Ive read so many horrible ways people judge theyre odds in this game lately so i will make a short guide to it so people hopefully can understand how it works.
First let us do a quick example.
If you throw a dice (1d6) you will get a number from 1-6, each of those have the odds 1/6 to appear or 16.67% chance.
Now what is the odds of getting a 1 in 2 throws in a row (
Rolling 2d1:
(1, 1) )?
I have the feeling a lot in here would say 16.67%, this is ofcourse wrong.
Reason is that first you throw one time, odds for it hitting a 1 is then 16.67%, but if you want it in the second throw aswell you have to make the assumption you also got it in the first. The way to calculate the odds for two throws in a row hitting a 1 would then be to multiply the odds from throw one, and throw two: 16.67% x 16.67% = (1/6) X (1/6) = 1/36 = 2.77%
Now over to AA50 odds:
A bit general odds:
Calculating odds of a battle with more units is highly complex as killed units are removed in every round someone score a hit. This makes it almost impossible to make a code that gives the exact odds for an attack. To counter this battlecalc constructors have used simulations of the actual combat you type in and give you an estimate for the odds. As you can see if you try the same battle in tripleA more times you will see a different result every time if the battle is semi complex like Karelia turn 1 attack of 3inf 1art 3fgt vs 5inf 1art. Also these battlecalcs only show the odds of winning the battle with 1 attacking unit surviving, it also says whats the most likely result of attacking units surviving but this is NOT what the odds are refering too. The odds stated refers to one unit surviving. The odds of more units surving will allways be lower then the odds of 1 unit survivng.
To calc the odds for 1 unit survivng the program runs the battle maybe 100 times. Then it adds up all the battles with 1, 2, 3…etc units survivng and dividing it by 100. And you then get an estimate for the odds that youre attack will survive.
Now over to more battles succeeding at once:
Sometimes in AA50 you are in the need off more then one attack succeeding at the same time, like if you hit the eastern front with Germany you want to take 3 territories for the bonus. So what you are interested in is the odds for ALL three attacks suceeding at the same time.
To calculate this you first need to determin the odds for each of the attacks individually, then as in the dice example over you need to multiply the odds for each off the three attacks to determin the attacks combined will succeed.
Now look on a German opening Turn 1.
Assault on Leningrad
Combined odds
Now to the interesting part, each of these attacks have no lower then 79.2%. Lets see what happens if we set as a demand that all 6 attacks have to suceed for us to have a successfull turn.
Odds for all these 6 attacks will succeed in same game turn 1: 0.772 x 0.963 x 0.981 x 0.944 x 0.84 x 0.753 = 0.435 = 43.5%
As we can see we will only have a chance off 43.5% of these 6 attacks to succeed at the same time. If you are looking for a solid gameplan you would like to get the combined odds of MUST accomplish goals maybe up to 90-95%. Though in some situations where the risk / reward ratio is great you might be willing to do an 20% individual odds attack, just becouse it wont hurt you much if you fail, but benefit you great if you succeed. In those cases you should NOT add that single attack into the multiplum you check to find if the must succeed attacks have high enough odds of succeeding for you to be willing to do it.
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Plz give me feedback on this guide, i would like to improve it so people wont start more threads based upon flawed statistics as im a geek who gets provoked by people using maths the wrong way to deffend theyre tactics or opinions. I might also add a paragraph about why this makes low luck a whole different game then dices in AA50 in the future, but that will have to wait.
Also english is not my native language so spelling errors do occur, and i would love if someone would help me correct it :)
Honestly I dont get a grip on your math.
You say if I roll 1 dice I have 16 % odds of success, but if I roll one extra dice I only got 2 % odds of success. What if I roll 3 or 4 dice in a row ? That would be 100 % odds of no success ? And you get provoced by flawed statistics and people who use math the wrong way ? ( had to edit last sentence)
Maybe it was bad explained. I was not talking about getting ONE hit one two dices, was talking about getting TWO hit on two dices.
Thats 16% for 1 hit on 1 dice (assuming you need a 1)
So for 2hits out of two, its 2,77%.
The example is applicable to cases where you need to win 2 out of 2 battles to be successfull.
No offence Adler, but it was pretty clear to me!
I feel it would be worth saying how to calculate the odds of getting x hits out of y dice. I’ll use 2 dice having to land at a one as an example.
There are 2 possible ways to have one 1 showing. Die A is a 1 and Die B is something 2-6 or Die A is something 2-6 and Die B is a 1.
You then find the odds of each event happening.
Die A is a 1: 1/6 x 5/6 = 5/36
Die B is a 1: 5/6 x 1/6 = 5/36
And you then add the two probabilities to find the total.
5/36 + 5/36 = 10/36 = 5/18 = 27.8% chance that you will land one hit.
I hope this helps.
I will add that to the article if you allow Butcher, also will add that thats the probablilty for exactly 1 hit with two dices, and not for 1 or 2 hits. (for the not math educated that might be a bit blurry, not sure though)
Yeah, I guess I should have elaborated. You would just add the probability of 1 hit to the probability of 2 hits to find the probability of 1 or 2 hits.
For attacks that you need at least one hit out of x dice, you should just subtract the probability that you will get no hits from 1. For example, let’s say that 2 inf and 1 fig are attacking 1 inf.
Odds that the attackers miss all hits in 1st round: 5/6 x 5/6 x 1/2 = 25/72
1 - 25/72 = 47/72 = 65.3% chance the battle is over after the first round.
Subracting the odds of a miss from 1 can help speed up a FTF game.
And Pin, you can put anything I post in this thread in an article. I posted in it because it reminded me of my Government teacher, “For all you mathletes out there…”
One question I have that would love to see added to this would be if I am making 6 attacks at various odds, what are the odds of each outcome, for example, 50% win all 6, 70% win atleast 5, 85% win atleast 4, etc, or 50% of the time I will win all 6, 10% of the time I will win 5, etc. I’m guessing its a bit complicated.
It is a more complicated process.
Say the six battles have odds of 95%, 83%, 89%, 97%, 92%, and 80%.
To find the odds of all six, multiply them together
0.95 x 0.83 x 0.89 x 0.97 x 0.92 x 0.80 = 0.501 = 50.1%
To find the odds of say, five battles working, you find the odds for every way the set of battles could go (there are six ways to win 5 battles out of the six)
0.05 x 0.83 x 0.89 x 0.97 x 0.92 x 0.80 = 0.026 = 2.6%
0.95 x 0.17 x 0.89 x 0.97 x 0.92 x 0.80 = 0.103 = 10.3%
0.95 x 0.83 x 0.11 x 0.97 x 0.92 x 0.80 = 0.062 = 6.2%
0.95 x 0.83 x 0.89 x 0.03 x 0.92 x 0.80 = 0.015 = 1.5%
0.95 x 0.83 x 0.89 x 0.97 x 0.08 x 0.80 = 0.044 = 4.4%
0.95 x 0.83 x 0.89 x 0.97 x 0.92 x 0.20 = 0.125 = 12.5%
So now you add the probabilities together
2.6% + 10.3% + 6.2% + 1.5% + 4.4% + 12.5% = 37.5%
That is the probability that you will win exactly 5 of your attacks.
To find the probability that you will win 5 or 6, you must add the probability of winning 5 to the probability of winning 6.
50.1% + 37.5% = 87.6%
You have an 87.6% chance of winning 5 or 6 of your attacks.
Thanks mate, because I feel that atleast for axis turn 1, that some risks should be taken, and was wondering what type of risks exactly were being taken.
Say the six battles have odds of 95%, 83%, 89%, 97%, 92%, and 80%.
To find the odds of all six, multiply them together
0.95 x 0.83 x 0.89 x 0.97 x 0.92 x 0.80 = 0.501 = 50.1%
I love this, man :-)
Lets say this is the 6 standard battles in Germany Turn 1.
If I do all 6 battles in G1, I will only have 50 % odd of success, right ?
So I split them, 3 battles in G1 and the other 3 battles in G2, right ?
G1 0.95 x 0.83 x 0.89 = 70 % odd of success
and
G2 0.97 x 0.92 x 0.80 = 71 % odd of success
So obvious it pays off to split all the attacks out over multiple turns, right ?
no it doesnt, if you still have a goal of having all 6 to succeed, then the odds is still 50.1%, as you need to multiply 70% and 71% to make sure both T1 and T2 succeeds.
But if you succeeded with the 3first in turn 1, then you have 71% chance of getting all 6 done by attacking the last 3 in T2.
You must end with one number when finding the odds of one outcome.
I cannot stress this enough. Breaking the attacks into two sets does not change anything. You will still be multiplying the same numbers together.
What you are looking at, Adler, is one outcome because you want all six to succeed.
And there is absolutely no coincidence to this resembling G1 attacks… :roll:
I feel it would be worth saying how to calculate the odds of getting x hits out of y dice. I’ll use 2 dice having to land at a one as an example.
There are 2 possible ways to have one 1 showing. Die A is a 1 and Die B is something 2-6 or Die A is something 2-6 and Die B is a 1.
You then find the odds of each event happening.
Die A is a 1: 1/6 x 5/6 = 5/36
Die B is a 1: 5/6 x 1/6 = 5/36And you then add the two probabilities to find the total.
5/6 + 5/6 = 10/36 = 5/18 = 27.8% chance that you will land one hit.I hope this helps.
Hey Butcher, thanks to you and Pin for the explanations - I’m not great on maths and you both have a good way of explaining this stuff. Just one thing to point out though is the typo above, I guess the probabilities being added at the bottom are actually 5/36 + 5/36 as opposed to the 5/6 you typed.
Just thought I would point this out as it had me scratching my head for quite a while before digging out my calculator and before I realised it was typo! I may not be the only one!
Yeah, it is supposed to be 5/36 + 5/36, and I have fixed it. Sorry for any confusion.
Yeah, it is supposed to be 5/36 + 5/36, and I have fixed it. Sorry for any confusion.
Yeah, now you have fixed it, but after I got confused
Also consider for a moment you are really discussing probability instead of odds (which are often interchanged despite they are different).
Probaility will give you a percentage of how often the event will occur given infinite rolls (example 16.67%). Odds will deal with possible chances for an event happening versus changes against. (example 1 in 6).
If you have 6 infantry attacking, you can expect 1 to hit for the 6 rolls but only use it as a guide line for you are rolling 6 dice one, not infinite times.
Another point to consider is the possible deviation (something I call the “wild”). It is the possibility of how many additional hits you might achieve. Example: The odds of 6 infantry attacking 2 armor will yield the same odds however, it is far more likely for the infantry to score many (a max possibility of 6) hits versus the armor (a max possibility of 2). This becomes especially poignant in larger battles.
I have played long enough to see the craziest outcomes that could not be repeated in over 100,000 rolls. Also consider it is your roll against their roll so the outcomes increase dramatically.
Anyone else get a big chuckle out of the fact that Pin’s example rolled two 1’s?! :lol: :lol:
Well, it is pretty likely given they were d1’s :wink:
Well, it is pretty likely given they were d1’s :wink:
LOL…i didn’t even see that! :lol: :lol:
