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These are mistakes which almost everyone makes, and, while often not a big deal in the spur of the moment, when trying to mathhammer something out they can lead to wrong conclusions.

The first problem is that people miscalculate expectation. In a binomial distribution (which is what one normally uses for die rolls) the expectation is n*p, where n is the number of rolls, and p is the probability of success. This is all well and good in theory, but we quickly run into a problem... The problem is that sometimes you factor impossible events into your distribution. If I have 10 marines shoot at a carnifex, in our model they have the possibility of doing 10 wounds, which is impossible, all wounds after the 4th would be truncated, thus you can't actually use a straight up binomial model and you can't really use n*p (though n*p will usually be close).

This leads us into the second problem, which is that people use expectation even though they really shouldn't and are interpreting it incorrectly. It pains me to count the number of times i've heard, "well, i'm expected to do at least 4 wounds to X, so half the time I'll do 4 or more wounds to X." UNTRUE! You are talking about the median not the mean (aka expectation) of the distribution! The median number of wounds is the number (not necessarily unique) such that half the time you will do better, and half the time you will do worse. This solves our first problem because in the truncation example, unless the median of a distribution is in the truncated, it will not be changed by truncation.

The mean minimizes the square sum of the error terms, while the median minimizes the absolute sum, so Expectation is much less robust to outliers (ie if a gun has a 1/1000 chance of doing 5000000 wounds to the target, it's going to totally screw up the mean but do almost nothing to the median)

Now, even with the median people can still misinterpret in a different way! I also hear people say "If I shoot with 10 guys at the Falcon and my expected number of vehicle destroyed results is exactly 1, then the expected number of guys I need to kill the Falcon in one round of shooting is 10." WRONG AGAIN! In order to determine what the expected number of guys needed to kill the Falcon is you need to look at a geometric distribution. You figure out the probability to kill a carnifex with 1 guy, then plug that in for p, and find its expectation (which you probably dont want anyways...) or find the median (which is more useful).

Anywho, all this said, the median is a huge pain in the ass to calculate and usually must be done by simulation, and outliers are usually not that fair out or that heavily weighted (like 100 bolter shots doing 100 wounds to a carnifex) so on the fly expectation is ok, but when people say "I mathematically proved that X is better than Y" they better have their statistics right... (or i'll melee them with my bolt pistol)

Example of how to apply this:
----------------------------------

Say we want to apply the above to shooting a ML with a marine at a falcon.

For a single ML we have:

2/3*1/3*1/3 = 2/27 = p = probability a single shot will destroy a falcon

Now we want to figure out the number of shots we need to take in order to have at least a 50% chance of destroying the falcon, so we look at a geometric distribution on wikipedia, it tells us that the median is the smallest integer greater than

-log(2)/log(1-p)

so we plug that in and get 9, thus in order to have a 50% chance of destroying a non-holofield falcon, we need to shoot 9 shots.

If we used expectation instead of the median, we would have gotten the usual answer of 27/2 or 13.5, which may have a useful interpretation (especially if you are considering total number of shots to kill 2 or 3 falcons), but is not the "50% interpretation" that most use it as.

anyone else have any common mathhammer mistakes to write up?

enjoy!

(ps anyone here play in the southbay area? mountainview-ish?)

When you apply mathhammer ( that other game people play ) to 40k. You forget about the dice gods and how horrible they can be to you sometimes.

Best thing to kill a falcon with everything on it....

HQ--312

1 Captain, Bolt Pistol, Power weapon,
command squad 2x Missle Launchers, Vet, Power weapon.

Chaplin, J-pack, Bolt Pistol, Melta Bombs.

ELITES--315
Vet squad, Lascannon, Skill: TH
Vet squad, Lascannon, Skill: TH
Vet squad, Lascannon, Skill: TH

TROOPS--515
Tact, Lascannon
Tact, Missle Launcher
Tact, Missle Launcher
Tact, Plasma Gun
Tact, Plasma Gun
Tact, Plasma Gun

FAST ATTACK--383
Assault squad, Vet, Power Fist, Flamer
Assault squad, Vet, Power weapon, Melta Bombs, Flamer
Assault squad, no packs, Vet, Power weapon, Flamer

HEAVY SUPPORT--325

Dev squad, 2x Heavy Bolters
Dev squad, 2x Heavy Boltters
Dev squad, 2x Missle Launcher



You can play mathhammer ( that other game ), I will play 40k.

its all just some fuzzy math to me

anyone else have any common mathhammer mistakes to write up?

