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Mathhammer - Odds vs Averages


This is an article i wrote for the more advanced players who use mathhammer frequently.


When a player works out mathhammer, 99% of the time they will use averages rather than the actual odds.

My goal is to show people that averages aren't always the best thing to go by, and that people should use the Odds instead.

For the people that are confused...

Averages - The average number of models you will kill.
Odds - The chances of you killing a model(s).


Lets take an example....

Lets say you have a 10-man unit of Tactical Marines who are rapid firing on a single Terminator.
Lets assume that 9 of them have Bolters, and that the Sergeant has some other weapon (lets ignore his weapon for the purposes of this example).

Now, first we will use the common mathhammer that 99% of people use... Averages.

9 Marines will have 18 shots
12 hit
6 wound
1 failed save


Now is where the problem comes in...
People will assume they should be able to wipe him out at this stage, which is a VERY POOR assumption to make.
Quite often they will be dissapointed when they dont kill it, and say things like "dam, he should have died".


To prove why averages are a bad guide, lets work out the same situation with the actual Odds...

For a single bolter shot to kill a Terminator, the chances are 2/3 to hit * 1/2 to wound * 1/6 failed saves.
(2/3)*(1/2)*(1/6)= 1/18.
If there is a 1/18 chance for a single bolter shot to kill a Terminator, then there is a 17/18 chance for a single bolter shot NOT to kill a Terminator.

So for the chances of 18 bolter shots NOT to kill a Terminator, the chances are...
(Odds for single shot to miss)^(Number of shots)
(17/18)^18 = 35.7%

Therefore, there is a 64.3% chance to kill the Terminator with the 18 shots, and a 35.7% chance that you wont.





To further prove the advantages of Odds vs Averages, lets look at another example....

Lets take a squad of IG Veterans with 3 Meltaguns, and a Platoon Command Squad with 4 Meltaguns.
Lets assume that they are shooting at a T4 Independant Character with no Invulnerable save, with the hope of killing him outright.
Assuming you will only need the 1 Meltagun to hit (we will ignore the To Wound roll for simplicity purposes), which are you better off with?

The main thing here is to compare the BS with the number of shots.

First of all, lets use Averages to see what will happen....

3 shots at BS4
Average of 2 hits

4 shots at BS3
Average of 2 hits

Now people will say "wait, there is no difference between them, the odds are the same"..... WRONG!

Lets look at it using Odds....

(Odds for single shot to miss)^(Number of shots)

3 Shots at BS4
(1/3)^3 = 3.7%
Therefore, there is a 96.3% chance that at least 1 Meltagun will hit, and a 3.7% chance that they will ALL miss.

4 Shots at BS3
(1/2)^4 = 6.3%
Therefore, there is a 93.7% chance that at least 1 Meltagun will hit, and a 6.3% chance that they will ALL miss.


Yes, thats correct, you are actually almost TWICE AS LIKELY to miss 4 BS3 shots than you are of missing 3 BS4 shots, even though they will both score 2 hits on "average".





So in conclusion, averages are still ok to work out a majority of mathhammer as quick as possible.
But when it comes to needing that 1 wound, destroy that 1 tank, or anything else that is crucial to the outcome of the game, by using Odds rather than Averages you can calculate a far more accurate probability to maximize your tactical movement and target priority as much as possible.

I hope you found my article an interesting read, and hope that you will consider calculating Odds rather than Averages when the game is on the line.
 

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Anyone using mathhammer should be beaten with a funmallet in my opinion
...But, I ENJOY calculating chances T_T

Frankly, people who assume the average will work is kidding themselves.
Plan for the worst, hope for the best, that's the way to play.
If something NEEDS to die, make damn sure it will; if it dies before you're done, good for you, shoot other things.
 

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*goes to fetch the mallet*
you were warned:biggrin:
Really, one time I calculated the exact chance that a SM Sergeant with a Power First would destroy a standard Defiler in CC, and I think he was charging in the hypothetical...

Anyway, it took me hours, and I'm pretty sure (can't quite remember) that the chance simplified to 1 in ~7, but I could be completely wrong in that memory...
There was definitely a ~7, it was 7.126 or something.

Yes, I enjoy calculating chances >_>
 

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Hmm well Deflier is WS 3 so each attack has 66% chance of hitting, Glances can't destroy it so we only care about pens, 33% chance of a pen. Of those pens only 5's or 6's can destroy it so again 33% chance. So roughly a 7% chance with each hit. With his 3 attacks on the Charge about 22% I would guess.

Aramoro
 

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Hmm well Deflier is WS 3 so each attack has 66% chance of hitting, Glances can't destroy it so we only care about pens, 33% chance of a pen. Of those pens only 5's or 6's can destroy it so again 33% chance. So roughly a 7% chance with each hit. With his 3 attacks on the Charge about 22% I would guess.

Aramoro
I also factored in the chance that the Defiler would kill him before he got to attack.
And I am absolutely certain that it was 1/~7 now.
 

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I had made the assumption that the Sarg would have ablative wounds. If not things don't look good for our hero, Deflier has 3 attacks, hitting on 4's killing on 2's. So what a 72% chance to murder him before he gets to strike.

Aramoro
 

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I'm afraid I don't understand your math. I'm not saying it's wrong, as it may very well be just over my head.

