Mathhammer - The guide

Discuss tactical and strategic development for 40K/Tau.
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PeterLageri
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Post#19 » Nov 14 2007 01:15

C*RP!

Doctor Stupid here at the keyboard found his brain again.

I've used the chance of success as only 1/3 (5 or 6) and not as I should have 1/2 (4,5 or 6)

....

:::(

I'll go a bang my head against the wall..............

....

...

Sorry, onlainari, your calculation is correct...........

:smile:

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Peter Lageri
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Peter Lageri

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PeterLageri
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Post#20 » Nov 14 2007 12:13

One thing though.........

How did the 8/9 get in the "1 - {1 - [P(success)]}^# shots = 1 - {1 - [8/9*2/3*1/2]}^4 = 1 - [1- 8/27]^4 = 0.755
" equation?

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onlainari
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Post#21 » Nov 14 2007 08:04

Chance to hit = 8/9.

2/3 + 1/3*2/3

No problems.
100/19/20 w/d/l
Tournament: 21/4/4 w/d/l

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PeterLageri
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Post#22 » Nov 15 2007 02:57

Why + ?

I would say:

To hit with one shot of a Missile Pod has 3/6 to succeded.
If hit the hit will have an effect on 4/6 since S7 against AV10 gives a glancing hit or better on 3,4,5 or 6.
Since the target is open topped, skimmer and moved fast all effective hit are downgraded to glacing hit and have +1 dice roll modifier. The goal is to destroy the target so in this case it is enough to have an immobilized result. That happend on 4+ due to moving fast, open topped skimmer. So 3/6.
In total 3/6 * 4/6 * 3/6 = 36/216 = 1/6

4 S7 shots would then have a 1/6 + 1/6 + 1/6 + 1/6 = 4/6 of at least one of them succeeding.

You are way ahead of me. Could you please explain your calculations in more detail?

Obviously I don't get it.

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onlainari
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Post#23 » Nov 15 2007 05:22

PeterLageri wrote:Obviously I don't get it.

(Were it not for this bit I quoted I probably would have been quite rude).

The deathrain has a ballistic skill of 4. The missile pod is twinlinked.

2/3 chance to hit with first roll. 1/3 chance to miss. Given a miss, 2/3 chance to hit.

2/3 + 1/3*2/3 = 8/9

8/9 chance to hit.

4/6 = 2/3
3/6 = 1/2

For example: Two deathrains shoot at an open topped AV10 skimmer that's moved fast.

Correct way to do it:

1 - {1 - [P(success)]}^# shots = 1 - {1 - [8/9*2/3*1/2]}^4 = 1 - [1- 8/27]^4 = 0.755

The mistake: Probability of killing the skimmer is not:

P(success) * # dice = (8/9*2/3*1/2) * 4 = 1.19 ~ this is wrong.

I hope it's clear now.
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PeterLageri
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Post#24 » Nov 15 2007 07:28

Yes, It's much clearer now.

Although it somewhat saddens me that you even considered to be rude.

All I try is to understand.............

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onlainari
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Post#25 » Nov 15 2007 09:57

PeterLageri wrote:Although it somewhat saddens me that you even considered to be rude.

PeterLageri wrote:4 S7 shots would then have a 1/6 + 1/6 + 1/6 + 1/6 = 4/6 of at least one of them succeeding.

Ok it's just that I specifically said in the first post that you do not add them together.

The mistake: Probability of killing the skimmer is not:

P(success) * # dice = (8/9*2/3*1/2) * 4 = 1.19 ~ this is wrong.

Which is what you did there.

Sorry for considering it though.
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Taipan
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Post#26 » Dec 02 2007 07:58

Question; When calculating the effects of twin-linking, do you multiply the original average hits by 1.5, or do it more complicated? Cos here is my breakdown of the boost you get by twin-linking;

Twin-linked MP at BS3; 2 x twin-linked shots

2 x 0.5 = 1

A re-roll with either succeed or fail, so its 50/50. So that you can just multiply by 1.5 the original hits, to represent the hits you already made (the 1) and the re-roll (the 0.5).

Hence; 1 hit becomes 1.5 when factoring in twin-linking.

