G40 VANN FORMULAS RESULTS

[Imperious Leader]

Thats why you use the Larrymarx formula to solve these issues.

[baron Münchhausen]

@Baron:

So, as a rule of thumb, if the enemy can bring 2 aircrafts in addition to the same number of Infantry picket, then you are toasted. Right?

2 Fgs & 3 Infs^2 * A9/5 = 25*1.8 = 45

3 Infs^2 * D2 = 18

It is more than double value of picket stack.
45 vs 18 = 5 : 2 ratio

But, if there is only 1 additional plane in attack, it is workable for defender?
(1A3+3A1)^2 *A6/4 = 16 * 1.5 = 24

24 vs 18 = 4 : 3 ratio

So, whenever the attacker get only 4:3 ratio or less, the defender should try it.

Is it a right way to apply your formula?

@Imperious:

Thats why you use the Larrymarx formula to solve these issues.

For now, this is Kreuzfeld formula that I’m applying here.

Someday, there maybe a way to use both but for now it is two different beasts.

And the Baron-Larrymarx formula does not help, not accurate or too many calculations to get to something.

Kreuzfeld formula is simpler for a F-2-F game, and few numbers.

[Imperious Leader]

Kreuzfeld formula.

As long as its not named after a member… better.

[baron Münchhausen]

@Imperious:

Kreuzfeld formula.

As long as its not named after a member… better.

For now, it is the case all Vann, Kreuzfeld, Baron and Larrymarx are members.

If you have an idea for a more suitable name for Baron-Larrymarx, I’m opened to it, as long as others agree.

I cannot deny Vann, Larrymarx then me filiation (or line of succession) to this Baron-Larrymarx formula.
But Kreuzfeld bring his own much earlier. And it is a different beast, for a different purpose.

[larrymarx]

You guys are forgetting something important: SIZE MATTERS

The back and forth nature of A&A battles means that very slight advantages will snowball into huge advantages when you scale up the number of units. To demonstrate, I pulled up a battle calculator and tried throwing infantry against half their numbers in tanks (stacks with equivalent values).

Infantry     Tanks       Attacker wins     Avg swing    Total value     % swing
   2               1            55%                   0.1                12               0.8
   4               2            60%                   1.2                24               5.0
   8               4            66%                   4.2                48               8.8
 16               8            74%                  11.6               96              12.1
 32             16            83%                  29.4              192              15.3
 64             32            92%                  68.3              384              17.8
128             64            98%                 147.5              768             19.2

You start with what is nearly a fair fight at the beginning and wind up with an overwhelming advantage as you draw out the battle and increase the number of turns over which the advantage is played out. In contrast, if the battles are only for one round, the % swing rises only to 4.166% (1/24) and then stays there no matter how large the stacks are.

This, combined with my earlier assertion that units are almost always better in mixtures means we really should be looking at how to determine the relative worth of stacks and mixtures of units as opposed to individual units. Units mean nothing by themselves - they always must be analyzed in context.

I propose analyzing different mixtures and quantities of units to determine their worth in attacking vs. stacks of units we already know to be strong on defense - infantry, destroyers, fighters and carriers (because they carry fighters). A compete analysis will increase the size of the defender’s stack if the attacking units are slow because the defender will “see it coming” and have more time to turtle up.

[CWO Marc]

@larrymarx:

The back and forth nature of A&A battles means that very slight advantages will snowball into huge advantages when you scale up the number of units.

And speaking of formulas, the point mentioned by Larrymarx ties in nicely with Lanchester’s Square Law, a differential equation which models the attrition effects of modern combat involving long-range weapons.  As I understand it, the equation basically states that the rate of attrition is proportional to the square of the number of weapons which are shooting, and therefore that any numerical advantage by one side will increase exponentially as the battle progresses.

[baron Münchhausen]

Yep Marc,
The original thread started with this Lanchester’s Square Law, then found a specific formula for stacks of A&A units.