Why are you using ^? Shouldn't you just be using multiplication?

I mean, in the original example, if each bolter shot has a 17/18 chance of not killing the terminator, which translates into each shot having roughly (by my math) a 5.55% chance of killing the terminator. 5.55% * 18 shots = 100%, which is basically just saying that 1/18 * 18 = 1. I don't see what bringing it to the power of 18 proves.
 

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Odds and averages both have their place, it depends on the question. How likely a given weapon is to destroy a target (to know what kind of things you want to take), or a given model to take a wound (to know what kind of heat they can take) or similar things are odds. When you start rolling entire units against entire units with questions like 'how effective are dire avengers at shooting space marines v orks?' or more broadly 'what kind of offensive output can I reasonably expect from this unit?', then you're talking about averages which work out like a bell curve or normal curve (google it if you're unfamiliar with the term) as far as odds of getting numbers near those averages go.

You just need to realize that when using averages, you aren't guaranteed to get the average result. You will get within 1 standard deviation (again, if you're unfamiliar with statistics, look it up) of the average result ~68% of the time, within 2 standard deviations ~95% of the time, and within 3 standard deviations of the average result ~99.7% of the time.

Looking at the King of Cheese's numbers, they fall in pretty well with this idea.
Therefore, there is a 64.3% chance to kill the Terminator with the 18 shots, and a 35.7% chance that you wont.
Those seem to be about the numbers that the result will fall within 1 standard deviation of average. Not a coincidence. :wink:

Here's a normal curve so you get an idea of what I mean. Basically the fatter the section of the graph (with the average in the middle denoted by the greek letter sigma), the more likely an individual random result will fall there.

 

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I'm afraid I don't understand your math. I'm not saying it's wrong, as it may very well be just over my head.

Why are you using ^? Shouldn't you just be using multiplication?

I mean, in the original example, if each bolter shot has a 17/18 chance of not killing the terminator, which translates into each shot having roughly (by my math) a 5.55% chance of killing the terminator. 5.55% * 18 shots = 100%, which is basically just saying that 1/18 * 18 = 1. I don't see what bringing it to the power of 18 proves.
No, you don't use multiplication.
To get the chance of, for example, two shots killing a Terminator, using the example chance of 10% per shot, you do:

10% chance for the first shot.
You get the remainder of the 100%, which is 90%, and then get 10% of that (the chance of the next shot doing it), which is 9% of the original 100%.
So you have a 19% chance of killing one Terminator with those two shots.

If you want the chance of killing two Terminators with two shots, you do the opposite.
You get the chance of killing one, 10%, and get 10% of that, which is 1% of 100%.
So a 1% chance of killing 2 Terminators with 2 shots.


If you have different chances, of for example 3 shots but you want the chance of killing 2 Terminators, it doesn't really matter how you order them.

It gets pretty complicated sometimes, but that's how I do it, the manual way :D
 

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I'm afraid I don't understand your math. I'm not saying it's wrong, as it may very well be just over my head.

Why are you using ^? Shouldn't you just be using multiplication?

I mean, in the original example, if each bolter shot has a 17/18 chance of not killing the terminator, which translates into each shot having roughly (by my math) a 5.55% chance of killing the terminator. 5.55% * 18 shots = 100%, which is basically just saying that 1/18 * 18 = 1. I don't see what bringing it to the power of 18 proves.
This is the point he was making - averages vs Odds. On Average, you should kill a terminator, because if each marine is doing an average of 0.0555 wounds, then 9 marines would kill an average of one terimator per turn. However, this is looking at what you might do over a number of turns - fire for 5 turns, and you might average 5 dead terminators, but sometimes you'll do 2-3 wounds, sometimes you'll do none.

However, if you are only aiming to kill a single terminator, your actual chances of doing so are only 65%, with a 35% chance that you'll do nothing. It's a useful distinction to know when you're looking at the potential of a unit taking out a threat.
 

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Ah, I see what the OP means now. I've actually heard this elsewhere, and the basic conclusion I've found is that while this method is indeed far more accurate, it's not very practical to figure out during the game. On the other hand, it is useful when deciding between units, such as in the 3 BS 4 shots VS 4 BS 3 shots example.

A good rule of thumb I use is that if you MUST destroy a target, you should throw twice as many shots/attacks at it that you would normally need to kill it, on average. It's not exactly accurate, but it's a lot easier to figure out.
 

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Plan for the worst, hope for the best, that's the way to play.
sort of, the thing which really needs to be taken into account in addition to the probabilities when making decisions is also the outcome when favourable and also when failed.

Two questions are essential -

1) what happens if I fail
2) what happens if I succeed

The key with good tactics is to balance these and appreciate your odds, for example:

A situtation which arises gives very poor odds of sucess, however if you succeed you will win your game, if you fail you loose a unit but still have other options.

If you play this by odds and you take the safe bet then you will miss your oportunity, you keep the unit and you loose nothing but you also miss a big potential game.

I am not saying throw caution to the wind, but sometimes take the short odds for the big gain. Ask "what have I got to loose":so_happy:
 

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Holy shit. My brain hurts after reading all of the numbers. I took statistics in college and I was never sure which hurt my head more... the statistics or the hangovers!

I generally use averages over probability. I do, however, recognize the difference between the two.
 
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