Easy, a very common one.
'1 dice is a 1/6 chance to wound, but I have 6 dice to roll so have a 100% chance to wound'

Sounds logical, but is entirely wrong.

Tarval wrote:When you apply mathhammer ( that other game people play ) to 40k. You forget about the dice gods and how horrible they can be to you sometimes.

Best thing to kill a falcon with everything on it....

HQ--312

1 Captain, Bolt Pistol, Power weapon,
command squad 2x Missle Launchers, Vet, Power weapon.

Chaplin, J-pack, Bolt Pistol, Melta Bombs.

ELITES--315
Vet squad, Lascannon, Skill: TH
Vet squad, Lascannon, Skill: TH
Vet squad, Lascannon, Skill: TH

TROOPS--515
Tact, Lascannon
Tact, Missle Launcher
Tact, Missle Launcher
Tact, Plasma Gun
Tact, Plasma Gun
Tact, Plasma Gun

FAST ATTACK--383
Assault squad, Vet, Power Fist, Flamer
Assault squad, Vet, Power weapon, Melta Bombs, Flamer
Assault squad, no packs, Vet, Power weapon, Flamer

HEAVY SUPPORT--325

Dev squad, 2x Heavy Bolters
Dev squad, 2x Heavy Boltters
Dev squad, 2x Missle Launcher



You can play mathhammer ( that other game ), I will play 40k.

dice gods making things go wrong is just another way of saying that people rely on expectation, when they should be looking at the whole distribution... the fact of the matter is that people who say "you play mathhammer, i'll play 40k" haven't looked at the math deep enough to see that it actually can be predictive. That's not to say that you necessarily should. We all play for fun, and that's great. However, to dismiss it because you haven't looked at it is not very open-minded.

That said, your list looks good, but it might have some trouble against 100+ orks :0)

I find a lot of people making the opposite mistake to the one you mentioned, nrs02004. Often the chances of overkill are extremely remote, so much so that they have no material effect on the final calculations. Consider a unit of ten marines with bolters firing at a carnifex; it’s technically possible to score five or more wounds so a straight np calculation will give a higher value, the odds of scoring five wounds or more is so ludicrously small calculating the straight average will be more than sufficient.

When calculating odds in the middle of a game it’s necessary to sacrifice a little accuracy in order to work with numbers you can calculate on the spot.

I’ve seen people neglect to do any maths at all during a game, and when I ask them about it they’ve listed all the complications that make a straight np calculation misleading. In most situations where overkill is unlikely, np is enough.

"on the fly math" without a calculator or a notepad can't be 100% exact, it needs to be mental so we can get a rough estimate to see if its "worth it" to shoot/charge a target.

A common math-hammer mistake is not understanding math at all, and then assuming that it will have no bearing on the game because you don't understand it. Praying to dice gods, switching dice that roll poorly for you, having cute girls blow on your dice before rolling - useless... actually, that last one is ok. Still, it's better to know the odds, know what odds are (an important, often overlooked step), and plan accordingly.

Matt-ShadowLord wrote:

anyone else have any common mathhammer mistakes to write up?

Easy, a very common one.
'1 dice is a 1/6 chance to wound, but I have 6 dice to roll so have a 100% chance to wound'

Sounds logical, but is entirely wrong.

yep have to roll one at a time (using a tree)
--(1/6)
-//
A---(1/6x5/6)
-\\----//
--5/6--(1/6x5/6x5/6)
-----\\-----//
----5/6x5/6
---------\\(and on)

first roll is 1/6 next roll must be 5/36 next roll is 25/216 next is 125/1296 next is 625/7776 and finaly 3125/46656
(7776/46656 + 6480/46656 + 5400/46656 + 4500/46656 + 3750/46656 + 3125/46656)
add them and you have the chance of one or more of them wounding 31031/46656 or 66.510...% but you still need to included the chance of getting 1-6 wound and that where it sucks quite a bit cause you can't combine them and you get a range of chances and i'd have to included a graph

... remember pure mathamatics can kill


ha i made a mistake as well ^_^ still that pure math

My personal favorite mathhammer error is presuming that you will lose the amount of guys your save says, with the successes coming first.

That is, I take 2 space marine saves. I figure I'm ok, since I only have a 1 in 3 chance of failing a space marine save. In truth, the most likely result (5/9) is that I lose one of them. In the same way, if you take 2 5+ saves, you are most likely to make one, rather than none.

In addition, people are biased towards success at 3+ saves, and failure at 5+ saves. If you make 2 5+ invulnerable saves, or fail 2 armor saves, (both 1/9, or more likely than failing a ld 10 test) they act like it's a huge deal.

