Converting Punch to Percentage; Question for Math Geeks!

[Zhukov44]

@Omega1759:

The probability of not hitting anything with 1 infantry, 1 armor and 1 fighter is one less the probability of missing them all.

Missing them all is:

5/6 (missing infantry)
x
1/2 (missing armor)
x
1/2 (missing fighter)

= 5/24, about 21% chance of missing them all.

So the probability of hitting at least once is 79%.

Good stuff.  Simple enough to do in my head.  Thank you!

[Omega1759]

@Zhukov44:

@Omega1759:

The probability of not hitting anything with 1 infantry, 1 armor and 1 fighter is one less the probability of missing them all.

Missing them all is:

5/6 (missing infantry)
x
1/2 (missing armor)
x
1/2 (missing fighter)

= 5/24, about 21% chance of missing them all.

So the probability of hitting at least once is 79%.

Good stuff.  Simple enough to do in my head.  Thank you!

For larger battles involving tracking more than one hit, it quickly becomes a lot of work to figure this out, hence the use of Monte Carlo simulators.

[Zhukov44]

@Herr:

Unfortunately, it doesn’t quite work that way. It depends on the units that make up your punch.

An easy way to compute the probability that you’ll get at least one hit, is to calculate the odds that you won’t hit anything and subtract that number from 1.
Chance that the inf doesn’t hit = 5/6; chance that the arm doesn’t hit = 3/6; chance that the fig doesn’t hit = 3/6. So the chance that nothing hits = 5/6 x 3/6 x 3/6 = 45/216. Subtract that from 1 and you’ll get 171/216, which is about 79%. So that’s the conversion for this punch of 7.

However, if you use 7 inf, the chance to hit nothing will be 5/6 x 5/6 x 5/6 x 5/6 x 5/6 x 5/6 x 5/6 = 78125/279936. Subtract that from 1, and your chance for at least 1 hit will be only 72%.

Best chances are 1 fig 1 tac: chance to hit nothing = 2/6 x 3/6 = 6/36, so there’s an 83% chance for at least one hit.

Generally, fewer units increase the chance for at least one hit. That’s what low luck does, by simulating a hypothetical unit that will always hit. The trade of is of course, that more units offer a perspective of a larger number of hits.

Edit: wow, five answers while I was typing. Lots of math geeks here. Anyway, it was in response to the original question.

That’s very interesting.  It seems counter-intuitive to me that the same punch could have different probabilities of getting 1 hit depending on the units involved.  Too much A&A :lol:?

Interesting that the probability of 2 inf getting a hit is actually 31%, of 3 inf 42%, with 4 inf approximately 52%.

[elzario]

@Omega1759:

… hence the use of Monte Carlo simulators.

Anyone have a working application that gives you “to hit” chance percentages when inputting types (and amount) of units?

I had thought of creating an app for it but why make something that already exists right?

[Spendo02]

Be interesting if there was a “cheat sheet” for different combinations of units getting at least 1, 2 and 3 hits.

We don’t allow the use of outside “inputs” once we start a game, but would be nice if you could commit some of the more standard combinations to memory such as 2 Inf 2 Art getting 1, 2 or 3 hits or 2 Armor and 2 Mech scoring 1, 2 or 3 hits.

[elzario]

@Spendo02:

Be interesting if there was a “cheat sheet” for different combinations of units getting at least 1, 2 and 3 hits.

We don’t allow the use of outside “inputs” once we start a game, but would be nice if you could commit some of the more standard combinations to memory such as 2 Inf 2 Art getting 1, 2 or 3 hits or 2 Armor and 2 Mech scoring 1, 2 or 3 hits.

I have a spreadsheet for each to-hit value (2 or less, 3 or less etc).  It gets pretty intricate when you start mixing to-hit values.

[Spendo02]

I know, I reached out to a statistics wizard I know and when he started getting into the calculations of it I basically said “nevermind” as it was too complicated to follow with my Business Stat 101 understanding of probability.

I generally just stick to counting up my roundels that would count as hits in the first round and count up the defender’s roundels and if I outnumber the defender’s roundels, its worth a single round of die rolls.  When the roundel tide turns against me in that battle, its no longer worth rolling dice and is time to retreat.

I make exceptions for when a counter-attack would cede the territory I’d be retreating to as I’d rather weaken the counter attacking force’s fodder than retreat and cede the territory with only a single round of die rolls likely to occur.

[elzario]

@Spendo02:

I know, I reached out to a statistics wizard I know and when he started getting into the calculations of it I basically said “nevermind” as it was too complicated to follow with my Business Stat 101 understanding of probability.

