' very hard to write maths expressions in normal text.... (In case you dont know what ! means, it is multiply all of the numbers below it down to 1, eg, 5! =5x4x3x2x1, 4!=4x3x2x1)

So, if you are taking 1 shot and you require 1 wound, and using your example of IG lasgun v marine, with p = 1/2*1/3*1/3=1/18, the calculation is:
1C1 * (1/18)^1 * (17/18)^0 = 1/18 =6%

Taking 2 shots requiring 1 wound is a bit harder as clearly 2 wounds is acceptable too - it's best to do 1-probability of failing to get 1 wound, ie getting no wounds. So the calculation would be:
1- (2C0 * (1/18)^0 * (17/18)^2) = 11%

As for a situation where you are taking say 10 shots and you require say 4 wounds, you need to repeat the process a little more, ie do 1 - probability of no wounds - probability of 1 wound - probability of 2 wounds - probability of 3 wounds. Giving, wait for it.....
1-(10C0 * (1/18)^0 * (17/18)^10) - (10C1 * (1/18)^1 * (17/18)^9) - (10C2 * (1/18)^2 * (17/18)^8) - (10C3 * (1/18)^3 * (17/18)^7) = 2%

You can do this more quickly using binomial distribution tables, but that's a bit tricky to explain over the internet, and more than likely your value of p isn't going to be on the table in the first place.

I hope this helps.... Good luck!

Um... I'll have to print that out to see if makes sense to me. Prob-stat was a long, long time ago.

But is my basic formula for the probability of 1 wound right?

Yes, but as you say, it gets more complicated the more you add to it.

Yes, the way you figured out the chance to get one wound is correct.

Well that's something...
I guess I ought to add in the chance of 2 wounds & 3 wounds to the odds of one wound to get the real picture.

to quote Barbie: math is hard.

To simplify things:

When trying to out the odds of scoring at least 1:

Let Phit## = probability of scoring a hit with ##

Then Phit = 1- (1-Phit1 * 1-Phit2 * 1-Phit3 * ...).

Simple.


For Pred Annhiliator (twin Las):
PhitLCa = 2/3 = 67%
PhitLCb = 2/3 = 67%
So PhitLCt = 1 - (1-67% * 1-67%) = 1 - (33% * 33%) = 1 - (11%) = 88%.

i.e. 88% of the time, SM will score at least 1 hit with a twin Lascannon.

For Pred Annihilator (full Las):
PhitLCt = 88% (as above)
PhitLCl = 67%
PhitLCr = 67%
So phitLC = 1 - (1-88% * 1-67% * 1-67%) = 1 - (11% * 11%) = 1 - 1% = 99%

i.e. 99% of the time, you will score at least 1 Lascannon hit.


The above formula generalizes to wounds, kills, etc. as long as you calculate each independent event separately *before* you calculate the combined probability.

i.e. Pwound = 1 - (1-Pwound1 * 1-Pwound2 * ...)
i.e. Pkill = 1 - (1-Pkill1 * 1-Pkill2 * ...)
...

Simple.

Kid_Kyoto wrote:(WARNING - this is a math question, be ready for formulas)

The method for calculating the chance of scoring a wound is well known

Chance of hitting * chance of wounding * chance of failing save

So IG lasgun vs Marine
1/2*1/3*1/3=1/18 (6%)

Marine bolter vs IG
2/3*2/3*1=4/9 (44%)

But when you get multiple shots it gets harder. It's tempting to just double it (2 lasguns=2/18 chance of killing a marine) but that's wrong. 3 Bolters do not have a 12/9 chance of killing a guardsman, you can always fail.

What you've calculated there isn't the probability, but the average number of kills.

The binomial method is more relevant when the target only has one wound, or whenever there is a significant chance of overkill, inflicting more casualties than the unit has wounds. If you’re looking at the chance of taking down a vehicle, wiping out an understrength unit, or maybe the odds of forcing a pinning check, then yeah, you should probably use the binomial.

But the rest of the time you really can keep it simple, as in most situations the chance of inflicting more kills than the unit has wounds is extremely improbable. You will get a perfectly sufficient understanding from calculating the straight average, and maybe also calculating the standard deviation.

Kid_Kyoto wrote:Well that's something...
I guess I ought to add in the chance of 2 wounds & 3 wounds to the odds of one wound to get the real picture.

to quote Barbie: math is hard.

Just to clarify something: the 1-(chance of failing to wound)^#of shots formula you applied gives the odds of causing at least one wound, i.e., the odds that not all of your shots will fail to wound... so you've already included the odds of 2, 3, etc. wounds.