Twin-linked MP at BS4 (Deathrain+ pattern); 2 x twin-linked shots


2 x 0.66 = 1.33 hits

Factoring in twin-link; 1.5 x 1.33 = 1.99

So, a Deathrain with a TA will almost always hit with both shots, on average.

Is that right?

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Post#27 » Dec 02 2007 09:11

You have 2 outcomes that lead to "hit" with twin-linking: hitting with the first dice, and missing on the first dice but hitting on the second.

The total probability to hit is thus the combination of these two: you hit with the first dice, or you miss with the first but hit with the second.

ie: probability to hit is (chance of hitting with 1st dice) + (chance of missing with 1st dice AND hitting with the second dice)

at BS4: the chance of hitting with the first dice is 2/3.
the chance of missing with the first dice is 1/3
the chance of hitting with the second dice is 2/3
the chance of missing with the first dice AND hitting with the second dice is (2/3 x 1/3) = 2/9
so you add the chance of hitting with the first dice (2/3) with the chance of missing with the first but hitting with the second (2/9)
2/3 + 2/9 = 8/9

An example at, say, BS2:
chance of hitting with first dice: 1/3
chance of missing with first dice (2/3) AND hitting with the second (1/3):
2/3 x 1/3 = 2/9
add the 2 together: 1/3 + 2/9 = 5/9

this is the probability of each hit at a given twin-linked BS of hitting. For an average number of hits, just multiply that by the number of shots.

ie: 8 gun drones each have a 5/9 chance of hitting. the squad in total will ON AVERAGE score 40/9 hits, or between 4 and 5.


On your example of a deathrain w/ TA, it is BS4. twin-linked BS4 has a 8/9 probability of hitting, so yes, it ALMOST always hits with both shots.

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Post#28 » Jan 30 2008 12:11

[[[ Sorry Onlainari, I have at best the mind for the basics, Multiplication/addition/subtraction/division, accounting, and basic geometry, When it comes to the mathemtaical abstraction stuff, like algebra/calc/trig, and any other ways of playing with numbers, i am very Ork-like. ;)

I have always posted my results through personal combat experience in battles and tournaments, and try as you may, an article like this as nicely done as it is merely makes my head hurt. :?

I will try to not get so angry at the math-hammer/mathhammer-probability posts that some of the guys are prone to make, but I am better at hands on, and fluff than math anyday! :P

I know that I have posted some stuff in past articles that seemed to make you upset, and after reading this I understand now, sorry if you took it like a directed attack on yourself, and or the same to any of the rest of the math heads. I hope at least this post helps clear up the issue on why I sometimes rebuff math hammer posts, it is not anything personal at you guys, it is my own issues in the fact that I have not gotten along with such mathematical model making ever. (HATED college semester algebra!)

This admittedly limits what kind of things I can do in RL as far as certain job improvements as well. art, model making, painting, writing, theses I have down in the hobby, what you guys do is like a complete foreign language to me that I cannot get my brain around.

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onlainari
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Post#29 » Jan 30 2008 01:37

(Ignoring the fact that I'm the guy giving the stats).

lenzabi wouldn't it just be useful for some guy to just give you the probability of killing 5 marines with a fish of fury? You don't have to work it out yourself, you just look at a list like this:

0 - 5.92%
1 - 17.76%
2 - 25.53%
3 - 23.40%
4 - 15.36%
5+ - 12.03%

And go "oh that's good to know".

All I'm trying to say in the article is that if you can do the stats, do them, and if you can't, understand them when someone does them for you.
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Post#30 » Jan 30 2008 01:40

[[[ Call me dense, but I will unload 31 shots, hit with about 15-20, and wound with say about 8-12. then they have to hope they save on those so as to not be all dead. they may save about half, or 2/3. that is the best I can do as far as the figures go. so, yeah, as long as i can max out the dice rolls I guess the stats will work. ;)

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Darkseer
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Re: Mathhammer - The guide

Post#31 » Dec 11 2008 04:58

This blog has surfaced recently and has some really good info about Mathhammer in Warhammer 40K.
Warhammer Tau My Tau Blog

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Ray
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Re: Mathhammer - The guide

Post#32 » Dec 11 2008 09:56

While understanding that this thread is over a year old, this is still the right place to post that link, IMO. Thanks for it too.

