# Tactics of Math



## NecronCowboy (Jan 8, 2009)

So I'm a big numbers geek, and since I'm starting to play 40K again I've put together a spreadsheet you can print out and bring to a game with you so you can make it easier to figure out the different expected results of certain actions.

I thought about doing this on a laptop or a blackberry, but I think that would be rude to figure out calculations using either during a game.

Here is how it works...

Say you have a unit of 5 SM shooting at another unit of SM 18" away.

So that's 5 shots with a bolter that need a 3+ to hit.

Go to the Hit & Wound table find the 3+, then look over to the 5 column, that means 3.33 of the shots should hit.

So let's round up to 3.5, then you need a 4+ to wound so find the 4+ and look over to the 3.5 column and you see 1.75 hits should wound.

Now look at the save table and find 1.5, the marines need 3+ to save so the combined fire should kill betwee .50 and .67 marines that turn.

Of course this just figuring out the expected result, but it will give you a good idea of what would be more effective etc...

****DON'T FORGET TO RENAME THE FILE EXTENSION FROM .ZIP TO .XLS, IT WOULDN'T LET ME ATTACK AN EXCEL FILE****

Hope this helps!!!!


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## Someguy (Nov 19, 2007)

NecronCowboy said:


> So that's 5 shots with a bolter that need a 3+ to hit.
> 
> Go to the Hit & Wound table find the 3+, then look over to the 5 column, that means 3.33 of the shots should hit.
> 
> ...


A few thoughts on this.

It's a good idea to have an understanding of probability maths. However, I find that it's better to work in fractions than in percentages. I find 1/3 easier to work with than 33%, though whichever is best for you is fine.

Also, I tend to work from the probability of a single attack wounding, then multiplying it by the number of attacks to get an expected result.

In this example then, each shot has a 2/3 chance to hit, a 1/2 chance to wound and a 1/3 chance that the marine doesn't save. Multiply 2/3 x 1/2 x 1/3 and you get 2/18, (or 1/9).

I prefer fractions because it shows me that I need to fire an average of 9 shots to drop a marine, which is clearer to me than finding that I have an 11% chance. If I have 5 shots then on average I kill 5/9 of a marine. Not very effective really, but the same odds you get, more or less.

The only objection I have with your method is the rounding. You shouldn't ever round figures part of the way through a calculation because it can skew your results.

Also, you have to be careful when looking at the probability of causing a single wound with multiple attacks (ie working out the chance of killing a single enemy) because it isn't the same as getting enough shots to cause one wound on average. In this example, 9 bolter shots will not always kill a marine. That's because there's a chance of killing 0 and a chance of killing 2 or more, which averages out as 1, but you can never actually be certain you will do anything. 

This most often applies when firing multiple heavy weapons at a vehicle, such as if you were trying to kill a land raider with 4 lascannon shots from a dev squad. The chance is not 4x as good as a single shot, it's actually quite a lot worse than that.


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## cco12 (Jun 30, 2008)

Wow man this sounds really intense.


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## NecronCowboy (Jan 8, 2009)

You are right fractions do make more sense, but my mind is deeply rooted in decimals unfortunatly.

I'm just using this as a way to compare units to each other and what they can be expected to do in a game, that's all.

I come from a poker background so it just makes sense to me to figure out the odds on everything, and although there is a huge element of chance if you put yourself in the best situation mathmatically, then in the long run you should make out like a bandit.

Thanks for your comments!


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## Steel Rain (Jan 14, 2008)

I love Mathhammer. When you work in probabilities, it actually improves your entire game plan. You can factor in the hit/wound chances for any unit you field and any unit you fight against.


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## H0RRIDF0RM (Mar 6, 2008)

Nice contribution +rep for you.


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## Pauly55 (Sep 16, 2008)

Someguy said:


> This most often applies when firing multiple heavy weapons at a vehicle, such as if you were trying to kill a land raider with 4 lascannon shots from a dev squad. The chance is not 4x as good as a single shot, it's actually quite a lot worse than that.


Isn't it? Explain to me how this isn't the case.


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## NecronCowboy (Jan 8, 2009)

Pauly55 said:


> Isn't it? Explain to me how this isn't the case.


The reason is if 4 lascannons shoot at one vehicle only one of the 4 can destroy it, since if the first LC destroys the vehicle then the other 3 can't since there is no vehicle to destroy.

So for example, correct me if I'm wrong if 4 lascannons were shooting at a vehicle and had each a 41% chance of destroying it, then all 4 together would have a .97% change of destroying it, not 41% x 4 = 164%. It's .41 * .41 * .41 * .41 = .971742 = 97%.


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## Pauly55 (Sep 16, 2008)

I don't think thats it. I do remember this being true from a stats class I took in college, but I can't remember why. 

If your gun has a 50% chance to destroy something, then I think for every shot after the first you take the chance it wont be destroyed and cut it in half, i.e. 

2 shots : 75%
3 shots : 87.5%
4 shots : 93.75%

The reason being, you never have a 100% chance of something happening when there is a possibility of it failing.


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## NecronCowboy (Jan 8, 2009)

Pauly55 said:


> I don't think thats it. I do remember this being true from a stats class I took in college, but I can't remember why.
> 
> If your gun has a 50% chance to destroy something, then I think for every shot after the first you take the chance it wont be destroyed and cut it in half, i.e.
> 
> ...


