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Misfratz
16-01-2008, 20:28
After seeing repeated abuses of mathematics similar to:

Orks with Shootas are effectively BS4 with a single shotI felt compelled to take remedial action.

I wrote a short program to simulate the "rolling to hit" for ten Orks with Shootas (at BS2), and ten "Orks are effectively BS4.." - producing the below graph.

http://i245.photobucket.com/albums/gg46/gwmisfratz/dakka.png

So, yeah, obviously the means are the same - but there's so much more to life than the average! Look at those tails!

In my opinion this is an example of GW getting things right. With BS2 and weapons with a high rate of fire, Orks have a much wider distribution. By contrast, Marines, Necrons, etc, have a higher BS, but normally fewer shots, giving them a less variable distribution - making them more reliable.

catbarf
16-01-2008, 20:51
Very good post, easy to understand and very concise without being wordy. A++. Brace for imminent 'Maths means nothing lolz' posts.

I think it's very flavorful and fluffy for Orks. I'm just not sure GW did that... they have a history of ignoring the maths when designing codexes. But hey, regardless of the reason, the result is the same: Orks are random. And that is alright with me.

Edit: One other thing to note is that while the averages are the same, more often will the Orks get less hits than Bolters- the peaks for the Shootas are lower, but this is compensated for by a higher possible hit number.

Imperialis_Dominatus
16-01-2008, 21:28
Math means nothing. Lolz. I once took down a Falcon with one lasgun shot, yadda yadda yadda....

Just kidding. Good post, I'd never put that much work just for the benefit of the questionably honorable audience of Warseer.

Mangobreeder
16-01-2008, 22:42
if you do a million iterations of the above example you get the following results
unfortunately i do not have any space to host pics if some one could plot the results in excel and post a jpeg id be grateful (im looking at you misfratz).

the results are also very interesting :) good work misfratz

# Hits ~hits at BS2 ~hits at BS3
0 11606 110
1 58108 1613
2 136944 10507
3 204690 42329
4 218042 111225
5 174624 200190
6 109161 251548
7 54952 215255
8 21959 120406
9 7393 40767
10 2015 6050
11 421 0
12 71 0
13 13 0
14 1 0
15 0 0
16 0 0
17 0 0
18 0 0
19 0 0
20 0 0
21 0 0


Comma seperate for ease of use

# Hits,~hits at BS2,~hits at BS3
0,11606,110
1,58108,1613
2,136944,10507
3,204690,42329
4,218042,111225
5,174624,200190
6,109161,251548
7,54952,215255
8,21959,120406
9,7393,40767
10,2015,6050
11,421,0
12,71,0
13,13,0
14,1,0
15,0,0
16,0,0
17,0,0
18,0,0
19,0,0
20,0,0
21,0,0

GrandReaper
17-01-2008, 00:09
The graph seems a little excessive. BS4 hits 2/3 of the time, BS2 hits 1/3 of the time (ie: half as often). Thus, with twice as many shots for BS2 the means are identical (2/3 of a hit per ork per turn of shooting).

On the other hand, twice as many shots will be guaranteed a wider distribution, and thus more variance, just because more values are possible.

It's a pretty graph tho!

Maxis Lithium
17-01-2008, 00:49
When looking at BS values and number of shot, I have always felt that BS2 with a re-roll was basically the same effect as BS3. The feefctivnett of an Ork with a single shot shoota (from the old book) was just as effective against a T4 target as a Guardman with a Lasgun. the difference came when you took the Lasgun out of the Guardsman's hands and gave him a Rocket Lancher. THEN you saw better fireworks.

With 2 shots at BS2, I was bassically putting them as affective as a singel shot with BS3, on avaerage. As usual, Ork shooting was most effective in numbers. Quantity over quality of guns is what you wanted. An ork with a Rokkit Launcha, against a T4 target, was less valuable then a bunch more with shootas.

This does nicely to confirm my feelings in regards to Ork Shooting.

GO LOGICS!

Chaos and Evil
17-01-2008, 01:21
On the other hand, twice as many shots will be guaranteed a wider distribution, and thus more variance, just because more values are possible.


Actually, with a higher sample size, data will tend towards the average.

Atypical results are far more common with smaller sample sizes; It is the Orks who are more reliable, BS4, single shot models (Marines) are going to present with atypical results far more commonly than our high volume of shots Ork boyz.


That's pretty much the first rule of statistics...

sabre4190
17-01-2008, 01:29
While I do agree with your math, remember that they effectively an assault 1 unit. When compared to other races, this does put them behind even though they are cheap.

Nevertheless, you do raise an excellent point. If you factor in sheer numbers, orks can shoot well. This is great work, and it is kind of fun to look at considering im in a statistics class right now!

Kirasu
17-01-2008, 01:52
I dont know.. The math is right but this only matters between 13-18" as marines get the same # of shots when both are within 12"

starlight
17-01-2008, 02:19
But then you have to factor in Assaulting after Shooting, giving Orks another potential advantage (depending on who gets the Charge).

19-24" - Marines get free shots
13-18" - Orks get two shots to Marines one
1-12" - Both get two shots, but Orks can Assault after shooting

Over several turns.....

onnotangu
17-01-2008, 03:03
down side is..Orks can't survive the return fire very well. with the low armour saves. sure they are going to get in range but there will not be many left.

starlight
17-01-2008, 03:22
But outnumbering Marines 2.5:1 allows them to survive that. ;)

The Gothic Me
17-01-2008, 04:14
The poor old guardsmen don't have much going for them here, the Orks have them beat in shooting and CC.

Misfratz
17-01-2008, 18:57
Actually, with a higher sample size, data will tend towards the average.

Atypical results are far more common with smaller sample sizes; It is the Orks who are more reliable, BS4, single shot models (Marines) are going to present with atypical results far more commonly than our high volume of shots Ork boyz.


That's pretty much the first rule of statistics...

Well, but my graph shows fatter tails for the Orks, with a sample size of twenty (BS2) compared to a sample size of ten (BS4). So, you're wrong.

Now if you were to compare a sample of twenty Orks shooting (forty shots) with ten Orks shooting, you would find that the larger sample would be closer to the mean (The mean would also be twice as large but you could divide all the answers by two - ie normalise them - to compare the size of the spread directly). So you're right.*

The point is that two shots at BS2 is not equivalent to one shot at BS4. They are like apples and oranges (or Orks and Marines). You can't compare them in such a simplistic way.

*In a relative sense (relative to the number of shots fired) the spread is smaller for the Orks, but it is the absolute spread that is of more concern, which is clearly wider foir the Orks.

Mangobreeder
17-01-2008, 19:31
removed dodgy stats

Misfratz
17-01-2008, 20:45
My dear friend mangobreeder, you need to debug your code. You're firing too many shots at BS4 - you can't get 11 hits with 10 shots!

My results were stable to repeat iterations.

It's good to see the effort though!

hucklongfin
17-01-2008, 20:52
Never tell me the odds... Han Solo.

Mangobreeder
17-01-2008, 20:57
corrected. BS4 is clearly superior

http://i268.photobucket.com/albums/jj13/mangobreeder/untitled1a.jpg

catbarf
17-01-2008, 22:06
Are you factoring in that BS2 gets twice as many shots? Your chart is way off, the averages are the same yet BS4 is superior in every regard on your data.

Mangobreeder
17-01-2008, 22:24
yes i am

I took 1,000,000 sets of 20 shots. then counted the number of 5 & 6's.
the chart shows how many 5 or 6 there were in each set of 20.

I the took 1000000 sets of 10 shots, then counted the number of 3,4,5,6
the chart shows how many 3,4 ,5 or 6 there were in each set of 10.

Grogugluk
17-01-2008, 22:49
That doesn't quite work Mangobreeder.


According to your chart out of 20 shots with BS 2 the orks never hit more than 10 times out of a million attempts... seems kinda fishy.

Mangobreeder
17-01-2008, 22:56
there was some hit, im simply ommited them becase they would not show on the graph
the data table is here

http://i268.photobucket.com/albums/jj13/mangobreeder/untitled3a.jpg

the probability of getting 20hits at BS2 is 0.00000008603916%
even the probability of getting 14 hits at bs2 is 0.00006272254744%
and thats the highest result

the probability of getting 10 hits at bs4 is 2.6%

catbarf
17-01-2008, 23:57
Only one roll of 14 for one million rolls seems quite fishy. By my calculations, there should have been quite a few more.