The problem with Mathhammer is that it does not take into account that my dice hate me, and roll accordingly.

On a serious note, I wonder how much math I actually do during a game. I do not think I do all that much, but I think I instinctively know what weapons have the best odds to kill certain things, and I know to use the right tools for the right job (For example: Without running the math, I know that plasma guns are a good way to kill a Carnifex). I do basic mathhammer during a game commonly during saves since that is the easiest time to do it, and you can know how many should die.

Some of the big ones:

Confusion of expectation and probability: Probably the biggest one in erroneous analysis; figuring expectation is easy, figuring probability (or the typically more useful median value, as noted above) is more difficult. Adding probability values doesn't yield a probability value, but instead an expectation.

Skewed evaluation of 3+ (or better) and 5+ (or worse), as mentioned above

Overvaluation of single events, especially combined with undervaluation of multiple events: Figure the actual value of a Lascannon attack on AV14 compared to an Assault Cannon attack on AV14. Then ask a random player what they think about it.

Overvaluation of reliability of sequences with individually reliable elements: When looking at something like a Lascannon on armor or a Power Fist in assault, players seem to expect results from high-S (and thus highly reliable on wounding/penetration rolls) as if they never miss. I also see people greatly overvaluing incoming high-S/low-AP fire for much the same reason.

Failure to incorporate side effects: A marine firing a plasma gun over 3 turns kills 1.0 marines per round on average. A naive calculation will yield 1.1, but that doesn't take into account Gets Hot's chance of removing future shots.

Failure to incorporate resilience in upgrade effectiveness valuation: How many times have you seen a "is X upgrade character worth taking?" discussion that compares upgrade cost per point to base model cost per point without considering the cost per point of wounds on the base models?

Blackmoor wrote:The problem with Mathhammer is that it does not take into account that my dice hate me, and roll accordingly.

On a serious note, I wonder how much math I actually do during a game. I do not think I do all that much, but I think I instinctively know what weapons have the best odds to kill certain things, and I know to use the right tools for the right job (For example: Without running the math, I know that plasma guns are a good way to kill a Carnifex). I do basic mathhammer during a game commonly during saves since that is the easiest time to do it, and you can know how many should die.

What you--and I'm willing to bet the vastest majority of 40k players--are doing is heuristic decision-making. We hold to general tried-and-true principles applied in specific situations rather than creating an in-depth analysis of every situation. You could also call it 'grass is green' reasoning. Heuristics generally work very well in games because so much is constant. In general, we know exactly what models do and what modifiers are acting in their favor or against before having to commit to a decision. This makes heuristics, which are a more simplistic form of reasoning, much more effective in-game than in real life because variables are minimized. It's for this reason that we target Terminators with lascannons; we "know" that lascannons are good at killing Terminators. Even then, we're not really running the numbers and considering that we'll convert a 5/6 success into a 1/3 success, we're thinking more along the lines that Lascannon > Terminator. It's a simple, somewhat sloppy, and somewhat effective method, much like the 40k gaming system itself.

I think a lot of people, when making quick battlefield math, go more along the lines of.

AP2 > SV3 S7 > T4 thus I will do better shooting a marine with a plasma gun than I will a bolt gun AP5 < SV3, S4=T4.

While its a given that volume of shots can make up for lack of strength and AP, its a question of resource management and what you need to kill first. Experience is the only thing that will tell you you need to kill the 10 assault marines long before you need to kill the 10 tac marines. The math on what to use to kill the 10 marines is the same. But which will hurt you worse can only really be determined by experience.

That's the largest math mistake I see.

For a single ML we have:

2/3*1/3*1/3 = 2/27 = p = probability a single shot will destroy a falcon

Now we want to figure out the number of shots we need to take in order to have at least a 50% chance of destroying the falcon,

For multiple ML shots you have to use the complementary probability of each shot and multiply them. This gives the overall probability for missing. The complement is that what you actually want. I'm too lazy to do it for you...

... remember pure mathamatics can kill

Lol, not really, but it can bust your head. I'm a pure mathematician.

nrs02004 wrote:If we used expectation instead of the median, we would have gotten the usual answer of 27/2 or 13.5, which may have a useful interpretation (especially if you are considering total number of shots to kill 2 or 3 falcons), but is not the "50% interpretation" that most use it as.

So what does the 13.5 represent? What is the chance of the 13.5 shots getting at least 1 "destroyed" result? While I see the difference between the math involved, I'm still not quite sure what the difference between median and expectation is. Could you go into some further explination as to what they both really mean.