Tell me about it!  Its crazy to think that something so simple in practice becomes so convoluted in execution.  I always think that estimations within probabilities are as good as accurate calculations.  Such as a 72% chance vs a 70% chance is really the same thing in practice even though in theory it can be something entirely different.

[elzario]

Here is the spreadsheet I did up a while back that can do a bit of chance workups.

Excel: https://docs.google.com/file/d/0B_qD08g2r8jXR19xNHpJa19VZVU/edit?usp=sharing
Google Docs: https://docs.google.com/spreadsheet/ccc?key=0AvqD08g2r8jXdDFFZTk1TWRROWJVaHlxenhGM2prbVE&usp=sharing

EDIT: After looking at the numbers aren’t correct.  I havent broken down formulas or anything so not sure how far out it is.  Good thing I don’t use it anymore!

[Cow]

Compare 6 infantry to 2 fighters. The 6 infantry have more variation, up to 6 hits possible, the 2 fighters have 2 hits possible. The standard deviation for 6 infantry is much greater than the standard deviation of 2 fighters, we can also find test this out to confirm that.
~

Generally speaking, you want more dice rolls than bigger dice if their “attack value” totals are the same.

As far as the likely hood for scoring 1 hit, we can compare this using statistics. 6 infantry vs 2 fighters to come up with some sort of general conclusion.
~

There are several ways of computing the math on this.

Tree Diagram and count out the likelihoods of all events in which 1 infantry confirms a hit.
or
you can use the formula.

Would you like me to do the math on this? If you want me to do the math on this you have to state if we are counting events of more than 1 hit or not. That changes the formula I use.

You will appreciate having more rolls for a number of reasons, not just in regards to hits, but in regards to cannon fodder factor as well. Attacking with pure infantry is reliable, but people always comment on it not scoring any hits or when it scores lots of hits (extreme outcomes)…I am not sure why, but I think people pay more attention to it more than say 2 subs losing to 1 destroyer.

Low luck does make having more rolls less significant and reduces it to cannon fodder factor issues.

[Cow]

To compute for no hits (5/6)^6

0.334897977
~
1-0.334897977= 0.665102023
~
6 infantry have a 66.5% chance of scoring 1 hit or more
~
Now we shall computer 2 fighters.

(3/6)^2

0.25

1-0.25 = 0.75%

Two fighters have a 75% chance of scoring 1 hit or more (25% odds for scoring 2 hits, but the 6 infantry have better odds for scoring more than 2 hits).
~

Sure you are getting more consistency (less deviation from the mean). Is that what you want? Depends on what you are attacking or defending against.

Say you are fighting just 1 unit, then yes you want 2 fighters to kill that instead of 6 infantry, since you just want 1 hit.

In bigger scale battles, would you attack 3 armor with 2 fighters or 6 infantry? You would take 6 infantry all day everyday. You would even take 6 infantry over 3 infantry and a fighter if you are attacking 3 armor as well.
~

The general rule of thumb when working with equal ipc of units is 2 infantry in front of every armor unit (as far as attacking goes, for defending infantry is most cost efficient). There is a reason why people buy 2 mech for every armor.

You can apply similar rules of thumb with naval units as well. There is a reason why the a couple sub dd carrier 2 fighter is the default middle of the road buy, it is strong on defense.
~
So for small skirmishes the question is do you want more than 1 hit or 1 hit?

I am not sure what the ratio of infantry to artillery should be. It is difficult to pan out exactly how much inf you will have after all the skirmishing for the main event, which is attacking Russia.

Generally speaking I like 1 arty for every 2 infantry/mech (then an armor for every 2 mech, the mech and armor I pick up later to make a timed attack at Russia, which is usually rounds 6-8… it is a timed attack so I would buy them starting round 2 or 3).

[Degrasse]

You might look up Lanchester’s Law and use that for combat estimating.

http://en.wikipedia.org/wiki/Lanchester's_law

[MajorJuggler]

You need to understand probability and statistics to comprehensively answer this question. As a point of reference, I covered the prerequesite material around the 3rd year of my undergraduate engineering degree, so the call for “Math Geeks” is well titled here. :-)

Some pertinent terms you’ll want to google/wiki/library: probability density function (PDF) cumulative density function (CDF), mean, standard deviation, and convolution.

@elzario:

Anyone have a working application that gives you “to hit” chance percentages when inputting types (and amount) of units?
I had thought of creating an app for it but why make something that already exists right?

