Welcome to Librarium Online!
Join our community of 80,000+ members and take part in the number one resource for Warhammer and Warhammer 40K discussion!
Registering gives you full access to take part in discussions, upload pictures, contact other members and search everything!
Move! move! move! analysis
i have just build a quick empirical study (after failing to get a mathematical formula, i admit ) about the effectiveness of this order, and the expected outcome of this order once it has been issued.
sample size: 500.000 triplets of d6
experimental results
max value
(x)events
(n)probability
p(x)=n/sizex * p(x) 1 2253 0.004506 0.004506 2 16248 0.032496 0.064992 3 43656 0.087312 0.261936 4 85718 0.171436 0.685744 5 141297 0.282594 1.41297 6 210828 0.421656 2.529936
Expected value: 4.96
(sum of: x * p(x) )
(this means that on average you will move 5" while running using this order)
Standard deviation:
2.2468
this value is the average distance of the dice result from the expected value.
this means that, on average, your run! distance will be between 2.72" (3") and 7.2" (of course the max is 6" but this means that the dices will likely be on 6" rather than 3")
Comparison with a standard running phase without orders
If you choose to run, you have to move d6" instead of (3d6 take the highest)"Expected value: 3.5"
x p(x) x*p(x) 1 1/6 1/6 2 1/6 2/6 3 1/6 3/6 4 1/6 4/6 5 1/6 5/6 6 1/6 1
std deviation: 1.7078
so, on average, your run phase without any order will be between 1.8" (2") and 5.2" (5")
Conclusion:
It won't happen often that you will need to run for your lives with imperial guard, since you'll either be in a ride or well entrenched, but, if you need that last rush for whichever reason, at least now mathammer gave you a rough extimation on how far you can expect to run.
Running on your own is really aleatory, you might achieve the sprint of your life or even stumble on your feet.
Running when you are ordered to, instead, can be quite satisfactory. More often than not you will run the whole 6" and 7 times out of 10 you'll run at least 5".
remember that if you run you can't shoot or assault this turn! (if you havent't the Fleet special rule, at least)
PS: if anybody can help me gathering a correct formula, i'd appreciate that. I have problems about events being counted more than once.
Thank you!
Last edited by Frost89; August 29th, 2011 at 12:51.
I don't use Vendettas or Valks...if Guardsmen were meant to fly, the Emperor would have given them wings. (343rd Mordian)
The battle for Hidaxes SubSector (Listwar)! 184th Cadian blog: 184th.blogspot.com
Just off hand the standard div is blatantly off, you should get a smaller deviation when using more dice cos it'll normalise the result and make the outcomes less spread out.
So I checked the 1st (massive) dataset and got this:
stddiv1.jpg
Check you did your sums right:
Calculating Variance and Standard Deviation - YouTube
It's been about a decade since I did any stats but I'll think on how to work out the real probabilities and come back to you. It should be possible. Maybe with factorals.
Copy, Improve, Innovate
I use Move move move only really when i feel that i NEED to get somewhere, i knew i'd get an average of about 5" running, so basically if i'm out in the open just from an exploded chimera, I can't really do anything and i need to get to some cover or to run to the last objective just so my melta/plasmavets don't die.
@Korona : i have used this formula to get std deviation 975337690693448192c8f1365b7982d4.png and therefore i did the following:
(1- 4.96)^2 + (2-4.96)^2 +....
then: /6
and then sqrt... where is the error?
I don't use Vendettas or Valks...if Guardsmen were meant to fly, the Emperor would have given them wings. (343rd Mordian)
The battle for Hidaxes SubSector (Listwar)! 184th Cadian blog: 184th.blogspot.com
edit:
That formula looks off, check the vid I posted. Maybe it's just condensing things but f.e. the denominator (what you divide by) should be N(N-1) not just plain N. Since your results are off I think that might be the issue.
Anyway for the rest - see the next post
Last edited by Korona; August 30th, 2011 at 23:18.
Copy, Improve, Innovate
OK I've cracked it.
1st thing I forgot to mention - the SD is a numerical representation of how spread out your results are. It's not a direct number. You need to multiply the average score by it and it'll give you the max outlier. Subtract the difference from the average and you get the min outlier.
Anyway
You wanted a general formula to work out probabilities. Here it is:
(1/216) + (3* ((n-1)/6) * (1/6) * (1/6))) + (3*((n-1)/6) * (n-1)/6) * (1/6))) = P
where n = the dice score in question and P is the probability that it is the high score.
Now I'm not so cruel as to post that without giving a why too so here it is:
If you think of the ways you can get a set value, there are 3 basic ways you can get it.
All 3 dice are the exact value
2 dice are the exact value and 1 is lower
1 die is the exact value and 2 are lower
Anything else gives you a bust.
So we can represent the ways you can get each like this:
Exact = E
Lower = L
E E E
L E E
E L E
E E L
L L E
L E L
E L L
The odds of getting that value is the sum of all 7 possible combinations
Ok so let's take 3 as an example -
To find the odds for a move of 3 inches we need to get the odds for each of the possible ways we can make a high score of 3:
There's a 1/6 chance of rolling a 3 so E E E is 1/6 * 1/6 * 1/6 or 1/216
There's a 2/6 chance or rolling less than 3 (2 or 1) so
L E E 2/6 * 1/6 * 1/6
E L E 1/6 * 2/6 * 1/6
E E L 1/6 * 1/6 * 2/6
L L E 2/6 * 2/6 * 1/6
L E L 2/6 * 1/6 * 2/6
E L L 1/6 * 2/6 * 2/6
Now it doesn't matter which order you multiply probabilities (or any number) so you can write the 3x combos for each block of results as 3 identical sums:
(2/6) * (1/6) * (1/6)
(2/6) * (1/6) * (1/6)
(2/6) * (1/6) * (1/6)
and
(2/6) * (2/6) * (1/6)
(2/6) * (2/6) * (1/6)
(2/6) * (2/6) * (1/6)
we can multiply the generic sum by 3 rather than working each identical sum out individually and adding them at the end. That gives us the following format for each block as a whole:
3* (2/6) * (1/6) * (1/6)
and
3* (2/6) * (2/6) * (1/6)
Combining the whole thing and the total sum looks like:
(1/216) + (3* (2/6) * (1/6) * (1/6)) + (3* (2/6) * (2/6) * (1/6)) or 19/216
For a generic formula we can say the odds of rolling any score is still 1/6 so that is the same for any number (N)
We need to adjust the odds of rolling less than N though, since dice scores are whole numbers we can do that by taking N-1/6
Swapping in that generic value we get the generic formula from above:
(1/216) + (3* ((n-1)/6) * (1/6) * (1/6))) + (3*((n-1)/6) * (n-1)/6) * (1/6))) = P
We can check this is true by working out 6s next. We can take a shortcut with 6 because we can rephrase "what is the chance of getting at least one 6?" as "what is the chance of getting no 6s?" and subtract that probability from 1
The odds of getting no 6s is 5/6 ^ 3 = 0.578 1- 0.578 is 0.421. Multiply that by 216 and you get 91 so the odds of getting a 6 is 91/216
Now using our generic formula we get
E E E 1/216 0.00462
+
L E E (5/6 * 1/6 * 1/6) * 3 0.069
+
L L E (5/6 * 5/6 * 1/6) * 3 0.347
=
0.421 or 91/216
So we know the formula is right.
Using it we get the following values for the numbers:
1 = 1/216 0.004629
2 = 7/216 0.03240
3 = 19/216 0.08796
4 = 37/216 0.1712
5 = 61/216 0.2824
6 = 91/216 0.4212
As a final check we can add all the values to make sure we end up with 216/216 which we do.
Ooookay so now we have the true probabilities we can feed them into the standard deviation formula for probability distributions.
SUMMARY
- aka TL: DR version starts here:
I didn't know how to do that so I applied the method in this vid:
Standard Deviation Discrete Probability Distribution.mp4 - YouTube
and got near identical results:
standarddiv2.jpg
Sooo you can see, trial and error with a big data set gets you pretty close but I think it's good to do it the "proper" way.
Compare these results to a run move on a single die:
standarddiv3.jpg
#1 - The average score is higher, 4.9 vs 3.5
#2 - The standard deviation is MUCH lower, 1.14 vs 1.7 (I think a standard deviation of 1 means you always get the average so this is way way way more realiable)
You can really see that from the odds: You've got great odds of rolling 4-6 inches with the move move move order. It's surprising but adding a couple of dice makes the odds of rolling high MUCH more likely given you can ignore the bad ones. In fact the odds of rolling 3 or less is just 27/217 or 12.5% while the odds of rolling a max score of 6 is 42%!!. That should surprise you, at least - it certainly does surprise me!
Move move move!
PS
You do realise the odds of passing this order aren't 100% right? To be a completionist the next step is to work out the odds of passing/failing for different squad setups and get an average expected run move when you issue the order, but before any leadership checks have been rolled.
But that's beyond my paygrade
Last edited by Korona; August 31st, 2011 at 23:24.
Copy, Improve, Innovate
hahaha well my brain needed a workout and I used to love statistics. I'd actually run into almost the same problem before so it was nice to sit down and figure out how to do it.
But you gotta leave something else for the next guy!
Copy, Improve, Innovate
oh well at least i got the average correctly
you can get all the credits, @Korona
I don't use Vendettas or Valks...if Guardsmen were meant to fly, the Emperor would have given them wings. (343rd Mordian)
The battle for Hidaxes SubSector (Listwar)! 184th Cadian blog: 184th.blogspot.com