The probability to destroy a Falcon with holofield (never go out without it) not moving more than 12'' (no vector engines required) is
2/3 to hit * 1/3 to penetrate * 1/9 to roll 5,5 or 5,6 or 6,5 or 6,6, that is, 2/81.

Phoenix wrote:

nrs02004 wrote:If we used expectation instead of the median, we would have gotten the usual answer of 27/2 or 13.5, which may have a useful interpretation (especially if you are considering total number of shots to kill 2 or 3 falcons), but is not the "50% interpretation" that most use it as.

So what does the 13.5 represent? What is the chance of the 13.5 shots getting at least 1 "destroyed" result? While I see the difference between the math involved, I'm still not quite sure what the difference between median and expectation is. Could you go into some further explination as to what they both really mean.

The expectation value may be interpreted as follows. If N marines shoot krak missiles, where N is a very large number, they will destroy N/13.5 falcons (ignoring the accumulation of immobilized and weapon destroyed results). On average, it takes 13.5 marines with missile launchers to destroy a falcon.

If I want to probability that s shots will get at least one destroyed result, I take the probability p that a single shot will get a destroyed result and compute 1 - (1-p)^s. 1-p is the probability that a single shot will fail, (1-p)^s is the probability that s shots will fail, and so 1 - (1-p)^s is the probability that s shots will not all fail, ie at least one will succeed. For p = 2/27 and s = 13.5, we get a probability of about 65% of getting a destroyed result.

edit - as pointed out above, this example is purely academic, since it involves a falcon sans holofield.

wuestenfux wrote:

For a single ML we have:

2/3*1/3*1/3 = 2/27 = p = probability a single shot will destroy a falcon

Now we want to figure out the number of shots we need to take in order to have at least a 50% chance of destroying the falcon,

For multiple ML shots you have to use the complementary probability of each shot and multiply them. This gives the overall probability for missing. The complement is that what you actually want. I'm too lazy to do it for you...

That's actually not what you Have to use. It gives you similar information, but in a more roundabout way. The point is that i'm often not looking for the probability that X number of shots will destroy a falcon, I'm looking for the probability that the falcon will be destroyed after X shots and not before. That way, if I want to combine probabilities, I can. (In this particular example with the median, yes you could putz around and just plug in numbers until you got to 9 with your method)

and yes I forgot to factor in holofields, but they are totally unimportant regarding the math involved.

Interesting. Good to see that dispite my lack of stat classes (something I'm almost regret...almost) I've more or less done things correctly. While I haven't calculated out the probability and instead use the expectation values for things, I apparently had a decent grasp of what it has been that I've been calculating.

Good to know. I'll have to try adding in new stat calculations to my spread sheet and see how much it changes things.

Phoenix wrote:

nrs02004 wrote:If we used expectation instead of the median, we would have gotten the usual answer of 27/2 or 13.5, which may have a useful interpretation (especially if you are considering total number of shots to kill 2 or 3 falcons), but is not the "50% interpretation" that most use it as.

So what does the 13.5 represent? What is the chance of the 13.5 shots getting at least 1 "destroyed" result? While I see the difference between the math involved, I'm still not quite sure what the difference between median and expectation is. Could you go into some further explination as to what they both really mean.

Shirou gives a pretty good explanation. If you have some large number of shots, N, almost surely the number of destroyed results will be N/13.5.

It being the expectation does not really say anything about the probability of getting one destroyed result with 13.5 shots. As Shirou points out, the probability of getting one destroyed result is 65%.

If we look at the median (9), we can say: the probability of getting 1 destroyed result with 9 shots is 50%. This does not say what will happen if we fire some large number of shots (N).

In general expectation is for a large number of shots, probabilities/medians are for fewer shots

"and yes I forgot to factor in holofields,"

Everybody does that once.

By the way, there's no need to look up the geometric distribution to get that the median number of shots needed to take down the falcon is approximately 9. Wikipedia gives far more information that we actually need. In my previous post I showed that the probability of scoring at least one success in s tries is 1 - (1-p)^s. By definition, the median value for s occurs when this probably is equal to 1/2.

1 - (1-p)^s = 1/2

The solution for s is then

s = - log(2)/log(1-p)

which is the formula given in the original post.

shirou wrote:By the way, there's no need to look up the geometric distribution to get that the median number of shots needed to take down the falcon is approximately 9. Wikipedia gives far more information that we actually need. In my previous post I showed that the probability of scoring at least one success in s tries is 1 - (1-p)^s. By definition, the median value for s occurs when this probably is equal to 1/2.

1 - (1-p)^s = 1/2

The solution for s is then

s = - log(2)/log(1-p)

which is the formula given in the original post.

