How strong does infantry defend?
The first answer might be, that it defends twice as good as it attacks. But that’s not corect, even though it’s defend value is 2 and it’s attacking value is 1, the actual number of units plays an important role. A hit by an attacking infantry takes away more strength of the oponent than a hit by a defender.
In battles of low numbers border effects are pretty difficult. If you attack one infantry with two, there’s a probabilyty of getting two hits, which won’t help you. So lets’ focus on a largescale attack, which might occur in a battle about Karelia.
if n infantries attack m infantries,
hits by the attacker:
n1
hits by the defender:
m2
The forces are of the same strength if the ratio priour to the fight is the same as the ratio after the fight:
(n/m)=(n-m2)/(m-n1)
mn - nn = mn - 2mm
nn = 2mm
n = 1.414… * m
hence the defending strength of infantry is aproximatly 1.4 time the attacking strength. Border effects of low numbers or other units present, might change this, but this number certainly gives you a clue, how many infantries Germany or Russia needs for a fight.
[Edited according to C_F’s post]
Whatever happened to kissing the dice and hoping for the best! J/k very nice, only one thing. When you said:
hence the attacking strength of infantry is aproximatly 1.4 time the defending strength
I think you meant that the defending strength was 1.4 times the attacking strength. Obviously infantry is better at defending than attacking, but everyone also knows that 10 inf attacking 5 inf will win almost every time. So you need 1.4 times as many attakcers as defenders when it is just infantry. I tried it on an odds grapher and it is just about dead even, very helpful, thanks.
Meijeng, I’m not sure exactly what that means n=1.41M? Does that mean you would need 41% more infantry attacking the same number in order to have a 50/50 chance? Ah yes, just ran it through a battle sim and that’s what it means. Nice bit of math.
BB
Normally your army consists of several units, the attacker has some infantry, with attacking value 1, and some tanks or planes, with value 3, the defender has units with value 2, tanks and infantry, and units with value 4, planes.
I tried to extend my aproach to other kinds of attack and came up with the following question.
If you have n tanks and you want to attack m infantries, how many infantries do you have to take at least with you if you don’t want to loose tanks? (With a probability of 50% you don’t loose any tank)
This number I will denote by f(n,m).
By scalability
f(an, am) = af(n,m)
and hence
f(n,m) = mf(n/m,1) = m*g(n/m)
I didn’t suceed to make much analytic progress, so I decided to use a battle simulator to determine g(n/m) for certain values of n/m empirically and to fit a function to the resulting data.
The result is:
g(n/m) = 108 * 10^(-n/m) + 33
I admit that this function is not very handy to be calculated while actually playing, but it might be handy if someone decides to program an AI.
Special values of g(n/m):
g(0.0) = 1.414
g(0.1) = 1.18
g(0.2) = 1.00
g(0.5) = 0.68
g(1.0) = 0.47
How to use these values:
Example 1:
Suppose there are 17 infantries in Karellia and you would like to attack with 9 tanks. The ratio 9/17 is slightly higher than 0.5 so lets considre g(9/17) to be 0.66 or 2/3. If you multiply this with 17 you get approximatly 12. This means, that if you take 12 infantries with you, you will win with a probability of 50% without loosing a tank.
Example 2:
Suppose you got 10 tanks and there are less than m<50 infantries in Karelia.
10/m > 10/50 = 0.2
g(10/m) < g(0.2) = 1
As long as there are less than 50 infantries in Karellia, if you take as many infantries with you as there are in Karellia the chance for winning the battle without loosing a tank is higher than 50%.
The function doesn’t take into account the cost effectiveness nor effects of different movement rates. Nonetheless it’s a rather interesting excercies. Once you start to get into more subjective values like assigning values to movement rates it becomes a real nightmere. Way to many variables to try and optimize. Time to take out the trusty ole genetic learning algorithm, my favourite machine learning tool. :-)
BB
There is a mistake in the original formula.
The number of hits is not m2 or n1, because that implies that each m removes 2n a turn, which is impossible (only heavy bombers role more than one dice, after all).
Rather, the proper values are:
m2/6 and n1/6
In English, this means that six defenders, on average will hit twice per round of combat (which is something that should be intuitive to most a&a gamers), and it also takes six attackers to hit one defender per round.