Kantur
18-01-2008, 00:05
Ahh, an insight into Warseer it seems...If there's something wrong or seemingly wrong with your figures, don't prepare for "Maths means nothing!", prepare for "Are you sure about..." until it's good enough for "Maths means nothing!" ;)

Mangobreeder
18-01-2008, 00:22
Only one roll of 14 for one million rolls seems quite fishy. By my calculations, there should have been quite a few more.

im suprised there was 1

if you look at the math /me prepares to be shot down

odd to hit at BS 2 = 2/6

we simplify this down to 1/3

then get the probability of getting 14hits

you need

1/3 * 1/3 * 1/3 * 1/3........so on 14 times

that gives you 1 in 4,782,969

3^14 = 4,782,969

so im surprised there was even 1 set of 14 hits to be honest

going to bed now ill take this up again in the morning

catbarf
18-01-2008, 00:57
im suprised there was 1

if you look at the math /me prepares to be shot down

odd to hit at BS 2 = 2/6

we simplify this down to 1/3

then get the probability of getting 14hits

you need

1/3 * 1/3 * 1/3 * 1/3........so on 14 times

that gives you 1 in 4,782,969

3^14 = 4,782,969

so im surprised there was even 1 set of 14 hits to be honest

going to bed now ill take this up again in the morning

Out of 20 shots. The math you appear to have done is the chance of rolling fourteen dice and every single one hitting. That could explain why it seems so weird.

Mangobreeder
18-01-2008, 01:00
no the guy above comments on he was suprised that 14 hits only came up once, i was commenting on the probability that 14 hits would come up. if you do 20 shots hitting it goes into the billions.

my math is correct :)

catbarf
18-01-2008, 01:13
no the guy above comments on he was suprised that 14 hits only came up once, i was commenting on the probability that 14 hits would come up. if you do 20 shots hitting it goes into the billions.

my math is correct :)

No, it is not. The math you presented was for the chance of getting 14 hits out of 14 dice. This is incorrect, you are calculating 14 hits out of 20 dice.

starlight
18-01-2008, 01:22
I don't even know advanced math, but I'm not accepting that the chance of getting 15, 16, 17, 18, 19 *or* 20 combined is basically zero...:(

The chance may not be high, but it's not zero (or zero as represented here). The simple reality is that there should be more *curve* to the graph for randomly generated numbers with fixed limits (ie dice).

CommisarMolotov
18-01-2008, 02:05
RAAAAAAGH! Skull HURTS! HULK SMASH!!!

Kymar
18-01-2008, 02:37
RAAAAAAGH! Skull HURTS! HULK SMASH!!!

:wtf: agreed.

As for the debate over 14+ hits possible on BS2...

Sadly the good Mangobreeder ONLY rolled 1 million dice sets. :p

As he showed, the chance of getting 14 out of 20 BS2 hits is 1 in 4,782,969.

Though as many people who play the lotto will still not believe:

1 million tries
at a 1 in 4.7 million chance
DOES NOT EQUAL 1 in 4.7 odds of a success!

Its still a 1 in 4.7 million chance for each and every set of rolls.

Even if Mangobreeder had set his tests for 4.7 million sets of rolls, the likelihood of achieving even ONE set of rolls where 14 of the 20 BS2 shots hit, is still very small.


PS- Good work Mangobreeder. :)

Mangobreeder
18-01-2008, 07:39
if you interested the odds of rolling 20 hits are....

1 IN 3,486,784,401

to put in simply that roughly the odds of winning the lottery 250 times

the odds are not impossible.

tell you what. after work (work in Tax) and after karate. ill rerun for 20million rolls and lets see what happens.

i getting excited.

bosstroll
18-01-2008, 08:13
My Math-Fu is better then your Math-Fu ;)

bdo
18-01-2008, 08:32
quite interesting... not really exciting but interesting :)

now i know what i can say to my mate who is about to start orks when he hits with 20 of 20 shots... with his luck heŽd couldŽve won the lottery 250 times... i think he owns me money :skull::eek:

Mangobreeder
18-01-2008, 09:54
PS- Good work Mangobreeder. :)


credit should goto misfratz, it was his idea and inspiration which icked this all off

i cant wait to get home and work on it how sad it that :)

sadly im doing a SX1 return at the moment

Sir_Turalyon
18-01-2008, 11:11
@ Mangobreeder: you only calculated chance of hiting with 14 of 14 dice. Probability of 14 succeses in 20 Bernoulli trials with probability of success of (1/3) is not (1/3)^14, but

(20!/(14!*6!)) * (1/3)^14 * (2/3)^6. [chance that your 14 dice hit * chance other 6 will miss * number of ways you can pick 14 dice out of 20 ]

It simplifies to (4*17*3*19*10) * (2/3)^6 * (1/3)^14, then to (4*17*19*10) * 64 / 3^5 * [1/3^14] = 826880 / 243 * [1/3^14] = 3402,798 * [1/3^14].

So actual chance of 14 out of 20 hits is 3402 times greater then your calculations showed... more like 1/1406 then 1/4782969 .

Also, rolling milion of pseudo-random virtual dice (I assume you did not roll them in person, but made your computer do it) is not a good way of testing probabilities. There are formulas for such things...

catbarf
18-01-2008, 11:20
I just did the math, and the odds of rolling 14 are .9%. With one million rolls, you should have rolled far more than one. In fact, my Java program rolled 914 of them out of one million rolls. Coupled with the small but still present possibility of rolling higher than 14, there should have been at least some.


@ Mangobreeder: you only calculated chance of hiting with 14 of 14 dice.

I said this, he seems to have ignored me.

marv335
18-01-2008, 11:56
meh.
Lies, Damned Lies, and Statistics.

I actually saw someone roll for a unit of 15 slugga boys and all hit, wiping out a 6 man marine unit (really upset the ork player, he planned to assault, and sweep into another unit)

Mangobreeder
18-01-2008, 12:10
@ Mangobreeder: you only calculated chance of hiting with 14 of 14 dice. Probability of 14 succeses in 20 Bernoulli trials with probability of success of (1/3) is not (1/3)^14, but

(20!/(14!*6!)) * (1/3)^14 * (2/3)^6. [chance that your 14 dice hit * chance other 6 will miss * number of ways you can pick 14 dice out of 20 ]

It simplifies to (4*17*3*19*10) * (2/3)^6 * (1/3)^14, then to (4*17*19*10) * 64 / 3^5 * [1/3^14] = 826880 / 243 * [1/3^14] = 3402,798 * [1/3^14].

So actual chance of 14 out of 20 hits is 3402 times greater then your calculations showed... more like 1/1406 then 1/4782969 .

Also, rolling milion of pseudo-random virtual dice (I assume you did not roll them in person, but made your computer do it) is not a good way of testing probabilities. There are formulas for such things...


I stand Corrected. do you got any links to where can find out more

Imperialis_Dominatus
18-01-2008, 15:00
meh.
Lies, Damned Lies, and Statistics.

I actually saw someone roll for a unit of 15 slugga boys and all hit, wiping out a 6 man marine unit (really upset the ork player, he planned to assault, and sweep into another unit)

Lesson: Don't always shoot. Sometimes I deliberately hold back those two meltas and eight bolt pistols just so I can see that Powerfist at work... if the enemy survives 27 meelee attacks before that. :D


RAAAAAAGH! Skull HURTS! HULK SMASH!!!

Indeed.

Sir_Turalyon
18-01-2008, 15:30
I stand Corrected. do you got any links to where can find out more

I took formula straight from probability handbook (in polish, polish authors, but there should be book titled "Introduction to probabiity theory" in every langueage ). These slides seem to cover basics of Bernoulli trials:

http://www.stat.psu.edu/~resources/ClassNotes/hrm_10/sld001.htm

in particular on fourth page is formula I used:

http://www.stat.psu.edu/~resources/ClassNotes/hrm_10/sld004.htm

There is wiki on Bernoulli trials, but I didn't found formula for k successes in n attempts there.
http://en.wikipedia.org/wiki/Bernoulli_trial

catbarf
18-01-2008, 19:48
meh.
Lies, Damned Lies, and Statistics.

I actually saw someone roll for a unit of 15 slugga boys and all hit, wiping out a 6 man marine unit (really upset the ork player, he planned to assault, and sweep into another unit)

So what? An unlikely result doesn't break the model. Do you really think we're trying to tell you exactly what you'll roll?

dabiggrotsboss
18-01-2008, 23:00
As if an ork could understand a chart.

Louda, fasta, betta. Enough said.

WAAAGH!

floyd pinkerton
18-01-2008, 23:07
@dabiggrotsboss- hey! that's mine! nice avatar:D

hmm, this is making me consider orks even more

dabiggrotsboss
19-01-2008, 00:44
@dabiggrotsboss- hey! that's mine! nice avatar:D

hmm, this is making me consider orks even more


Well, I have a Yorkshire terrier, how coincidental is that!


Go WAAAGH!

susu.exp
19-01-2008, 02:50
Actually, with a higher sample size, data will tend towards the average.

Atypical results are far more common with smaller sample sizes; It is the Orks who are more reliable, BS4, single shot models (Marines) are going to present with atypical results far more commonly than our high volume of shots Ork boyz.

That's pretty much the first rule of statistics...

Ah, the law of big numbers and Orks... Well, not quite. Since everybody has supplied Monte Carlo tests, which show that this is incorrect, IŽll add the probability theoretical backing.

Shooting in 40k follows binominal distributions. For binominal distributions the Expected value
E = n * p
where n is the number of trials and p the probability of a successful trial.
The Standard Deviation
SD = (n*p*(1-p))^0.5
Now the law of Big numbers basically says, that as n->infinite, SD/E -> 0
SD/E = [(1-p)/(n*p)]^0.5 = 1/n^0.5 * constant
However in the example, weŽve taken precautions to keep E constant (2n*1/2p).
If E is kept constant, we can replace p with E/n and get:

SD= (n*E/n*(1-E/n))^0.5 = (E*(1-E/n))^0.5 = (E-EČ/n)^0.5
And
SD/E=(1/E-1/n)^0.5
Here for n->infinite SD/E -> (1/E)^0.5
In other words: If we keep E constant and vary p, the law of big numbers does not apply!

The first rule of statistics requires a constant p.

starlight
19-01-2008, 03:03
...and all that means....? :eyebrows:

dabiggrotsboss
19-01-2008, 03:37
...and all that means....? :eyebrows:

Exactly. Nada. Not-a-ting. Nothing.

It's all a whole lotta nothing.

More WAAAGH! Less complicated formula.

susu.exp
19-01-2008, 03:43
IŽll give you an example.
30 Space Marines shoot. (n=30)
2/3 of them hit (p=2/3)
Then the expected number of hits is
E = P * n = 2/3 * 30 =20
The standard deviation is the expected spread. Roughly 2/3 of your results will fall between E+SD and E-SD.
In this case SD = (n*p*(1-p))^0.5 = (30*2/3*1/3)^0.5 = (20/3)^0.5 = ~2.6
So 30 marines will usually hit between 17 and 23 times.
If you divide SD/E you get the deviation as a percentage. They spread around the mean by +-13%.
If you increase the numbers of shots, this percentage will decrease:
60 marines:
E = 40
SD = (1/3*2/3*60)^0.5 = 3.0
Between 37 and 43
SD/E = 7.5%
300 Marines:
E = 200
SD =(1/3*2/3*300)^0.5 = 8.2
Between 192 and 208
SD/E = 2.7%

If infinitely many marines fired SD/E would be 0.

Alternatively we can keep E constant letŽs say we want 100 hits.
We shoot 600 times at 6+
n = 600
p = 1/6
E = 100
SD = (1/6*5/6*600)^0.5 = 9.1
SD/E=9.1%
At 5+ we need half the number of shots:
n=300
p=1/3
E=100
SD= (1/3*2/3*300)^0.5 = 8.1
SD/E=8.1%
Less shots, but the spread also decreased.
on 4+
n=200
p=0.5
E=100
SD= (1/2*1/2*200)^0.5 =7.1
SD/E=7.1%
On 3+
n=150
p=2/3
E=100
SD = (1/3*2/3*150)^0.5 = 5.8
SD/E = 5.8%

So in this case the spread becomes bigger the more shots are fired. People always remember the law of big numbers: The spread gets smaller as you take more shots. They forget the other bit: The spread gets bigger if the chances for a single success decreases. And it turns out that this effect is larger than the numbers effect.

Sir_Turalyon
19-01-2008, 11:38
...and all that means....?

That larger number of shots fired by Orks does not make them more likely to score average (1/3 of shots fired) number of hits, but more likely to score some really odd results, like massive hits or misses?

In other words: massed Orks shootas are random, you just pull trigger, shout WAAAGH and see what happens rather then expect what'll happen before pulling the trigger?

I'll have to take Maple and calculate actual probabilities of set numbers of hits...