[baron Münchhausen]

I made it on AACalc then I revised numbers by applying this formula derived from above Stack formula:
√(P2 / P1) = N1 / N2

To get break even (meaning 50% vs 50% odds) number of unit ratio for different  Power relative to each other:

Power  1
vs 2 = 14 units :10 units  (140:99)
vs 3 = 19:11 (189:109)
vs 4 = 2:1

Power 2
vs 1 = 10:14  (99:140)
vs 3 = 11:9 (109:89)
vs 4 = 14:10 (140:99)

Power 3
vs 1 = 11:19 (109:189)
vs 2 = 9:11 (89:109)
vs 4 = 15:13 (149:129)

Power  4
vs 1 = 1:2
vs 2 = 10:14 (99:140)
vs 3 = 13:15 (129:149)

So, if you want to see if your on the right side, find the average Power and compare your number  of units with the ratio above.

For example, you have 3 Tank and 1 Inf vs 2 Inf 1 Art.
Avg less than 3 compared to 2.
4 units vs 3 units 4:3

The table for 3 vs 2 say 9:11, so, if you drop to 2 tanks vs 3 units during combat resolution, you know  you are below 50% odds of survival. You may then retreat, or try your luck.

[larrymarx]

This is very interesting, and I think it validates the cost formulas: a quadrupling in strength is required to justify a doubling in cost.

I also like that you’ve removed the factor of cost for the above ratios. It’s always better to bring more than you need to win the fight, and if you know what break even is for a fight, you can focus on whether or not you can bring more than that to the battle. For this purpose, unit cost might be a factor or it might not.

With large battles, as I demonstrated earlier, it just takes a slight advantage to be certain of victory. If it is a large battle, then winning it is probably more important than preserving your units with high values. If it’s a small battle, cost matters more because you want your good units left for future battles, but in the case of small battles these formulas and ratios don’t really matter anyways because anyone can look at the units and tell which side will win.

[baron Münchhausen]

@larrymarx:

This is very interesting, and I think it validates the cost formulas: a quadrupling in strength is required to justify a doubling in cost.

I also like that you’ve removed the factor of cost for the above ratios. It’s always better to bring more than you need to win the fight, and if you know what break even is for a fight, you can focus on whether or not you can bring more than that to the battle. For this purpose, unit cost might be a factor or it might not.

With large battles, as I demonstrated earlier, it just takes a slight advantage to be certain of victory. If it is a large battle, then winning it is probably more important than preserving your units with high values.

If it’s a small battle, cost matters more because you want your good units left for future battles, but in the case of small battles these formulas and ratios don’t really matter anyways because anyone can look at the units and tell which side will win.

I agree.

Would you agree Larrymarx, about this formula being renamed ENIGMA formula ? To consider all contributors without having a too cumbersome name.

Here is the table based on Baron-Larrymarx formula completed on effective cost vs combat points ratio:
For all 1 hit units, you use : 36 Power/(cost^2) = offense or defense factor* based on cost
For 2 hits and 3 hits unit : 36 Power/(cost^2) {1+[(nb hit -1)/11.62] }= offense or defense factor* based on cost

To get the cost of a 1 hit unit for a given factor of reference: √(36*Power of unit / Offence or defence Factor)= Cost.
For a 2 hits unit for a given factor of reference:
√(36*Power of unit {1+[(nb hit -1)/11.62] } / Offence or defence Factor)= Cost.

For combined arms and multiple units you have to average both combat points per unit and cost per unit.
Then you can add it into the formula.

@Baron:

I made it on AACalc then I revised numbers by applying this formula derived from above Stack formula:
√(P2 / P1) = N1 / N2

To get break even (meaning 50% vs 50% odds) number of unit ratio for different  Power relative to each other:

Power  1
vs 2 = 17 units :12 units (170:120) 1.42
vs 3 = 19:11 (189:109)                 1.73
vs 4 = 2:1                                     2.00

Power 2
vs 1 = 12:17  (120:170)                0.70
vs 3 = 11:9 (109:89)                     1.22
vs 4 = 17:12 (170:120)                  1.42

Power 3
vs 1 = 11:19 (109:189)                0.58
vs 2 = 9:11 (89:109)                    0.82
vs 4 = 15:13 (149:129)                1.15

Power  4
vs 1 = 1:2                                    0.50
vs 2 = 12:17 (120:170)                 0.70
vs 3 = 13:15 (129:149)                 0.87

So, if you want to see if your on the right side, find the average Power and compare your number  of units with the ratio above.