For the odds of exactly one wound (or any other exact number), you'll need that binomial distribution formula (or any equivalent mathematical procedure).

It does? Oh good, then that's good enough for me.

Math make mind hurt.

Sorry for bringing this topic up again but what about multiple attacks with different weapons?

Calculating the average number of wound the “normal” way is easy but how do I add the multiple weapons into the Binomial Distribution thingie?

Something like what is the probability of X wounds on target Y with 8 Lasguns, 1 Heavy Bolter and 1 Grenade launcher. Or in other words pretty much a standard IG squad. Thats 12 attacks but they don't all have the same probability to score a wound.

Do I have to modify this “nCw * p^w * (1-p)^(n-w)” formula to get what I want or do I calculate the probability for each weapon type individually and add/multiply them together in some way and then run em trough the function?

Or is there an entirely different way to handle this?

I would probably have a line for each weapon, then add the %'s together.

Unfortunately, Lormax, you are wrong.

For example, if you had a 10% chance of causing a wound with a particular weapon and you lined up 10 of them to attack, you wouldn't get 10 x 10% = 100% chance of hitting. And if you had 11, you certainly wouldn't have 110% chance of hitting!

To be completely honest, although the figures for different line ups of different weapons CAN be worked out, I would suggest that it's too complex for explaining over the internet. In my day job, I'm a Maths and Stats teacher, who has lectured at universities about statistics - and whilst the maths involved is not particularly far above the level I've already discussed in my earlier post, the level of complexity is way beyond what I would go through with maths students in my lessons. I realise that this isn't the answer you wanted, but it's the best way out.

As for what you can actually do, I would suggest using the method I gave earlier for each group of like-equipped models. Remember that mathhammer can only go so far. It's probably enough to work out "My 8 lasguns give me a probability of causing a wound of..., my grenade launcher ... and my bolter..."

Please feel free to ask for more help with this, but it does get needlessly complex. Even given my background in the subject, personally I don't use any mathhammer when planning my lists beyond basic GCSE (exam taken by 16 year olds in the UK, for our foreign dakkites) maths. I don't even go as complex as the maths posted previously!

There are two different things which are typically calculated Probability and Expectation.

Expectation is easiest - this is the adding up of the percentages. You can expect 5 Marines rapid-firing their bolters into unprotected Guardsmen to kill (10 shots * .66 hits * .66 wounds) 4.44 of them.

But the Probability of them scoring at least one kill is where you use the formula JohnHwang gives.

Both are useful on the 40k tabletop.

InyokaMadoda:

A quick question since you apparently have the credentials to answer it and you gave the initial formula I used:

Let n = the number of shots / attacks fired
Let w = the number of wounds you would like
Let p = the probability of scoring a wound (as calculated in your formula)

The formula is then nCw * p^w * (1-p)^(n-w), where nCw is the expression for combinations (n!/(w! x (n-w)!))

What kind of numbers should be entered for n, w, and p? I would assume that n and w would be positive integers (whole numbers like 1, 2, 15, etc for those of you not up on your math lingo) and p would be a decimal in the 0 to 1 range. Is that correct? After “programming” the binomial distribution into excel I got some really odd results if I used decimals for n or w. I'm guessing that's due to them being used in the factorial (!) portions of the equation. Just want to try and make sure I did this correctly if I'm going to end up using it to make 40k decisions

No offence, but this is a pure waste of time
no matter how well done a formula is done it will not predict even a few shots.
the dice have thier own mind

its much easier to just play.

for instance.
working on probability you can roll a D6 6 times, all formula's say there is an equil prob. of rolling each number, so you should roll 1,2,3,4,5,6.
have a quick go, roll 6 dice and see what happens

JD21290 wrote:No offence, but this is a pure waste of time
no matter how well done a formula is done it will not predict even a few shots.

While it won't do that great of a job at predicting a few shots, it will do a great job at predicting a lot of shots.

working on probability you can roll a D6 6 times, all formula's say there is an equil prob. of rolling each number, so you should roll 1,2,3,4,5,6.
have a quick go, roll 6 dice and see what happens

Actually, the chances of rolling that number combination on 6d6 is rather low. It's extreamly low if you need to roll them in that order. However, if you roll 600d6 and then count up how many times you rolled each number (and you are using well balanced dice) you'll find that you get very close to 100 of each. Go ahead and try it, it really does work (again assuming you have well made dice).