I just spotted my error and fixed it before I saw your post.


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## Othiem (Apr 20, 2008)

NecronCowboy said:


> So for example, correct me if I'm wrong if 4 lascannons were shooting at a vehicle and had each a 41% chance of destroying it, then all 4 together would have a 102.5% change of destroying it, not 41% x 4 = 164%.


While you may get diminishing returns on the increase in expected value by tossing more lascannons at it, the situation is not dire. The more shots you toss in there, the smaller the standard deviation gets. This is why tons of orks shooting at BS2 is so scary, the results are must more predictable and the player can set up their shooting phase confident that they have the right amount of fire power in each location.


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## NecronCowboy (Jan 8, 2009)

Othiem said:


> While you may get diminishing returns on the increase in expected value by tossing more lascannons at it, the situation is not dire. The more shots you toss in there, the smaller the standard deviation gets. This is why tons of orks shooting at BS2 is so scary, the results are must more predictable and the player can set up their shooting phase confident that they have the right amount of fire power in each location.


You bring up a very critical point, the more dice you roll, the better the chance of randomness being smoothed out. Excellent point!!!


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## Revelations (Mar 17, 2008)

Steel Rain said:


> I love Mathhammer. When you work in probabilities, it actually improves your entire game plan. You can factor in the hit/wound chances for any unit you field and any unit you fight against.


Until you actually try to put it into practical application. Just because probabilities say it will probably happen, doesn't mean it's going to. Try playing a game with mathhammer instead of dice, turns out it isn't very fun at all. And I rather enjoy the extreme cases of everything happening; "What? That F'in grot killed how many Terminators??!!" 

But sure, I'll agree it's a useful tool for plotting.


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## Morgal (Sep 26, 2007)

I like decimalls 10% means i have to shoot 10 times to kill one.
but no american people may find that mroe as we don't use 3/16 as a stardard unit of measure.

also mathhammer says my las cannons and missle launchers should kill three trucks a turn...yet they do nothing, even with sharpshooters!...%^^$ math


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## NecronCowboy (Jan 8, 2009)

Revelations said:


> Until you actually try to put it into practical application. Just because probabilities say it will probably happen, doesn't mean it's going to. Try playing a game with mathhammer instead of dice, turns out it isn't very fun at all. And I rather enjoy the extreme cases of everything happening; "What? That F'in grot killed how many Terminators??!!"
> 
> But sure, I'll agree it's a useful tool for plotting.


If you put your troops in a position where they will destroy an enemy with an expenctancy of 75% then you are going to get much better result more often than not than if you place them in a position where they will destroy an enemy with an expectancy of 25%.

So you are missing the point here. If you want to play me in a game where if you roll a 5+ you get $1, and if I roll a 3+ I get $1, then yeah you might come out ahead after the first few rounds, but play me all day and I'll be rolling in money.


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## Revelations (Mar 17, 2008)

NecronCowboy said:


> So you are missing the point here. If you want to play me in a game where if you roll a 5+ you get $1, and if I roll a 3+ I get $1, then yeah you might come out ahead after the first few rounds, but play me all day and I'll be rolling in money.


I'll bet my wieghted dice against your mathhammer. 

But you must have skimmed my post because I didn't miss the point. For plotting reasons, it's usefull, but I won't go beyond that. I won't rely on mathhammer for my tactics because while statistics are sound, I disagree with the principles of mathematical probabilities. (Or rather, my mind thinks more mechanically then mathematically)

Kind of a wierd thing to say, but I can best demonstrate this via the Monty Hall problem. For those of you that aren't familiar with this, you are given 3 doors. Behind 1 door is a prize, the other two doors have nothing. You pick 1 door, but before it's opened, someone else opens up another door showing nothing and then offers you the chance to switch. The question comes in whether or not it's better to switch to the other door. Now *math* says that due to probability and statistics, it's always better to switch doors. *I* disagree with this because regardless of what the math says, the prize has never moved.

Before I get into an even further dive into math and the like, I'll simply point out that while the Math may be sound, I'm not going to determine my choice based on it. Mathhammer, like many other forms of probabilities, tend to avoid specific situations, which 40K is rip with. So again, as a basis, it's helpful, but I disagree with using it beyond that.


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## MaidenManiac (Oct 2, 2008)

Revelations said:


> Kind of a wierd thing to say, but I can best demonstrate this via the Monty Hall problem. For those of you that aren't familiar with this, you are given 3 doors. Behind 1 door is a prize, the other two doors have nothing. You pick 1 door, but before it's opened, someone else opens up another door showing nothing and then offers you the chance to switch. The question comes in whether or not it's better to switch to the other door. Now *math* says that due to probability and statistics, it's always better to switch doors. *I* disagree with this because regardless of what the math says, the prize has never moved.


Thats one of the most gay things folks tried to point out in school. Im also very much stuck with your answer!



Revelations said:


> Before I get into an even further dive into math and the like, I'll simply point out that while the Math may be sound, I'm not going to determine my choice based on it. Mathhammer, like many other forms of probabilities, tend to avoid specific situations, which 40K is rip with. So again, as a basis, it's helpful, but I disagree with using it beyond that.