I'm not sure on what the distinction is between natural rolling and practiced rolling that the writer specified but I believe it has to do with large sample pools and small ones.

small pools (under 5ish dice) tend to show skewed results on a roll-by-roll basis. This is because there is a greater chance that a result not fitting within the standard distribution will occur.

Example:
Your Firewarriors take 5 hits and want to continue to breath. To pass your armour saves you need to roll 4, 5 or 6. Here's what you get:
1, 2, 6, 3, 2, 3

So while your firewarriors run away like sissies, you wonder to yourself, "how could this happen? The average roll is 4 and half of them (2.5) should be fine!"

"Well," says yourself...or some other voice in your head, "If you keep rolling those dice up to a million times..you will reach values which are statistically correct."

"Thanks self," I say.


A lot of people get hung up on expecting their dice to be on average every single time. This is just not the case with any random result. If we were to only sample 5 members of population too determine the extent of a disease, we would most likely be totally wrong in our findings.

Also, I apologize for the slightly off-kilter dialogue. I have an exam tonight for biochem and I guess the studying has made me crazy.
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Soji
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Re: Mathhammer - The guide

Post#33 » Dec 11 2008 10:43

Practiced Rolling is cheating by controlling the dice roll. Depending on the way you hold the dicesand throw them, you can force a result to appear, or even easier, stop a result from appearing. At the very least increasing/decreasing the probability of a given result. A good cheater can pass a 4+ save 100% of the time, and even a beginner can pass a 2+ save over 95% of the time.

Natural Rolling is an acquired movement, that some players get after throwing lot of dices, and repeating what they did when they got good results. "Ho, I got a 6 ! I'm going to throw it the same way !" If you do it often enough, you start getting more than 1/6 of 6. Some people who do it aren't really aware they are cheating, they're simply being superstitious.
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synchronicity
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Re: Mathhammer - The guide

Post#34 » Dec 11 2008 10:08

Also, for anyone interested (if your self conscious about your rolling like myself), this study examines the "Games Workshop" and "Chessex" dice to roll ones more than average probability says they should.

Now, I'm not saying anything stated in this article is fact, but it IS interesting to see the difference between large pools of the rounded, pipped "Chessex" dice compared to sharp-edged "Vegas" dice.

In the end, one could argue it's not a big deal, since a good majority of Warhammer players use the same type of dice (rounded edges) anyway. So we're all at a disadvantage! :P
The Tau aren't Communists, they're simply Utilitarians.

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Soji
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Re: Mathhammer - The guide

Post#35 » Dec 12 2008 09:05

Well, it skew the results and consequently make all mathammer results incorrect. The probability of failing a Terminator save is not 17% but 23%, so You don't need 6 wounds from firewwarriors to kill them, but 4/5. That mans that AP2 is not as good as its supposed to be. That is very interesting.
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Kies'el
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Re: Mathhammer - The guide

Post#36 » Dec 14 2008 10:34

That study was potentially flawed on at least one account.

first fact: their corespondency with the ceasar's palace casino stated that the way the dice were rolled was a deciding factor.
second fact: their machine did NOT roll dice in the same manner that they are rolled on the tabletop. (they put each die in a box on a hydraulic table)

as further evidence, I just conducted a limited experiment with a brand new box of chessex dice i just opened.
I first rolled the entire pile several times, setting aside each dice that rolled a 6 or a 1 into two separate piles untill all dice had been separated.
(I was carefull to not look at the dice when rolling, and shook them to randomize the faces so that I wouldnt be controlling the results.)
I then rolled each pile in the same manner, creating two more piles for each.
Piles where the dice had rolled the same number of 6s as opposed to 1s were combined (creating 3 piles, one where the dice had rolled 2 6s, one where they had rolled 1 of each, and one pile where they had rolled 2 1s)
I then repeated this process 17 times, and ended up with this distribution. (the highest ratiof of 6s:1s was 13:4 and the lowest was 4:13)