So I think part of the problem here is that you may not fully understand what the numbers mean. The things we are discussing here are probability, which is the chances of a certain event happening. Again going back to the dice, if you roll a d6, you have a 1/6 chance of rolling any particular number. So what are your chances of rolling a certian combination of numbers (a roll to hit of a certain value or hight, a roll to wound of a certain value or hight, and a roll to fail an armor save of a certain value or lower)? That is the sort of thing we are talking about here. So with this I can figure out how likely a unit is to damage another unit. It doesn't tell me that the first time I try, I'll fail and the second time I try I'll succede or anything like that, it just gives a % chance of it occuring which can then be used as a guidline for play. If GW had used this sort of thing more, we wouldn't have had some really broken things in 40k like holo fields and 4th edition assault cannons.

i didnt mean in a run through order, just all 6 numbers showing up in 6 rolls.

ive tried the experiment (using GW's dice, so not that well made)
and its not allways balanced, dice are all about luck, not probibility.

if you work out that you need 20 shots with a lascannon to destroy a landraider (just picked a number) the chances of killing it in around 20 shots is vague, and doing it in dead on 20, that would rarely happen.

i just stick to the idea of shoot something untill it dies (still got ork blood in me) as thats the only sure way to say it can be done, just because a die has an equil prob. of rolling each number doesent mean it will do.
look at battle reports from any WD mag going back as many years as you wish, you will notice in each that there are some stupid rolls, either things like shooting a fex with 5 lascannons and doing nothing, then shooting it with a bolter and causing an unsaved wound.

next time you play look for these moments, its allways to do with luck.
hence why most games involving dice are used for gambling (other than kiddies board games and such)
if you can predict a rough idea what the dice will be then gambling wouldnt exist, why would they do something when they know that they could lose too much money from if someone does this?

Ok, so there's quite a few things to respond to here, so I'll do my best to answer all questions:

Phoenix wrote:InyokaMadoda:

A quick question since you apparently have the credentials to answer it and you gave the initial formula I used:

Let n = the number of shots / attacks fired
Let w = the number of wounds you would like
Let p = the probability of scoring a wound (as calculated in your formula)

The formula is then nCw * p^w * (1-p)^(n-w), where nCw is the expression for combinations (n!/(w! x (n-w)!))

What kind of numbers should be entered for n, w, and p? I would assume that n and w would be positive integers (whole numbers like 1, 2, 15, etc for those of you not up on your math lingo) and p would be a decimal in the 0 to 1 range. Is that correct? After “programming” the binomial distribution into excel I got some really odd results if I used decimals for n or w. I'm guessing that's due to them being used in the factorial (!) portions of the equation. Just want to try and make sure I did this correctly if I'm going to end up using it to make 40k decisions

n and w must definitely be integers (whole numbers) for this to work. Remember that they are the number of attacks and wounds. The factorial function will not work with decimals and excel will just go screwy because it doesn't understand maths in that way that it should, it just processes formulae without thinking about it. The value of p should be a decimal between 0 and 1, with 0 being impossible and 1 being definite. For anyone not too comfortable with maths but wanting to work on this, for example a 1 in 4 chance is converted to a decimal by doing 1/4 = 0.25 chance. Good luck phoenix with your calculations!

JD21290 wrote:No offence, but this is a pure waste of time
no matter how well done a formula is done it will not predict even a few shots.
the dice have thier own mind

its much easier to just play.

for instance.
working on probability you can roll a D6 6 times, all formula's say there is an equil prob. of rolling each number, so you should roll 1,2,3,4,5,6.
have a quick go, roll 6 dice and see what happens

Personally, I agree with you that it's just easier to get out there and play the game. I have a very good understanding of statistics and numbers and so could happily work out all of the probabilities of all my attacking units against each type of opponent if I wanted to. However, I find that this takes the fun out of the game for me though and I'd rather just not let myself think about it whilst playing. I do this hobby to escape my day job, so why bring it in? However, there are a lot of players out there who take the game very seriously and are looking to make the best list possible of their chosen army. This may mean that they wish to optimise those percentages for a tournament or something. I say good luck to them. If that's the way they want to play the game, that's fine by me. It's not my style, but if it floats their boat then go ahead! Even an optimised list can lose with unlucky dice rolls or poor generalship, but when you are getting serious about winning you want to give yourself the best chance. Just remember that it takes all sorts to play this game and I always say that, provided you play fairly, everything is acceptable.