Whilst its damn good to have decent insight in probabilities whilst playing WHFB/40k one should defo not "demand" that to happen all times. Under a normal 40k game you will most likely roll 300+ dice(bit depending on army ofc) and that gives opportunities for weird low-chance things to happen. Normally you should not roll 5 ones on 5 dice, but it does happen:angry:
Normally you shouldnt fail LD10 with reroll, but that does happen too:angry:
Normally you shouldnt make 10 wounds on SMs(before saves) with 6 Sonic Blater armed NMs:biggrin:
Normally you shouldnt kill a Leman Russ MBT with your 1st LC shot, but that happens too:scare:

Ive come to 3 conclusions during my quite many years with these games:
1: Anomalities happens more frequent then what you can imagine, and they always tend to pop up on important rolls!
2: The less math you know the better you roll. In a more rude way. If youre stupid and fail to realize that your "splendid plan" is a 1/216 chance of hitting home its more likely to do so. The God of Random favourized stupid people:threaten:
3: The best way of making your plans strike true is to use lots of dice, just like mentioned by the "Ork Shooty-boys" example. If you roll enough dice the God of Random ends up fisting himself when he tries to interfere, forcing him(painfully) to acknowledge his own existance and thereby he stops his own fiddling participation:laugh:


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## Chaosftw (Oct 20, 2008)

FYI.... your ruining this game by worrying far too much about probability.

but thats just my opinion.

Cheers,

Chaosftw


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## Someguy (Nov 19, 2007)

Maths is not about intuition, it's about fact.

With the monty hall problem, you have a 2/3 chance of winning if you switch and a 1/3 if you stick. You should switch.

Here's why. Initially, you have a 2/3 chance of picking one of the 2 wrong doors. Now, if you do that, the host guy *must* pick the other wrong door to show you, because he can't pick your door and he can't pick the right door or he would be giving the game away.

So you should assume that the reason the host didn't pick the 3rd door is because it's the correct one. 2/3 of the time that is the case, while 1/3 of the time it isn't because you happened to pick the right door at the start.

You don't know that the 3rd door is the correct one, but the fact that the host didn't pick it is usually because he couldn't. Usually you will have picked one of the wrong doors at the start and he will have been forced to show you the other one. Therefore you should switch and go for that door.

Thankfully, we don't have to do anything as complicated with our mathammer. However, it is an example of how probability calculations are not always intuitive, particularly when multiple variables are introduced.

For example, over a very large number of shots, a unit of 20 ork shoota boyz will not kill as many marines as you would expect if you calculate the average number of kills from 20 shoota boyz firing at 10 marines.

Tihs is because the average number of kills is in fact made up of all the possible outcomes of rolling the 40 shots to hit, wound and save, added together. Yet actually, some of the luckier results for the orks that get included in the calculation, cannot happen. Theoretically, all 40 shots could hit, wound and fail to be saved against, but even if they did you would only kill 10 marines, not 40, because there are only 10 marines.

Actually the average number of kills is 2.11, so the chance of getting more kills than the unit's size is pretty remote when shooting at big squad. You get wasted kills much more easily on smaller units.

This reduction in the chance to get many kills is not matched by any change in the chance to get below average amounts of kills. It's still perfectly possible to kill 0 or 1 when you "should" kill 2, so you always have the chance to do badly but don't always have the chance to get lucky.

Which brings us back to firing 4 lascannons at a tank. You can only kill the thing once, but the dice may potentially come up with more than one kill. Any 2nd, 3rd or 4th kills are wasted.

The way you calculate the chance to do one thing with multiple attempts (like killing a land raider with 4 lascannon shots) is by going through in stages, one stage for each attempt. Let's say you are trying to do something with a 1/3 chance of success and you have 3 attempts. Intuitively it seems like a 100% success rate, but of course we know that isn't going to happen when the dice start rolling. The calculation goes as follows:

Take your initial 1/3 chance. When the second attempt happens you only look at when the first try failed, and add 1/3 of that. Ie, 1/3 of the initial 2/3 that failed.

To work this out you have to expand the fractions. We have to do this because we can't divide 2 by 3 and get a whole number. So, instead of talking about 1/3, we have to talk about 3/9, which is the same thing. You are allowed to multiply (or divide) the top and bottom of a fraction by the same number to get amounts you can work with.

So we have an initial 3/9 chance of success and 6/9 chance of failiure. 1/3 of those failiures will succeed on the second attempt. That's 2 of the 6 fails from the initial 9 results. We now have a 5/9 chance of success. 3/9 was from the first try, and 2/9 from the second. There's also a 1/9 chance that both attempts would have succeeded, but that wouldn't be any better than a single success.

Now the 3rd attempt. You still have 4/9 failiures, so you try again. Again a third of the 4 failiures are going to become successes now, and we need to enlarge everything again to get numbers we can divide by 3. 4/9 becomes 12/27. We lose a third of that, from the successful 3rd attempts, and we get an 8/27 chance of failing or a 19/27 chance of success. 

So for example, the chance of a marine failing one of 3 armour saves is 19/27 (or slightly above 2/3). That's also the chance of killing 1 marine or more from a unit if you make them tke 3 saves. However, the average result of 3 marines taking 3 saves is that one of them will die.