For example, you have 3 Tank and 1 Inf vs 2 Inf 1 Art.
Avg less than 3 compared to 2.
4 units vs 3 units 4:3

The table for 3 vs 2 say 9:11, so, if you drop to 2 tanks vs 3 units during combat resolution, you know  you are below 50% odds of survival. You may then retreat, or try your luck.

It is my first shot working with Table feature of the Forum.
If someone can do better, I will appreciate.
With this small 5 x 5 table, you get in a glimpse what is the 50%-50% break even according to the ratio of units and the average power of a given stack.
This table may also be written in decimal instead of a ratio, below. But ratio are better to understand how each ratio is paired to another one which is simply reversed.

A more complete table would include average power: 1.5, 2.5 and 3.5
But, in F-2-F, you can always round up the average enemy’s power, so you do a safer battle to be sure you are above break even ratio.

X means the ratio is 1:1. I did not want to overcharge this table with obvious infos.

| Power
1
2
3
4 | 1
X
12:17
11:19
1:2

| 2
17:12
X
9:11
12:17
| 3
19:11
11:9
X
13:15
| 4
2:1
17:12
15:13
X

| Power
1
2
3
4 | 1
X
0.70
0.58
0.50

| 2
1.42
X
0.82
0.70
| 3
1.73
1.22
X
0.87
| 4
2.00
1.42
1.15
X

For instance, reading from left row to the right, if you have an average power of 1.3 for 17 units and the defender has 2.2 average power for 10 units.
You may cross-referenced the 1 row with the 2 column, saying you need  17: 12 ratio, or 1.42 more units than defender.
So, your 17:12 ratio or 1.72 more units  give an above 50-50% odds of winning.

Of course, during battle, ratio of units and average power may changes, especially when fodders are done.
So, you may decide at critical moment to recheck your odds of success.

For example, the defender may have only 3 Tanks left, while attacker only 4 Infs and 1 Tank.
This give a 5: 3 units ratio. And the 1 (A7 / 5= 1.4) row compared to 3 column, says: 19 to 11 or 1.73.
If looking the 2 row, it says 11:9 or 1.22.
So, assuming this rounding up or down, it reveals you are above or below the break even point.
Thus, it is still near 50%-50%.
In fact, AACalc says:
Overall %*: A. survives: 70.6% D. survives: 24% No one survives: 5.5%

But, Punch formula might be better to anticipate results?
A7 for 5 hits vs A9 for 3 hits.

Stack formula is more revealing IMO:
251.4=  35  vs 93= 27 |

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[baron Münchhausen]

X means the ratio is 1:1. I did not want to overcharge this table with obvious infos.

| Power
1
2
3
4 | 1
X
12:17
11:19
1:2

| 2
17:12
X
9:11
12:17
| 3
19:11
11:9
X
13:15
| 4
2:1
17:12
15:13
X

| Power
1
2
3
4 | 1
X
0.70
0.58
0.50

| 2
1.42
X
0.82
0.70
| 3
1.73
1.22
X
0.87
| 4
2.00
1.42
1.15
X

This table can also be memorized with 5 basic ratios and you can reverse at will, or 6 if we include 1:1 ratio:

1:2 (4 vs 1), 11:19 (3 vs 1), 12:17 (4 vs 2 or 2 vs 1), 9:11 (3 vs 2), 13:15 (4 vs 3)

OR same order but from reverse ratio:
2:1 (1 vs 4), 19:11 (1 vs 3), 17:12 (1 vs 2 or 2 vs 4), 11:9 (2 vs 3), 15:13 (3 vs 4) |

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[larrymarx]

Sure, enigma works fine as a name.

For the tables you are developing, why don’t we call them Lanchester tables? They are very clearly a derivative of his work.

[larrymarx]

Can we steer this discussion into an actual application towards developing strategy?

Take a German push into Russian territory. Germany needs to be concerned with a stack of units in Moscow whose average power is slightly higher than 2. Russia is going to be building a lot of infantry and artillery so there is very little skew - the exception would be if UK / US have landed a bunch of fighters there.

Germany is going to bring a bunch of tanks and its air force to the battle, but it will also have infantry attacking at one. The average power will likely be higher than the Russian stack’s, and there will be a significant skew as well.

It makes intuitive sense that you should just try to match the number of units that they have over there, and your greater power as well as the skew in your favor will win you the battle. Can we use the Lanchester tables to be a little bit more precise in our assessment of this strategic objective?