Nice.

However (for a reason to use a geometric distribution) say we wanted to both kill a falcon and a wraithlord with our 10 missle launchers (all from different squads). To come up with the probability of killing both we can sum (over all n) [the probability of killing the falcon after exactly n shots (and not before) times the probability of killing the wraithlord with the 10-n remaining shots]. One can definitely do this by hand without referring to the "geometric distribution" but essentially that's what you're using (the product of a geometric and a binomial)

shirou wrote:By the way, there's no need to look up the geometric distribution to get that the median number of shots needed to take down the falcon is approximately 9. Wikipedia gives far more information that we actually need. In my previous post I showed that the probability of scoring at least one success in s tries is 1 - (1-p)^s. By definition, the median value for s occurs when this probably is equal to 1/2.

1 - (1-p)^s = 1/2

The solution for s is then

s = - log(2)/log(1-p)

which is the formula given in the original post.

So let's see if I understand this.

Lets find out the median of the number of lasgun shots by gaurdsmen it'll take to take down a marine.

So chance of hitting * chance of wounding * chance of failing their armor save, correct? So 1/2*1/3*1/3 = 1/2 according to my spread sheet.

So tofind the mean would be s = -log(2)/log(1/2) which this is gives us 1 3/7 according to my spread sheet. I that right?

nrs02004 - I know that the probability distribution is essentially what I'm using, I'm just trying to keep things simple. I mean, let's be honest. Most people who already know about distributions aren't going to need help with their mathhammer, and those who don't know about distributions probably aren't going to learn probability theory just to play this game.

Tresson wrote:

So let's see if I understand this.

Lets find out the median of the number of lasgun shots by gaurdsmen it'll take to take down a marine.

So chance of hitting * chance of wounding * chance of failing their armor save, correct? So 1/2*1/3*1/3 = 1/2 according to my spread sheet.

You must be using a Microsoft product. 1/2 * 1/3 * 1/3 = 1/18.

So tofind the mean would be s = -log(2)/log(1/2) which this is gives us 1 3/7 according to my spread sheet. I that right?

Using this formula we are computing the median, not the mean. The median is the number of attempts needed to have a 50% chance of success; the mean is the number of tries needed, on average, to achieve one success. Since the probability of getting a success with one try is 1/18, the mean is just 18: you need, on average, 18 lasgun shots at BS3 to kill a marine. The median is given by the -log(2)/(1-p), with p = 1/18. That gives about 12, so you need approximately 12 lasgun shots at BS3 to have a 50% chance of killing at least one marine.

PS. You should check the way you're entering things into your spreadsheet. Not only was your multiplication wrong, but computing -log(2)/log(1/2) should have given you exactly 1.

yeah, i know shirou. I just felt silly for saying "look up the median on wikipedia" when it is such a simple computation. The however was more for the people above saying "dont bother with a distribution, just use the complement."

Tresson, this is basically what I said.
Keep in mind, math is locally trivial.

Excellent material so far nrs02004. I don't want to hijack the thread, but I was wondering if you could show us an example using Striking Scorpions.

I have this hunch that, if a Striking Scorpion Exarch does not get bonus attacks from its Mandiblasters and charging, that the Scorpion's Claw is one of three live options for assaulting Guard, Orks, and Space Marines (WS3, T3, I3, Sv5+|W4, T4, I2, 6+|WS4, T4, I4, Sv3+), rather than being the 'no-brainer' option per cost. The other options are the Biting Blade, and the Chainsabres.

If the Exarch assaults, and doesn't get the bonuses for the Scorpion's Claw, the numbers would be something like:

Scorpion's Claw

A2 S6 AP2 Pts15

Biting Blade

A3/4 S4-(7 v 8) AP- Pts5

Chainsabres

A4/5 S3 AP- Pts5 Re-roll to-hit and to-wound

Scorpion Chainsword & Shuriken Pistol

A5 S4 AP- Pts0

Scorpion's Claw

A2 S6 AP2 Pts15

Biting Blade

A4 S4-(7 v 8) AP- Pts5

Chainsabres

A5 S3 AP- Pts5 (Re-roll to-hit and to-wound)

Can you show how to check the veracity of this hunch?

Nice thread. Brings back memories of P. Chem and the brain exploding math I had to learn back then. Also makes me sad cause I don't recall how to do any of it now only 5 years later...

At least I can still compute basic probabilities and expectations and grasp the topic at hand. I guess I personally never worried too much about whether I was computing the median or the mean, unless I was dealing with multiple shots against a single tank.