#of sixes: 13 - 12 - 11 - 10 - 09 - 08 - 07 - 06 - 05 - 04 - 03 - 02 - 01 - 00 (the + shows the expected mean)
# of dice: - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 19 + 17 - - n = 1
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 08 - 22 - 06 - - n = 2
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 04 - 16 + 13 - 03 - - n = 3
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 03 - 10 - 13 - 09 - 01 - - n = 4
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 02 - 05 - 13 + 11 - 04 - 01 - - n = 5
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 05 - 10 - 10 - 18 - 03 - - - - n = 6
- - - - - - - - - - - - - - - - - - - - - - - - - - - - 01- 08 - 12 + 07 - 08 - - - - - - n = 7
- - - - - - - - - - - - - - - - - - - - - - - - - 01 - 05 - 10 - 09 - 08 - 03 - - - - - - n = 8
- - - - - - - - - - - - - - - - - - - - - - 01 - 03 - 06 - 10 + 08 - 06 - 02 - - - - - - n = 9
- - - - - - - - - - - - - - - - - - - 01 - 01 - 04 - 11 - 11 - 05 - 05 - - - - - - - - - n = 10
- - - - - - - - - - - - - - - - - - - 01 - 02 - 10 - 10 + 06 - 04 - 03 - - - - - - - - - n = 11
- - - - - - - - - - - - - - - - 01 - 02 - 05 - 08 - 12 - 01 - 05 - 02 - - - - - - - - - n = 12
- - - - - - - - - - - - - - - - 02 - 04 - 05 - 11 + 07 - 03 - 02 - 02 - - - - - - - - - n = 13
- - - - - - - - - - - - - 02 - 02 - 05 - 07 - 08 - 06 - 04 - - - - 02 - - - - - - - - - n= 14
- - - - - - - 00 - 01 - 01 - 04 - 07 - 10 + 04 - 06 - 01 - 01 - 01 - - - - - - - - - n = 15
- - - - - - - 01 - 01 - 03 - 04 - 09 - 07 - 04 - 05 - - - - 02 - - - - - - - - - - - - n = 16
# of dice: 02 - 01 - 05 - 05 - 11 + 03 - 05 - 02 - 01 - 01 - - - - - - - - - - - - n = 17

the curve looks like this
3.2.1..0.9.|.8..7.6.5.4
O O O O X | O O O O O
O O O O X | O O O O O
O O O O X | O O O O O
O O O O X | O O O O O
O O O O X | O O O O O
O O O O X | O O O O O
O O X X X | O X O O O
O O X X X | O X O O O
O O X X X | X X X O O
X O X X X | X X X O O
X X X X X | X X X X X

the number of dice that rolled 6s to 1s at a ratio of 9:8 or higher was 24
the number of dice that rolled 1s to 6s at a ratio of 9:8 or higher was 12
obviously, something is going on here. (no wonder I seem to do so well with the CIB :P )

now I don't know about you, but that pretty conclusively backs up the "intuitive" guess that the heavier side 1 pip side of the dice will end up on the bottom more than the lighter 6 pip side.
36 dice rolled 17 times(not including 2-5 results) isn't the largest sample size in the world, but even so, a 2:1 ratio is a bit much to ignore.

either there is something different about my dice, or there was something wrong with their dice rolling machine.
I would be willing to bet money that their machine was inducing some kind of abberation in the results by rolling the dice the same way every time.
Eventually a pattern would have set in, where the exact same rolling motion would keep looping back to the 1 result at a set frequency. switching to the different syle of las vegas dice may have reduced the frequency below that of the ability of the study to detect, or eliminated it all together if it was being caused by the pips. (sharpening the corners of the GW dice reduced it some but not entirely.)
Essentially, while the vegas dice may still be more random, the 1 may not necessarily be true most common result rolled on the chessex

(please note however, that my test only checked the the ratio of 6 results to 1 results. My dice are not ridiculous magic dice that roll only sixes and ones. each time I rolled a pile, I removed any 1s and 6s and set them aside, and then re-rolled the rest. All my dice came from the same chessex box, and were MagmaTM w/blk Vortex DiceTM 12mm d6)

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