[baron Münchhausen]

@larrymarx:

Sure, enigma works fine as a name.

For the tables you are developing, why don’t we call them Lanchester tables? They are very clearly a derivative of his work.

I’m fine with it too: Lanchester tables for these 2 tables above.

[baron Münchhausen]

@larrymarx:

Can we steer this discussion into an actual application towards developing strategy?

Take a German push into Russian territory. Germany needs to be concerned with a stack of units in Moscow whose average power is slightly higher than 2. Russia is going to be building a lot of infantry and artillery so there is very little skew - the exception would be if UK / US have landed a bunch of fighters there.

Germany is going to bring a bunch of tanks and its air force to the battle, but it will also have infantry attacking at one. The average power will likely be higher than the Russian stack’s, and there will be a significant skew as well.

It makes intuitive sense that you should just try to match the number of units that they have over there, and your greater power as well as the skew in your favor will win you the battle. Can we use the Lanchester tables to be a little bit more precise in our assessment of this strategic objective?

Using Stack formula, also based on Lanchester’s square law.
Maybe you have to calculate what will be the Metapower of Russia in R5, R6 or R7.
Then try to compose an army which can get higher metapower, assuming units already on board and additional purchases able to reach Moscow on R6, G6 or G7.

Then, it tell how much you can invest for Naval.

Maybe, predict the ratio to reach, like keeping always above 9:11, ratio.
Meaning for each @2 Russia built, Germany have to built an @3, to stay 1:1.

This link OP might provide an applying.
But the formula suggested might not be exactly accurate.
Is it total power or average power?

http://www.axisandallies.org/forums/index.php?topic=39526.msg1640769#msg1640769

[baron Münchhausen]

Lanchester’s Table for Axis and Allies 2nd Edition

I made it on AACalc then I revised numbers by applying this formula derived from above Stack formula:
√(P2 / P1) = N1 / N2

I completed the table to include 2 hits, @4 for battleship.

X means the ratio is 1:1. I did not want to overcharge this table with obvious infos.

| Power
1
2
3
4
4, 2hits | 1
X
12:17
11:19
1:2
5:16

| 2
17:12
X
9:11
12:17
4:9
| 3
19:11
11:9
X
13:15
10:19
| 4
2:1
17:12
15:13
X
5:8
| 4, 2hits
16:5
9:4
19:10
8:5
X

| Power
1
2
3
4
4, 2hits | 1
X
0.70
0.58
0.50
0.31
| 2
1.42
X
0.82
0.70
0.43
| 3
1.73
1.22
X
0.87
0.53
| 4
2.00
1.42
1.15
X
0.63 | 4, 2hits
3.20
2.30
1.87
1.60
X

This table can also be memorized with the main 5 basic ratios and you can reverse at will, or 6 if we include 1:1 ratio, the @4 with 2 hits for BBs might not be very relevant in F-2-F:

1:2 (4 vs 1), 11:19 (3 vs 1), 12:17 (4 vs 2 or 2 vs 1), 9:11 (3 vs 2), 13:15 (4 vs 3)

OR same order but from reverse ratio:
2:1 (1 vs 4), 19:11 (1 vs 3), 17:12 (1 vs 2 or 2 vs 4), 11:9 (2 vs 3), 15:13 (3 vs 4)

For gameplay, you can easily replaced a 11:19 or 19:11 with 4:7 or 7:4.
And 17:12 or 12:17 with 3:2 or 2:3, in fact the real number is √2 and 1/√2. 1.5 vs 1.42 and 0.7 vs 0.75.
The ratio is not too different, just less accurate.
But in game, it is easier to calculate it with 1 digit number.