Nonetheless, the answer works out to be the same thing:
n/m = (n-m/3)/(m-n/6)
n(m-n/6) = m(n-m/3)
nm - nn/6 = mn - mm/3
nn/6 = mm/
n*n = (6/3)mm
n = 1.414… * m
There is no error in the original formula as the ratio 1:2 =1/6:2/6
The 1/6 and 2/6 might make it a bit easier to follow but it’s not in it’s ‘simplist form’ which is a required trait to get full marks in acedemia.
BB
There is no error in the original formula as the ratio 1:2 =1/6:2/6
There is an error, it just happens to work out to the same thing in the end.
Meijing said:
hits by the attacker:
n1
hits by the defender:
m2
The mistake is that he is saying that two times as many hits defending versus attacking is the same as two hits for every defender, and one hit for every attacker. This is not really true - a defender can only possibly hit once, and averages hits only a third of the time.
You can easily demonstrate the flaw of the formula for yourself - just try it out with the same number of attackers and defenders, and see what happens, OK?
As for the simplicity barb, one should never sacrifice correctness to get a simpler looking formula, either. By your logic, if I had $4 and you had $2, that is the same as me having $2 and you having $1. Perhaps the second is simpler, but neither of us would call it the same!
Yes, there is an error, my fault. Works out to the same end, because it only depends on the relation between the hits. While thinking about it, I made the simplification of 1 and 2. But you’re right, I should haven been more carefull as I wrote it down.
To answer:
If you have n tanks and you want to attack m infantries, how many infantries do you have to take at least with you if you don’t want to lose tanks? (With a probability of 50% you don’t lose any tank)
If n is attacking infantry, m is defending infantry, and t is for attacking tanks, then the formula:
nn + 3tn = 2m*m will give you the values you need so that n/m on average stays the same each turn of battle.
This is derived from the formula:
n/m = (n-m/3)/(m-n/6-t/2)
the extra term, -t/2, makes sense because each round m is going to be reduced by half the number of tanks, on average
If both formulas offer the same output given the same input then they are equivalent formulas. If x=2y then 2x=4y and 1/6x=2/6y.
If the formulas are the same then 1 cannot be wrong while the other is correct.
When answering exam quesiton 1/6x=2/6y gets you part marks and x=2y gets you full marks.
The question is: How many more attacking infantry do you need given Y defending Infantry or tanks to have a 50/50 chance of mutual destruction.
If Y=10 then X=14. Simple.
Your money example is silly, we’re talking about the ratio not absolute amounts. You’re comparing apples to oranges, a typical tactic used in sophistory.
shrugs
BB
There’s a problem with your formula, it is not sufficient, that the ratio m/n is conserved. The number of tanks will stay the same and hence in the second round the tanks will have a higher influence on the battlefield than in the first round. This formula would be correct if the number of tanks also decreases and the ratio m/t stayed constant.
Let m0, n0, t0 denote the numbers before the first round m1, n1, t1 the numbers after the first attack.
Further assume
| A0 = n0n0 + 3t0n0 - 2m0m0 = 0
or equivalently
| (n0/m0)^2 + 3(t0/m0)(n0/m0) - 2 = 0
and hence
| n1/m1 = n0/m0
therefore
| A1 = (n1/m1)^2 + 3(t1/m1)(n1/m1) - 2
| = (n0/m0)^2 + 3(t1/m1)*(n0/m0) - 2
but since
| m1 < m0 and t1 = t0
| => (t1/m1) > (t0/m0)
which implies
| A1 > A0 = 0
so after the next round the ratio between the infantries would change.
Nevertheless this formula gives an uper bound for n and hence is helpfull.
If you use more infantry for your attack then this formula indicates, then you are on the save side.
My error is not in the final answer, which is correct, but in the deduction of this formula. I stated that the attacker will loose 2*m troops in one round, which is wrong.
Now I know how my professors must feel when I point out an error in what they do.
Despite your use of big words like “sophistry,” you are not right. You can read what Meijing has to say about it.
The original contested formula is:
(n/m)=(n-m2)/(m-n1)
The claim is that (n-m2)/(m-n1) is the ratio of n/m after one round of fighting. Try the formula for yourself to see if it matches what you would expect to see in a real a&a game (see what it says when 5 inf attacks 5 defenders, for instance).
Then try out (n-m/3)/(m-n/6), and see how that works. As you can see, it does make a difference.
And Meijing - you are right, my second formula is wrong - and I don’t know how to fix it, either.
I was talking about the infantry only example. x=1.4Y, not sure why you guys thought I was talking about anything else. My x=2Y was merely an example
Like I said, x=1.4Y or 1/6x=1.4/6 Y or 2X=2.8Y
BB