Phoenix wrote:

JD21290 wrote:No offence, but this is a pure waste of time
no matter how well done a formula is done it will not predict even a few shots.

While it won't do that great of a job at predicting a few shots, it will do a great job at predicting a lot of shots.

working on probability you can roll a D6 6 times, all formula's say there is an equil prob. of rolling each number, so you should roll 1,2,3,4,5,6.
have a quick go, roll 6 dice and see what happens

Actually, the chances of rolling that number combination on 6d6 is rather low. It's extreamly low if you need to roll them in that order. However, if you roll 600d6 and then count up how many times you rolled each number (and you are using well balanced dice) you'll find that you get very close to 100 of each. Go ahead and try it, it really does work (again assuming you have well made dice).

So I think part of the problem here is that you may not fully understand what the numbers mean. The things we are discussing here are probability, which is the chances of a certain event happening. Again going back to the dice, if you roll a d6, you have a 1/6 chance of rolling any particular number. So what are your chances of rolling a certian combination of numbers (a roll to hit of a certain value or hight, a roll to wound of a certain value or hight, and a roll to fail an armor save of a certain value or lower)? That is the sort of thing we are talking about here. So with this I can figure out how likely a unit is to damage another unit. It doesn't tell me that the first time I try, I'll fail and the second time I try I'll succede or anything like that, it just gives a % chance of it occuring which can then be used as a guidline for play. If GW had used this sort of thing more, we wouldn't have had some really broken things in 40k like holo fields and 4th edition assault cannons.

JD21290 wrote:i didnt mean in a run through order, just all 6 numbers showing up in 6 rolls.

ive tried the experiment (using GW's dice, so not that well made)
and its not allways balanced, dice are all about luck, not probibility.

if you work out that you need 20 shots with a lascannon to destroy a landraider (just picked a number) the chances of killing it in around 20 shots is vague, and doing it in dead on 20, that would rarely happen.

i just stick to the idea of shoot something untill it dies (still got ork blood in me) as thats the only sure way to say it can be done, just because a die has an equil prob. of rolling each number doesent mean it will do.
look at battle reports from any WD mag going back as many years as you wish, you will notice in each that there are some stupid rolls, either things like shooting a fex with 5 lascannons and doing nothing, then shooting it with a bolter and causing an unsaved wound.

next time you play look for these moments, its allways to do with luck.
hence why most games involving dice are used for gambling (other than kiddies board games and such)
if you can predict a rough idea what the dice will be then gambling wouldnt exist, why would they do something when they know that they could lose too much money from if someone does this?

Phoenix is once again quite correct. The chances of rolling 6 dice and getting just one of each of the six scores is actually pretty low considering how 'sensible' a result it seems. I can't quite be bothered to work out the probability here but it would be something like 15 - 20% of the time.

The conversation from Phoenix and JD21290 is exactly what I was talking about earlier about the different approaches to the game. 'Fun' versus 'Optimising'. For what it's worth, I think all players do a bit of both. We all choose an army that we would like to play or that inspires us in some way (fun) and we all choose troops that we think have some way of winning (optimising). It's just a question of how far you go with each. I play orks and deliberately choose fun units like the shokk attakk gun, but it doesn't mean that I just load up on grots because that would give me no chance of winning, so I choose other things like boyz and lootaz. It sounds to me as if Phoenix is simply further down the road of optimising than either myself or JD21290, and is choosing troops based upon a different method of deciding whether he feels they are worth the points. Again, just to make it perfectly clear, we all do this. For example, if it cost me 3 points to upgrade a boy to a slugga, but 200 points to upgrade to a rokkit launcha, I would go with the slugga every time because I feel that the improved destructive capabilities of the launcha do not outweigh the price. Similarly if it was free to upgrade a mek to having a kustom force field, we would all do it because it optimises our chances of success. Yes, these are extreme and unrealistic examples, but hopefully they illustrate my point. In reality it tends to be more of the case that "weapon choice A gives me x% of success, whereas weapon choice B gives me y% of success. Which one do I feel is worth it?"

Ultimately, just remember that we're all different and have different approaches. Whatever you like to do with this daft game is your own choice!

While I do understand the probability equations (go-go-gadget-physics degree...) these are not actually that useful with relation to 40k (or any other game for that matter).

What people have called the 'expectation' is not only far easier to work out but also much more useful when gaming.

When deciding which units are going to attack which targets, you are basing your decision on two things. Priority and effectiveness.
Priority will be decided on circumstance - control of objectives, units which are near to giving up a KP, threats to your own army/plan.
Effectiveness is where math can help. If you are an IG player faced with a choice between a 10-man SM scout squad or a 5-man SM tac squad, which do you target?

