Math, probability, n-hits mixed force

[Guest]

Hi guys,

I wondered if anyone could point me in the right direction as how to calculate the following:

Given a mixed force of say _, what is the probability of getting x hits on a single roll (in terms of i,a,t,etc and x) ?

It would be exciting finding (something close to) an analytic function, i know stuff about binomials, probability distribution etc (i even have a physics degree…lucky me…), so i’m not easily scared by math.

Can anyone give me head start, maybe a great url or a formula? Many thanks in advance, have a great day._

[DTDeGrave]

AtroAtt,

One simple way is to multiply the number of hits you wish to get (in the first round), n, by 6; then ensure that your total sum of all attacking units adds up to at least that.

eg:  You wish to get 4 hits in a round, 4 * 6 = 24…so make sure that you have a total of 24 (or more) attack strength.  In this case, say we had 6 infantry, 2 Art., 4 tanks, and 2 fighters and one bomber available; 4@1, 4@2, 4@3 totals exaclty 24 and would be ‘enough’ according to the odds (but since this is probability and not for-sure-ability you would probably bring in some aircraft).

The main problem with this simple method is that you need to re-calculate each round of combat and you do not have the ability to reinforce mid-battle should your first round completely blank-out.

I hope this helps.

[Ruanek]

Triple A has a battle calculator that may help.

I generally just add up all the attack (or defense) values together and divide by 6.  So if I’m attacking with 5 infantry, 2 artillery, and 3 tanks, that’s 9 for the tanks, 6 for the artillery (added in the supported infantry), and 5 for the infantry, making 20, or 2.3 hits (rounded to 3).  That’s not as versatile as what you’re asking for, though.

It’s fairly simple to calculate the values for them all hitting or missing (just multiply the miss chances by each other, then subtract it from 1 if you want the chance for them all to hit instead of miss).  A general formula for any number of hits is a lot more complicated.  I’ll see if I can figure something out and if I do, I’ll post it.

[BadSpeller]

There is also these….

http://www.dskelly.com/misc/aa/aasim.html?&rules=4&attTacticalBomber=1&defFighter=1

http://frood.net/aacalc/

http://aacalc.nfshost.com/

Also there is an “Axis & Allies Anniversary Ed. Calculator” with the file name “AAProbabilityCalulator.jar” file.
Would someone with information about this please shed more light on this program?

[The Dessert Fox]

You could just play Low Luck dice.

[Ruanek]

@The:

You could just play Low Luck dice.

What is that?

[cts17]

Don’t worry about it. That’s the wuss’s way of playing.

[The Dessert Fox]

@cts17:

Don’t worry about it. That’s the wuss’s way of playing.

I agree but It makes it easier to work out how many hits you are getting

[Ruanek]

@The:

@cts17:

Don’t worry about it. That’s the wuss’s way of playing.

I agree but It makes it easier to work out how many hits you are getting

But what is it?

[ThomasJefferson]

it uses a mathematical algorith to determine dice roles which prevents you from rolling 18 4’s with your 20 tanks.

[Gallo Rojo]

@Ruanek:

@The:

@cts17:

Don’t worry about it. That’s the wuss’s way of playing.

I agree but It makes it easier to work out how many hits you are getting

But what is it?

it’s a way of reducing chance when playing.

works basically the way you described you calculate odds in your battles but applied as a rule and rolling just one dice to cover for the ‘rounded’ factor.

So: add all the attack values in your attacking force and divide it by 6. That’s the total number of hits you get. Then, you roll a dice for the rest.

Example:

You’re attacking with 5 infantry, 2 artillery, 2 tanks, and 1 bomber: 3@1 + 4@2 + 2@3 + 1@4 = 21 ; 21/6 = 3 with a rest of 3.
That means you have 3 hits, plus roll a dice @3 for a 4th hit.

hope that helps

[edfactor]

Hey guys, it looks like he is asking for the math to find various probabilities he isn’t asking about averaged dice or low luck dice.  He just wants to be able to find the various probabilities of x number of hits with a mixed force.

Now i am no math wiz so all i can do is show some brute force calculations.  But in the end i don’t know of a formula that you can plug in numbers and get the results i think the o.p. is looking for.