| Power
1
2
3
4
4, 2hits | 1
X
2:3
4:7
1:2
5:16

| 2
3:2
X
9:11
2:3
4:9
| 3
7:4
11:9
X
13:15
10:19
| 4
2:1
3:2
15:13
X
5:8
| 4, 2hits
16:5
9:4
19:10
8:5
X

| Avg Power
1
1.5
2
2.5
3
3.5
4
4, 2hits | 1
1.00
0.82
0.70
0.63
0.58
0.53
0.50
0.31
| 1.5
1.22
1.00
0.87
0.77
0.70
0.65
0.63
0.38
| 2
1.41
1.15
1.00
0.89
0.82
0.76
0.70
0.43
| 2.5
1.58
1.29
1.12
1.00
0.91
0.85
0.79
0.5
| 3
1.73
1.41
1.22
1.10
1.00
0.93
0.87
0.53
| 3.5
1.87
1.53
1.32
1.18
1.08
1.00
0.94
0.58
| 4
2.00
1.63
1.41
1.26
1.15
1.07
1.00
0.63
| 4, 2hits
3.20
2.64
2.30
2.00
1.87
1.73
1.60
1.00

| Avg Power
1
1.5
2
2.5
3
3.5
4
4, 2hits | 1
1:1
9:11
12:17
5:8
4:7
10:19
1:2
5:16

|
1.5
11:9
1:1
13:15
7:9
12:17
9:14
5:8
3:8
| 2
17:12
15:13
1:1
9:10
9:11
3:4
12:17
4:9
| 2.5
8:5
9:7
10:9
1:1
10:11
5:6
4:5
1:2
| 3
7:4
17:12
11:9
11:10
1:1
19:20
13:15
10:19
| 3.5
19:10
14:9
4:3
6:5
20:19
1:1
20:21
4:7
| 4
2:1
8:5
17:12
5:4
15:13
21:20
1:1
5:8
| 4, 2hits
16:5
8:3
9:4
2:1
19:10
7:4
8:5
1:1
http://www.axisandallies.org/forums/index.php?topic=39526.msg1686896#msg1686896 |

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[Dauvio Vann]

@Baron:

@larrymarx:

Sure, enigma works fine as a name.

For the tables you are developing, why don’t we call them Lanchester tables? They are very clearly a derivative of his work.

I’m fine with it too: Lanchester tables for these 2 tables above.

Lanchester tables it will be then. We are making some game changing history here boys!!!  8-) 8-) 8-)

[baron Münchhausen]

This is a different derivation from Stack formula and it probably allows to get the derivative formula for Enigma.
Indeed, all 50% vs 50% comparison always requires like x Cruiser A3 = y Fighter D4 meaning that Cruiser should cost y while Fighter cost x.

Assuming Kreuzfeld is a mathematician, I’m pretty sure he can be able to derive Enigma formula (former Baron-Larrymarx) from his formula.

@Kreuzfeld:

Yes.

TLDR Version: (number Attacker units needed for 50% chance) = (Number of Defender units) * SQRT ( ( Average Defender Strength) / (Average Attacker Strength))

My formula is exact and easy to prove correct when you attack with only units of one type, against only units of a different type ( like inf v inf, inf v tanks, inf v ftrs, etc). I am a mathematician, I can send you the proof if you don’t trust me.

If average Defence  power is 2, and average attack power is 1, then  you need Sqrt(2) (1.141 number of units to win the attack 50%

I have made some calculations. If you assume you attack with only units with strength 1, and the defenders have only units of strength 2,3,4, then the number of units needed to have 50% chance of taking the terr is as follows;
1 v2 -> (sqrt(2)  =) 1,41… This means that 141 infs will have less than 50%, while 142 has more than 50% against 100 infs defenders.
1v3 -> (sqrt(3) = )1.73…  This means that 173 infs will have less than 50%, while 174 has more than 50% against 100 tanks defenders
1v4 -> (sqrt(4) =) 2 . So 200 infs have exactly 50% chance of winning against 100 planes.
2v4- > (sqrt(2)  =) 1,41… This means that 141 art will have less than 50%, while 142 has more than 50% against 100 ftr defenders.

The main advantage of the attacking infs is that they lose less of their combatstrenght when taking losses, than defenders does. This is why they need fewer dice than the defendes.

Lets assume that the defender has the highest average strength (it the attacker has the highest, just switch it around)
The quick and dirty formula will then be :
strenghtratio = (Defenders Average Strength) / (attackers average strength)
Number of units needed for 50% to win for the attackers will then be:
#Numbers needed = SQRT(Strength Ratio) * (#Number of Defender units)

This will change depending on the “structure” of the strength, however it will not be a Huge change. The more diverse, the better the force is.  A force defening force of 50% inf and 50% FTRs  is better than a defending force of 100% tanks.  So depending on Who I judge to have the better designed force, I add some Strength to that side when I calculate the average strength