Surprisingly, this does have practical implications for list creation. Some weapons are more affected by this than others and you can take them in different ways as a result. Weapons for killing vehicles suffer much more from diminishing returns than anti-infantry weapons. 2 razorbacks with lascannons will kill about as much as a rpedator annihilator with 3 lascannons, despite having one less gun, and also have tactical advantages in being mobile, able to carry troops and so on - and best of all they can shoot 2 different targets. On the other hand, 3 razorbacks with heavy bolters will not do much better than a devastator squad with 4 heavy bolters (both getting the 8 hits on average) because the heavy bolters tend to be fired at infantry units where their full potential can be realised.

So, try to keep your melta guns and lascannons seperate and stick your AP weapons together. You will get slightly better results than if they are the other way round.


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## Steel Rain (Jan 14, 2008)

Bonus rep to you Someguy! You revealed the mathmatical reasoning for picking some weapons over others and the tactical application of such. What I propose is that we put our heads together and build a tactica for this. Present the math. Give the recommendations for army list choices. Then finally provide the preferred units to use said weapons upon. What do you say?


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## Revelations (Mar 17, 2008)

Someguy said:


> Maths is not about intuition, it's about fact.
> 
> With the monty hall problem, you have a 2/3 chance of winning if you switch and a 1/3 if you stick. You should switch.
> 
> ...


I've seen the Math. I've done it myself. It makes perfect mathematical sense. There are even studies which prove students do better on multiple choice tests if they switch answers in doubt. But the prize never moved, therefor my brain says no matter what math will not move the prize, so I should not either. 

The fact is simple; the Math is still only working with probabilities, not absolutes. Most people I've seen argue in favor of probabilities seem to think they are absolutes, which is where I disagree. There's the disconnect, I'm not arguing the math, I'm arguing it's use.


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## newsun (Oct 6, 2008)

Rounding prior to completing the math was brought up and if you are going to do it, do so such that the results you want are worse, this gives you a bit of 'padding'

Though yeah generally always do any rounding at the end.


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## dopey82 (Jul 8, 2008)

screw mathhammer!!! mathhammer doesnt explain how my 6 man terminator squad can charge 8 striking scorpions and loose 5 models before i hit. he did 10 wounds and i failed 5 two plus saves. one he shouldnt have hit and wounded that many times and two I shouldnt have failed 5 2 plus saves. what it all comes down to is chance and luck.


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## Steel Rain (Jan 14, 2008)

You didn't even bother to read the thread did you?


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## crack3rjack3d (Mar 11, 2008)

Nicely done, when I played Battletech, (Before it became "classic") I read an article on how to use the mathhammer to your advantage. In BT the math hammer could very well be the difference too. I think I shall look deeper into the mathhammer for 40k, perhaps a similar tactica could be devised. Mad props Necron Cowboy, time to start living with my 5th ed rule book...


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## Pauly55 (Sep 16, 2008)

The monty haul problem can be solved. The trick is that you _never_ had a 1/3 chance to pick the correct prize. It was always 1/2. Since the host always shows you a wrong door, you can just take it out of the equation.


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## Someguy (Nov 19, 2007)

It's 2/3, not 1/2.

The reason this seems odd is that it appears that only random stuff is taking place, but it isn't. The host is not acting randomly. He has to pick one of the wrong doors to show you.

2/3 times you have picked one of the wrong doors at the start. 100% of the time if you did that, the host *must* pick the other wrong door, because that's the only thing he can pick. You then switch to the other door. 

It isn't a 50/50 when you switch. It's the same 2/3 chance that you started with. If you picked the wrong door at the start, which you probably did, then only the correct door remains when the host takes the other wrong one away.

Don't take my word for it by the way. Google it and find out, if you care enough. 

Now, ignoring this, guys like Revealations are entitled to trust their intuition if they prefer, and 1/3 of the time they will be right. However I probably wouldn't be doing my job as tactics mod properly if I didn't recommend taking the 2/3 chance.

Likewise, quite a lot of people have mentioned a dislike of mathammer, and I can sympathise with that. Most people don't really enjoy this sort of stuff and there's only a pretty limited amount it can tell you. Personally I tend to advise against using mathammer in great detail, and certainly not to spend a lot of time workin out who is likely to win in this unit vs that unit type comparisons. However, it does help to have a rough idea what you can expect to happen before you do something, so that you know how much force to apply - or whether to run like hell.

One quite successful use is to work out the efficiency of your units versus different targets. This can help when choosing between them. 

For example, I found that the shooting from kroot is actually more efficient for shooting MEQs than the shooting from fire warriors. Even though the kroot take an average 12 shots to kill a marine compared with 9 from FWs, the kroot are so much cheaper that they work out better. It ends up 84 points worth of kroot vs 90 points worth of FWs. That isn't a huge difference, but I think it makes the fire warriors (who can do nothing other than shoot) look pretty stupid when compared with the kroot's 2 attacks in cc at strength 4, its ability to infiltrate or outflank and its 3+ cover save in woods. Fire warriors are still better vs anything with a 5+ save, as they bypass the armour but kroot don't, and they are a threat to things like light vehicles and the toughest MCs, that the kroot can't hurt. The range is an issue too, but I think that infiltration makes up for it. On the whole I prefer kroot.

It's rare to have such a direct comparison, and even in this case there are many other issues to consider. A melta gun is better at killing tanks than a lascannon, and usually cheaper. Does that mean it's always a better option? Maybe, but it's a whole lot harder to get it into a position to fire, and doing so often results in the death of the firer. Likewise it's very difficult to compare shooting with CC through mathammer. Usually it is just one of many things you should consider, and it is in no way a substitute for trying things out in games and learning tactics.