But here is some brute force calculations:
Take a 3 infantry attack as an example:

1st infantry rolls 1/6 hit and 5/6 miss

2nd infantry rolls and 1/6 hit and 5/6 miss, but your totals now are more complicated. 1/6 of the time your first roll will have hit and 1/6 of the time your second will have hit, 5/6 of the time your second will have missed. 5/6 of the time your first will have missed and 1/6 of the time your second will hit. So you get 2 hits 1/6 * 1/6 or 1/36 times, you get 1 hit 1/6 * 5/6 + 5/6 * 1/6 or 5/36 + 5/36 = 10/36 giving you totals of 2 hits 1/36 and 1 hit 10/36 and no hits 25/36 times.

3rd infantry rolls 1/6 hit and 5/6 miss now 1/36 times you will already have 2 hits so you get 3 hits now 1/36 * 1/6 = 1/216 you keep your 2 hits 1/36 * 5/6 or 5/216 times.  10/36 times you have 1 hit and 1/6 * 10/36 you get another for 2 hits 10/216 times, 5/6 * 10/36 is 50/216 times you stay at 1 hit.  You have no hits 25/36 times so you get 1 hit 1/6 * 25/36 or 25/216 times and 5/6 * 25/216 is 125/216 for still no hits.

You end up
3 hits 1/216, 2 hits 15/216, 1 hit 75/216, and no hits 125/216

Now add a tank into the mix and we start having some real fun.

A tank being simply 1 out of 2 hits gives us 432 new possibilities:

3 hits 2/432  (same as 1/216), 2 hits 30/432 , 1 hit 150/432, No hits 250/432

On half of these the tank hits and half the tank misses
1/432 get 4 hits 1/432 stay at 3
15/432 get a 3rd hit 15/432 stay at 2
75/432 get a second hit and 75/432 stay at 1 hit
Lastly 125/432 get 1 hit and 125/432 still have not hit anything

Giving you 4 hits 1/432 3 hits 16/432 2 hits 90/432, 1 hit 200/432 and no hits 125/432

Now as I said earlier I do not know of any formula to replicate this type of math, but if one exists I am sure someone here will let you know.

[DTDeGrave]

Hey guys,

Edfactor is bang-on with the mathematics here.

What you end up with is essentially a ‘probability tree’ with a double-branching for each unit being rolled, since each unit can only either hit or miss; hence two options or what is known as binomial probability.

These are relatively simple to figure out if all the units are attacking at the same probability (have the same attack value).  Since we are looking at mixed units, n of them, you could end up with some very large numbers of brances to have to add-up.

For example, if you have 10 units attacking, then you could have 210=1024 probability branches to add up…this volume of calculation is best done by some electronic processor; allowing your brain to worry about interpreting the odds and adjusting your strategy as required.

I’m sure many out there can suggest countless different options in terms of apps (PC or other format).

[A Former User]

If you only have one type of troop, it’s a binomial distribution:

Prob(k) = (n choose k) * p^k * (1-p)^(n-k),

where k is the number of hits, n is the total number of units, and p is the probability of a hit.

If you have different units, you need a product of binomial distribution for each permutation of possible events.

So, for the example of two tanks and two infantry, what is the probability of three hits? I’ll in words, then write it out:  You can have either 2 infantry hit, and one tank, or two tanks and one infantry.  Those are the only two ways that you can get two hits:

Prob(3) = (2 choose 2) * (1/6)^(2) * (5/6)^(2-2) * (2 choose 1) * (3/6)^(1) * (3/6)^(2-1) + (2 choose 1) * (1/6)^(1) * (5/6)^(2-1) * (2 choose 2) * (3/6)^(2) * (3/6)^(2-2)

Prob(3) = 1 * 1/36 * 1 * 2 * 3/6 * 3/6 + 2 * 1/6 * 5/6 * 1 * 9/36 * 1

Prob(3) = 18/1296 + 90/1298

Prob(3) = 108/1298

Now, if you have 3 INF, 2 TANKS, and 1 FGT, what is the probability of 2 hits?

Well, how many ways are there to make two hits?  I’ll write them out:

2 INF, 0 Tanks, 0 Fighters
1 I, 1 T, 0 F
1 I, 0 T, 1 F
0 I, 1 T, 1 F
0 I, 2 T, 0 F

An example of how one of these terms expands:

1 I, 1 T, 0 F:

(2 choose 1) * (1/6)^(1) * (5/6)^(2-1) * (2 choose 1) * (3/6)^(1) * (3/6)^(2-1) * (1 choose 0) * (3/6)^0 * (3/6)^(2-2)

If you want Prob(2), then you need to take the sum of the probabilities of all those possible ways to make 2 hits.

Clear as mud?