A beginners guide to statistics in AA50.

[Adlertag]

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)

[Pin]

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.

[Twigley]

No offence Adler, but it was pretty clear to me!

[Butcher]

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.

[Pin]

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)

[Butcher]

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…”

[bugoo]

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.

[Butcher]

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.

[bugoo]

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.

[Adlertag]

@Butcher:

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 ?

[Pin]

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.

[Butcher]

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:

[Twigley]

@Butcher:

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.
5/6 + 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!

[Butcher]

Yeah, it is supposed to be 5/36 + 5/36, and I have fixed it. Sorry for any confusion.

[Adlertag]

@Butcher:

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

[Octopus]

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.

[jsp4563]

Anyone else get a big chuckle out of the fact that Pin’s example rolled two 1’s?! :lol: :lol:

[Butcher]

Well, it is pretty likely given they were d1’s  :wink:

[jsp4563]

@Butcher:

Well, it is pretty likely given they were d1’s  :wink:

LOL…i didn’t even see that! :lol: :lol:

[Corbeau Blanc]

I think this great for judging dices odds on ONE battle.

What is wrong with your strat exemple, or should I say MY strat, is that :

• It does not reflect in any way a lowluck setting in which context the strat was written
• It does not consider possibilities like straffing
• It does not consider the casulaties taken by the ennemy
• It does not consider the importance of each battle but rather try to multiply the sums of all moves as if each battles are equally important ( which is simply UNTRUE )
• It is based on triple A simulator
• It does not take in account the fact you CAN choose the order in which you do your battles. Calling off an attack after a failure somewhere else is possible. You can retreat when there is no point…
• Are you willin to take risks, if not, did you read the alternative plan?

My goal in that strat is simple, Take and Hold Karelia.

There is only 3 Battles in there that are MUST win and they must be resolved in THAT order.
Using low luck, I will use triple A calculator like you did.

• 1st : Sinking the BB/TR ( Pull out a sub, and send the bomber if you are too scared with 85%. It will be 100% )
• 2nd: Taking Baltic States ( it’s 100% odds, not 96.3%… )
• 3rd: Taking Leningrad 92% odds ( it’s not 79%, sim it again )

So anywhere between 78.2% to 92% odds of success…

Also let’s be clear, while you would try to round up your odds in thoses 3 battles to 92% by sending bomber on the BB, my bomber would go to Leningrad making it 100% instead and I’d risk the whole thing at 85% ( 100% x 100% x 85% ). Why? Because i’d resolve the sea battle first and already would have a clear idea either or not go the whole ten yards aka straffing or taking it.

Still, let’s take a look at WHY it is 100% win sending the bomber. Your Bomber, fighter and one sub  fight against a BB.  If we go by theses ‘‘statistics’’, the battle is at 93-97% when in fact it’s 100% win Low luck. There is NO way around it.

Let’s do the “maths”: 4+3 = 7. That’s 1 automatic hit, and 1/6 for a possible second hit. PLUS 2/6 for the sub in yet another possible hit. Let’s say everything miss and the damaged BB hit back. The sub goes, another automatic hit goes in, the BB hit back again before sinking, the fighter goes. Worst case scenario, still 100% win.

Now, Why is Baltic States 100%? The fact is that in low luck you have a minimum and maximum number of hit you can do as well as the notion of automatic hits. 3 russian inf will invariably yield 1 hit vs the german 2 inf/3tanks. Even if you would not roll anything for germans and Russia would hit every round despite losses, it would go like this:

2inf/3tanks vs 3 inf
1inf/3tanks vs 2 inf
3tanks vs 1 inf
2 tanks win, worst case scenario.

It’s not odds, it’s a SURE thing as far as winning. The odds are there to estimate how many losses you will sustain in the attack.

The strat was written in the context of low luck, so I will not debate it outside of the context.

Edited: typo and missed sentence.