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## Steel Rain (Jan 14, 2008)

There is always going to be some element of luck involved. Not just probability. Every gamer has a lucky dice, unless you're my wife. Then you've got all unlucky dice and some that just may roll high on accident.


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## Othiem (Apr 20, 2008)

Steel Rain said:


> There is always going to be some element of luck involved. Not just probability. Every gamer has a lucky dice, unless you're my wife. Then you've got all unlucky dice and some that just may roll high on accident.


True. There's also a ton of luck when you play a game of Magic:TG. Yet we see the same people at tournaments every year. Poker is a game of luck, yet again the same people are in the major tournaments year after year. The skill in games of luck is understanding the odds and knowing how to work with them. The fact that in every decent game of chance the same people are the best competitively should tell you that it's more than just a matter of luck. Stats in warhammer suffer though since we're rolling D6. This means that instead of nice smooth distribution curves, we've got giant discrete chucks with a hard wall between passing a test, and failing it. This is why weird events like grots taking on terminators show up more than you might expect. And because whenever a grot takes on a termie and wins, somebody posts on a forum. Nobody speaks for all the dead grots slain by terminators =(.

For the people confused about the Monty Hall problem, think about it in terms of information. You start out knowing that there is 1 prize behind 3 doors, so you make a choice based on that information. After Monty opens a door, you have far more information about the problem. Does it not stand to reason that given more information, you can make a better choice? 

One last comment on Someguy's math on killing a tank with 3 lascannons. When dealing with a situation where the outcome is binary, i.e. tank is dead, tank is alive, I find it easier to do it mentally if instead of comupting the chance of success as Someguy did, you compute 1 - chance of failure. So a lascannon has a 1/3 chance succeed, it fails 2/3. Chance of 3 failing is (2/3) cubed = 8/27. Chance of success = 1-8/27 = 19/27. Same result, less mental math, nice for quick thinking on the tabletop.


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## dopey82 (Jul 8, 2008)

Steel Rain said:


> You didn't even bother to read the thread did you?


yes i did read the thread and my comment was inline with the discussion. we are discussing what units to choose and how to use them or equipe them based on there statical ability to perform. mathematically I should have not problem assaulting striking scorpions with assault terminators and since they died horrendously then the math failed. meaning one cant relay on statics in a game of random dice rolling.


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## Steel Rain (Jan 14, 2008)

There are other factors involved with your assault termies. Poor dice rolls, for one. Who strikes first is also important. If they kill a few of your termies before you can strike, and I'm assuming you are using the standard TH/SS combo, then that throws off your chances of winning the combat.


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## Wraithian (Jul 23, 2008)

dopey82 said:


> yes i did read the thread and my comment was inline with the discussion. we are discussing what units to choose and how to use them or equipe them based on there statical ability to perform. mathematically I should have not problem assaulting striking scorpions with assault terminators and since they died horrendously then the math failed. meaning one cant relay on statics in a game of random dice rolling.


You are talking about one occurance, however. Sit with some dice, a piece of paper, and the stats of both units. Roll out the combat on paper. Compare the results. I'm sure that, over time, you will see you are correct--assault terminators should have no problem dealing with striking scorpions. However, in that one occurance, you just happened to get an, "unlucky," roll.

Statistics are not about, "what can I do about this now," when applied to WH40K. They are more of, "this is the projected result I will tend to see after multiple occurances."

The word, "probably," and, "probability," are from the same root word. So yes, assault terminators will, *probably* molest striking scorpions, not guaranteed.


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## The Wraithlord (Jan 1, 2007)

One of my regular opponents is a multi GT winner and he applies stats and probabilities religiously in a tournament setting. In a fun game he will do goofy things because they are fun but when playing for keeps it is math hammer all the way and it generally works for him. And on the other hand you have myself. Never use it, never will, I just go by my gut when deciding what I can or can't do during a game. I don't do quite as well as he does obviously but I also win more than I lose by a fair margin. Math hammer has a place in figuring out the best way to build an army or what your chances are during a battle when faced with the decision to go against different units with one of your own but it is not everything and won't, by itself, win a game for you.

If used as a tool to show you possibilities for a single choice it will be helpful but math hammer does not take into account the next 3 turns and further aspects of your battle plan. Yes, math hammer tells you that those assault terms will most likely rape the Scorpions in front of them but it doesn't help you if killing that unit throws your entire battle plan off by leaving them stranded in front of something that will wipe them out in turn. I have seen guys run the numbers and do something based on those that makes absolutely no sense whatsoever and then get all bent because they feel they did the best possible thing by the math, never once realizing the math hammer didn't even come close to applying in the larger picture. By the math, 10 Assault Marines should tear through a 5 man basic CSM squad with ease but if doing so leaves you standing out of cover in front of my 10 man Thousand Sons squad in rapid fire range does it make sense to do so?

Math hammer is a tool but it is not the be all, end all that many think it is.


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## dopey82 (Jul 8, 2008)

The Wraithlord said:


> If used as a tool to show you possibilities for a single choice it will be helpful but math hammer does not take into account the next 3 turns and further aspects of your battle plan. Yes, math hammer tells you that those assault terms will most likely rape the Scorpions in front of them but it doesn't help you if killing that unit throws your entire battle plan off by leaving them stranded in front of something.


exactly a tool to show you possibilities but do you need a tool to tell you that assault terms should kill scorpions and of course just killing the unit and leaving the terms to be shot to death is a very poor tactic reguardless of any math.

really all im saying is that I think math hammer is far over rated do to the randomness of game play that changes from turn to turn. a series of good or bad roles is what desides a game.


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## Daneel2.0 (Jul 24, 2008)

Math-hammer is a tool. Like any tool, it has jobs that it excels at, and jobs that it isn't suited too. In other words, hammers make poor screwdrivers.

Used correctly statistical analysis and game theory can be applied to 40K just as they apply to other game settings. But they don't tell you the outcomes of individual games. Nor were they meant to.


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## Steel Rain (Jan 14, 2008)

What are you talking about? Hammers make great screwdrivers! They make great forks, knives, band saws, lawnmowers. You name it! You want proof, just ask the Imperial Guard! They use nothing but Sledgehammers!


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## the cabbage (Dec 29, 2006)

I might use a bit of mathhammer when building an army. 

However, once the battle starts I absolutely cannot resist trying for the hollywood moment. I have never refused a last crack a tank shock. I will normally charge when rapid fire could be a better option.

I rarely win howeverk:

On a side issue I wonder what mathhammer GW use when assigning points values. 

_Could all the answers to this last part bleating about your percieved nerfed unit please be put straight into the pointless griping thread thanks._


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## crack3rjack3d (Mar 11, 2008)

I would like to point out that the most useful math hammer approach I have ever seen was more like a series of per turn guidelines to enhance your statistical probability against the enemy player. For example, which units to move when and why, the appropriate way to screen an advance and many other numerical superiority questions. 

Mathhammer shouldn't break the enjoyment of the game, but it should be a tool in the force commanders arsenal. Just as kind men from history like Sun Tzu (Fun fact, fake name), Miyamoto Mushashi, and Carl von Clauswitz have pointed out in their classic tomes...

Doom Ye....


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## Djokovic (Dec 22, 2008)

http://www.heresy-online.net/forums/showthread.php?t=25703

If you are interested in how probability relates to warhammer, i.e. 'Mathhammer', you will be very interested in my thread.

Please look the section of this thread that covers the effectiveness of certain weapons vs. vehicles of different armor values. It is full of statistics, and I derive tactics entirely from mathematical equations.


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## NecronCowboy (Jan 8, 2009)

dopey82 said:


> screw mathhammer!!! mathhammer doesnt explain how my 6 man terminator squad can charge 8 striking scorpions and loose 5 models before i hit. he did 10 wounds and i failed 5 two plus saves. one he shouldnt have hit and wounded that many times and two I shouldnt have failed 5 2 plus saves. what it all comes down to is chance and luck.


See my problem with this is you go through a situation like that and then your brain might decide to not do that in the future due to bias.

So if afterwards that situation really annoys you, then before your inner warhammer start thinking terminators suck you can sit down and work out the probability that could have happened.

Hey maybe the probability of that was only 3%, in that case don't flinch at doing it again because 97 out of a 100 times you are going to come out ahead.


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## b.anthracis (Nov 18, 2008)

In all these probability calaculations, there is one thing you should always remember:" There is no statistic on the single case"


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## NecronCowboy (Jan 8, 2009)

b.anthracis said:


> In all these probability calaculations, there is one thing you should always remember:" There is no statistic on the single case"


That's like telling the blackjack dealer to hit you when you have 20, since you can't predict exactly what the next card is.

If you make the best case statistical choice everytime then in the longrun you will come out ahead.


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## ServiceStud (Mar 1, 2008)

MaidenManiac said:


> Under a normal 40k game you will most likely roll 300+ dice(bit depending on army ofc) and that gives opportunities for weird low-chance things to happen.


In other words: "One in a hundred"-shots occur roughly 6 times per game...


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## Pauly55 (Sep 16, 2008)

Someguy said:


> It's 2/3, not 1/2.


Ah, yes you are correct. I was wrong. I retract my previous statement. It helps me to imagine there are 100 doors rather than 3.


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## kiret (Oct 1, 2008)

This has been a really great read with lots of interesting points presented. It will give me one more tool to use when playing. In response to those who are talking about how you can't predict how every thing will occur and freak happenings on the field, all I have to say is, that's what makes the game fun. If you knew that 100% of the time your termies were going to stomp those Eldar then so much time of the game would be spent with the eldar avoiding the termies. It's that one in one hundred chance that makes the game so much fun. Think of how great the eldar player felt when he saw five termies drop. This is why I love table top games, because just as in real life the best laid plans are always subject to the vagaries of fate. :good:


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## Jack Jack (Nov 16, 2008)

Well I can't believe someone posted this AFTER I posted a thread called wounding probabilities in the technical help section. Anyway, I guess I'll post here my excel sheet that I posted there.


After having read the first two pages, it occured to me that no one was talking about the reliability of the averages, that is, how many times do your spread is correct. 

I designed my own excel file for that purpose that uses the binomial law. How it works is that the user must enter the numbers for his calculations in the yellow squares. How many dies he'll roll, on how much he'll hit, wound and if the opponent has a save, he must write on what will he save. because of how the formula works, he must enter 7 for the save if the save is denied.

*Here is the math logic, you can skip that part if your nose starts bleeding:*

Let's say that you fire *n* number of shots. that you hit on *a*, wound on *b* and they save on *c*. Then, each single shot will have an equal probability of hiting( *P(h)* ), which is given by 
*P(h)* = (7-a)/6​
then, each shot that hit will also have an equal probability to wound *P(w)* which is calculated quite similarly :

*P(w)* = (7-b)/6​
And each wound cand be saved using the same formula, but by changing the random values the probability to save( *P(s)* ) is therefore :

*P(s)* = (7-c)/6​
However, you don't want them to save, you want them to fail. As they can only save or fail and that they can't do both at the same time, we can say that the probability that they save (P(s) or not (P(ns) is exactly equal to one. Therefore:

P(s) + P(ns) = 1
P(ns) = 1 - P(s) = 1 - ((7-c)/6)​
But all these calculations are made for a single shot. What about when you get multiple shots ? For simplicity's sake (you may laugh) we'll consider that we are aiming at a squad with infinite wounds. That way, every shot is independant and as an equal probability of hitting the target, which means *these probabilities do not change with every passing shots*. So if you have a 2/3 chance of hitting and you missed with your 5 first shots, than your 6th shot still has 2/3 chance of hitting.

This allow us to use the binomial law. This law states that the probability of getting X (*P(x)*) success on a random experience in which the results are either success or failure whith N trial on a constant probability of success noted B(n,p) is 

P(x) = N!/((N-X)!X!)p^(x)(1 – p)^(n-x)​
Now, the following can be demonstrated :

Suppose X and Y, two random values and x and y, the results of these values.

If X is B( n, p ) and Y is B( x, q )

than Y is B( n, pq )​
Next, for our mathammer, 

Number of hits (h) is B(n, p(h) )

Number of wounds (w) is B( h, p(w) )

Number of failed saves (ns) is B( w, p(ns) )​
Which becomes:

NS is B ( h, p(ns)p(w) ) 
NS is B ( n, p(ns)p(w)p(h) )​
(Your nose should stop bleeding now)

The above formula gives the probability of getting any exact number of wounds. What is so great about Excel is that if you start from 0 and drag it to n, then it will automatically change any part of the formula you tell it to be revelant and you'll end up having a full table of your probability spread. If you ask it to calculate the cummulative, then you can have an interval of reliability. That is, an interval in wich you are 95% sure that your amount of unsaved shots will be in here. You can either widden it for an increased reliability, or decrease the reliability to narrow the interval. 

so to use the file I've attached, simply fill in the yellow squares and drag the table to its desired lenght.

PS. Necron cowboy, could you send me your filed uncompressed via hotmail ?

[email protected]

I can't open that file.


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## Phenatix (Feb 15, 2009)

EDIT: Above post is better than mine, which is an average.

I'm not sure if someone already said this but I think when you do:

BS4 + S4 (Bolters) vs T4 Sv3+ (SM) you get *(Shots) * 0.125 = wounds*

I actually made it a program on my calculator with the BS, S, T and Sv adjustable, cause I'm a nerd like that. 

Also, I am not a genius, so I may have an error, but I believe this is correct?

X = BS

Shots*(X/6) = Hits

Hits*(3/6) = Wounds (3/6 meaning you need 4s to wound, it goes: 1/6, 2/6, 3/6, 4/6, 5/6 for 6s, 5s, 4s, 3s, and 2s to wound)

Wounds-(Wounds*(4/6))= Kills (4/6 assumes a 3+ Sv right? And 1/6, 2/6, 3/6, 5/6 represent 6+, 5+, 4+, and 2+ correct?)

---

This is what I was using, I believe... Check anyone?


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## Jack Jack (Nov 16, 2008)

Jack Jack said:


> Let's say [...] that you hit on *a*, wound on *b* and they save on *c*. Then, each single shot will have an equal probability of hiting( *P(h)* ), which is given by
> *P(h)* = (7-a)/6​
> then, each shot that hit will also have an equal probability to wound *P(w)* which is calculated quite similarly :
> 
> ...


Just enter the minimum roll you need in these formulas and you'll get what you need, but are think you are indeed correct.

as (7-6)/6= 1/6 for bs 1, (7-5)/6= 2/6 for bs 2 and so on. as for the saves, 1 - ((7-2)/6) = 1/6 for a 2+ save (or you can say he miss on at most ones )

remember you want to substract the saves he *makes*, and for that you have to multiply by the saves he *misses.*


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## Someguy (Nov 19, 2007)

Phenatix said:


> I'm not sure if someone already said this but I think when you do:
> 
> BS4 + S4 (Bolters) vs T4 Sv3+ (SM) you get *(Shots) * 0.125 = wounds*


I'm not sure how you are doing it, but you are getting the wrong answer. 0.125 is 1/8, but the actual chance of killing a marine with a bolter is 1/9.

Personally I find that the calculations involved in 40k are simple enough to do in my head or, if not, they go into more detail than is all that helpful.

Don't lose sight of why you are doing all these calculations. The intended result is to improve your effectiveness in 40k, not invent a whole new game. As such, it's worth knowing stuff like whether a lascannon or a melta gun is more likely to kill a vehicle with armour value X, how much of a difference it makes if you take missile launchers instead of lascannons, how many guys you can expect to kill on the charge with your zerkers, and so on. 

Trying to model an entire game from beginning to end isn't really worth the time it would take, because the end result would be nothing like that. There are just too many variables in "real" 40k.


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## Eleven (Nov 6, 2008)

Revelations said:


> Kind of a wierd thing to say, but I can best demonstrate this via the Monty Hall problem. For those of you that aren't familiar with this, you are given 3 doors. Behind 1 door is a prize, the other two doors have nothing. You pick 1 door, but before it's opened, someone else opens up another door showing nothing and then offers you the chance to switch. The question comes in whether or not it's better to switch to the other door. Now *math* says that due to probability and statistics, it's always better to switch doors. *I* disagree with this because regardless of what the math says, the prize has never moved.


Perhaps I can't figure out what you are talking about, but in this scenario, at first you have a 33% chance of winning. After someone selects a door and it was the wrong one, you have a 50% chance with either remaining door. There is no magical math trick that makes it better to switch. Am I missing something?


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## maddermax (May 12, 2008)

Eleven said:


> Perhaps I can't figure out what you are talking about, but in this scenario, at first you have a 33% chance of winning. After someone selects a door and it was the wrong one, you have a 50% chance with either remaining door. There is no magical math trick that makes it better to switch. Am I missing something?


Yeah, you'd think so, but mathematically, there is actually a difference. Think of it this way:

There is a prize behind door A, but door B and C have nothing. 

There are 3 options: 

1)you choose Door A, door B(or possibly C) is opened, and you're given the option to change your choice - if you do change your choice, you lose.

2) you choose Door B, Door C is opened, and if you choose to change your choice, you win.

3) you choose Door C, Door B is opened, and if you choose to change your choice, you win.

Therefore, statisically, you have a 66% chance of winning if you change you choice, not a 50% chance for each door. Weird yes, and only works in this sort of competition, but interesting none the less.


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## Jack Jack (Nov 16, 2008)

If it can help you, imagine the monty hall with 1000 doors.

Once you've chosen your door, the guy opens 998 doors with nothing behind, because he knows where is the prize. you are left with two doors. the one you chose and the one with 99,9% probability.

Remember, the doors open by the host are NOT selected randomly. He choses all those who are not prized minus one and the door you chose must be included.

So your initial choice has 0,1% chance to be on target, (1 in a thousand) which means that there is 1 - 0,1% = 99,9% chance that you missed.

So all the doors you did not choose are opened and the prize is not revealed. Of course you could have chose the "good" door to begin with, but like I said, that only happen once in a thousand. So you have 99,9% chance that the prize is the door behind the door you did not choose, because each of the 0,1% of every door must be transferred equally to the other doors remaining when they are opened. 

Why is your door unaffacted? Because you did not know, when you picked it, which would be the 998 doors to be revealed.

The problem remains the same with three doors, although it's quite less spectacular.

Hope to have brought the problem from a different angle.


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## Eleven (Nov 6, 2008)

maddermax said:


> Yeah, you'd think so, but mathematically, there is actually a difference. Think of it this way:
> 
> There is a prize behind door A, but door B and C have nothing.
> 
> ...


Ah I see. that makes perfect sense then. thanks for taking time out to explain. So why is this problematic to the use of statistics? If anything, it proves that more advanced knowledge of statistics can increase your odds of winning even more.


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## Revelations (Mar 17, 2008)

Eleven said:


> Ah I see. that makes perfect sense then. thanks for taking time out to explain. So why is this problematic to the use of statistics? If anything, it proves that more advanced knowledge of statistics can increase your odds of winning even more.


It could be that I think more in the realm of logic and quantum mechanics (I know, wtf right?). Without getting into it, any mathematical argument and formula can be crafted to dictate your odds of winning, and I won't argue with it because the math is sound. But, if you came at me with a perfect system for gambling, I still wouldn't be motivated to hop a plane to vegas with you. 

You're still playing odds. That's works for a lot of people, and it doesn't work for others; like me. 

(And no, you don't need to try to convince me otherwise, I don't disgaree with the math )


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## Jack Jack (Nov 16, 2008)

No one who knows his math will ever try to go to Vegas as the house knows hers!

The thing is, if the number of trial tends toward infinite, then your results will ALWAYS be equal to the previsions made by the formula.

I won't try to demonstrate it here, because it's a little more complicated than it seems, however, you should know it's an important result in the theory of probabilities that's been proven times and times again.

In fact, if you said you don't believe in this, it's quite the same as if you said you were a creationist or that you did not believe that the theorem of Pythagora is true. ( If the sides of a triangle ABC are such that square mAB plus square mBC equals to square mCA, than it is a square triangle in ABC ).


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## NecronCowboy (Jan 8, 2009)

Sorry the monte hall problem does NOT in any invalidate probability, it only proves that sometimes people are looking at a probability problem incorrectly. I have complete faith in dice probability and in standard deviation. I do not however think you can figure out the probability of an entire game of 40K because the problem space is not discrete enough. This math is good for helping you make choices during a game, such as should this unit shoot at that tank, or at that infantry in cover?

Do NOT let monte hall blabering by anyone who happened to have watched the movie 21 and missed the point, to steer you from probability, because it does work, if it didn't then Vegas would have gone out of business years ago.


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## NecronCowboy (Jan 8, 2009)

Let me also add, that if you believe probability to be false or that it does not apply to you because your luck is somehow special then you are a fool.


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## Gandalf the Black (Jul 18, 2009)

Very, Very helpful. Thankyou and Very good!


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