To keep things simple, lets assume you have 10 guardsmen with flashlights at rapid fire range to both targets.
The number of kills you can expect on the scouts is 20*1/2*1/3*1/2 = 1.66 dead scouts
The number of kills you can expect on the marines is 20*1/2*1/3*1/3 = 1.11 dead marines

The correct decision here is to target the marines. You should get at least one kill on each target and may get a second. Although you are more likely to get the second kill on the scouts it will have no appreciable effect on the squad, whereas a second kill on the marines would force a morale check.

Now lets swap 1 of the flashlights for a heavy bolter.
vs scouts you now have:
18*1/2*1/3*1/2 = 1.5 kills
AND 3*1/2*2/3 = 1 kill for a total of 2.5 kills.
vs marines you now have:
18*1/2*1/3*1/3 = 1 kill
AND 3*1/2*2/3*1/3 = 0.33 kills for a total of 1.33 kills

The correct decision now is to target the scouts. You should get 2 kills, whilst a 3rd (on a 50/50 chance) will force a morale check. Against the marines you should get 1 kill, whilst a second (on a 1 in 3 chance) will force a morale check.


Whilst you would probably become quickly unpopular if you whipped out a pen and paper to calculate the odds each time you had a decision to make, you can use a knowledge of the likely effects of an action to help inform your decision.

Chimera_Calvin wrote:
Whilst you would probably become quickly unpopular if you whipped out a pen and paper to calculate the odds each time you had a decision to make, you can use a knowledge of the likely effects of an action to help inform your decision.

Couldn't agree more.

One of the reasons that some players are better than others is because they can either make these kind of probability calculations in their heads, or remember the results from working them out in a spreadsheet.

If you knew one of your units could not harm a particular enemy, because of low AP weapons of whatever, you would not shoot at it. Choosing between two other units, who have different chances to harm the enemy, is just an elaboration of this approach.

It comes down to making good informed decisions and taking measured risks. The ability to calculate and weigh probabilities and expected outcomes, both of what you can do to the enemy and what the enemy is likely to do to you, is a integral part of that.

JD21290 wrote:if you work out that you need 20 shots with a lascannon to destroy a landraider (just picked a number) the chances of killing it in around 20 shots is vague, and doing it in dead on 20, that would rarely happen.

This is very true, but again, its in how you use the numbers. If I know that it takes 20 las cannon shots on average to kill a land raider, but only 10 melta gun shots then I can use that information to more effectively direct my fire power. So if I'm looking to pop said land raider, I'll try to get my melta guns in range and spend more of my time shooting my las cannons at other targets. Conversly, if I really need to destroy that land raider, I will devote every las cannon I have to the task since I understand that its going to take a lot of shots to get the job done.

The other place this information is useful is when trying to figure out the best weapon for a particular target. So here is a question for you, which is better for taking down holo field falcons, a las cannon, an auto cannon, a scatter laser, or a melta gun? Do you know the answer? Without running numbers, I certainly don't. All the weapons function so differently that its difficult to figure it out without some more investigation of some sort. I suppose you could roll out 100 turns of shooting with each and see which one works better but without running numbers, its really hard to tell.

look at battle reports from any WD mag going back as many years as you wish, you will notice in each that there are some stupid rolls, either things like shooting a fex with 5 lascannons and doing nothing, then shooting it with a bolter and causing an unsaved wound.

Again, while this is true, would you rather have 5 bolters to kill a fex with or would you rather have 5 las cannons to try and do it with? You'd rather have the las cannons because they have a significantly higher probability of hurting the fex. So sure, the bolter can (and sooner or later will) hurt the fex, but over the course of many shots, the las cannons are going to be the ones really doing the damage...unless you manage to get a whole lot of bolters in there.

if you can predict a rough idea what the dice will be then gambling wouldnt exist

Funny enough, probability is why gambeling exists. Do you think casinos would still be in buisness if they lost money? No they wouldn't be. The probability of all the games you see in casinos favors the house over the players and over the course of many many games the house earns money on them all. Ever heard of a casino losing too much money on its tables? I certainly haven't. That's how they afford to put up all those huge buildings on the Vegas strip. Its all fuled by probability and peoples lack of understanding about it (or desire to believe that they can beat it). And while some people go home with lots of money in their pockets from winnings, a significantly larger portion go home empty handed. That's probability, not luck.