Yes, but I haven’t publicized it. I wrote my own battle simulator in MATLAB which returns EXACT values, not Monte Carlo estimates. It calculates (and graphically displays) the total probability density function of a battle of an arbitrary number and type of dice on offense and defense. (i.e. PDF of all attackers surviving, all the way through to all defenders surviving, and everything in between). For a math geek, it’s very cool.  8-) The Monte Carlo simulators get reasonably close results, but the tail ends (say less than 2% probability) need a high run count to converge to the correct answer. Viewing the net PDF is significantly more useful than a simple “win/loss” percentage.

@elzario:

Its crazy to think that something so simple in practice becomes so convoluted in execution.

Oh, the poetic irony of that statement! One of the required pieces (also what the original poster was asking about) is to calculate the PDF of an arbitrary number of dice being rolled. Using the mathematical function convolution, this subroutine only takes 7 lines of code in MATLAB.

PDF = 1;
    for i = 1:4
        while(DiceCounts(i) > 0)
            DiceCounts(i) = DiceCounts(i) - 1;
            PDF = conv(PDF,[6-i i]/6);
        end
    end

Unfortunately, getting the exact answer using this method isn’t something that you do in your head, especially for a large number of dice rolls. On the plus side, it only requires one convolution per dice roll to get the exact answer, unlike Monte Carlo simulators that may require 1,000-10,000 runs to converge to a statistically reliable result.

DiceCounts is a 1x4 vector, where each element contains the number of dice being rolled at each number. I.e. DiceCounts(i) is the number of units rolling at i. The output PDF is a 1x(n+1) vector containing the probability of getting the exact number of hits for each possible outcome, where n is the total number of dice rolled, and PDF(x) is the probability of getting exactly x-1 hits. Integrating the PDF results in the CDF, which can be used to calculate the probability of getting a minimum (or maximum) number of hits.

@Zhukov44:

How would I determine the % likelihood of getting 1 hit with a punch of 7?

So, to answer your question – it entirely depends on what units are attacking. 7 infantry attacking will yield a different PDF than a tank and a bomber, even though both have a total attack value (punch) of 7.

Lets use Cow’s example of 6 infantry attacking, and also 2 fighters attacking. Both clearly have a mean of 1.

6 infantry: mean = 1, std = 0.1704.
6 Infantry PDF (in %):            33.4898  40.1878  20.0939    5.3584    0.8038    0.0643    0.0021
6 infantry min# of hits (in %): 100.000  66.5102  26.3224    6.2286    0.8702    0.0664    0.0021

2 fighters: mean = 1, std = 0.1443.
2 fighters PDF (in %):            25    50    25
2 fighters min# of hits (in %): 100    75    25

The 6 infantry do have a slightly higher standard deviation than the 2 fighters, and from the above you can see that 6 infantry have a 66.5% chance to get at least one hit, and the 2 fighters have a 75% chance to get at least 1 hit.

However, it does not always follow that smaller numbers of dice rolls always result in lower standard deviation given the same attack power. For example, consider 8 infantry attacking, compared to 2 bombers attacking.

8 infantry: mean = 1.33, std = 0.1418.
8 Infantry PDF (in %):            23.2568  37.2109  26.0476  10.4190    2.6048    0.4168    0.0417    0.0024    0.0001
8 infantry min# of hits (in %): 100.0000  76.7432  39.5323  13.4847    3.0656    0.4609    0.0441    0.0024    0.0001

2 bombers: mean = 1.33, std = 0.1925.
2 bombers PDF (in %):            11.1111  44.4444  44.4444
2 bombers min# of hits (in %): 100.0000  88.8889  44.4444

2 bombers attacking have a higher standard deviation than 2 fighters, but you are still more likely to get at least 1 or 2 hits with 2 bombers than you are with 8 infantry. So, standard deviation is only one measure of variability, and not always the best measure.

Another example previously given: 1 fighter, 1 tank, 1 infantry (1@1, 2@3):
1@1 + 2@3: mean = 1.1667, std = 0.1735
1@1 + 2@3 PDF (%):        20.8333  45.8333  29.1667    4.1667
1@1 + 2@3 min # hits (%): 100.000  79.1667  33.3333    4.1667

In general, as the number of dice rolls increases, the net distribution ends up looking like a Gaussian distribution regardless of what the PDFs look like for the individual dice rolls. (Central Limit Theorem).

[Cow]

So basically when in doubt get a grab bag of units! :D

[variance]

@MajorJuggler:

You need to understand probability and statistics to comprehensively answer this question. As a point of reference, I covered the prerequesite material around the 3rd year of my undergraduate engineering degree, so the call for “Math Geeks” is well titled here. :-)

Geee I surely do wish I had me an education too so’s I could know them things  :roll: