An In Depth Analysis of ARB

Yo yo yo waddup, peeps. That sounded painful. So I've seen a lot of people doing the Math for the battle mechanics, but almost none factor in ARB. Lucky Bastard is seen as the worst strat, because of ARB. Fact is, it's one of the most hated mechanics in the game, not because it's unfair, takes away from skill, or is too hard to understand, but because people think it's plain useless. I set out to prove them wrong.

The mistake that so many people make when thinking of ARB is that it's less useful because you need that original 25% to get its effects. However, consider a battle of 16 infantry vs. 8 infantry. Assuming average attack/defence (average attack is half the attack value plus .5- remember this), that's going to have around 30 battles, and that means ARB will kick in about 7 times. Get it out of your minds that ARB has no power- once you have that first 25%, it's extremely strong.

The point of this is to find out the average attack added by values of ARB, and to show it's more powerful than most people think. For the purpose of simplicity, I'm using average attack to mean average attack/defence.

WARNING. LOTS OF TEXT AHEAD.



If you guys hate, are bored by, or suck at Math, just skip this section and go to results please; don't bother going "tl,dr hurr durr". Otherwise, you may find this interesting. This is pretty much all algebra, anyway, so it shouldn't be that bad.

As you guys probably know, whether you get to a certain threshold of ARB is determined by (let 'a' be ARB and c1(a) be the chance) c1(a)=.25(2/3)^(a-1). If you have 2 ARB, as infantry do, you'll have a 16.667% chance of getting 2 ARB. Below is a table of A values 1 to 10.

A	1	2	3	4	5	6	7	8	9	10
c1(a)	.25000	.16667	.11111	.07407	.04938	.03292	.02195	.01463	.00975	.00650

It is so tedious to create tables in BBcode... Anyway, the above is all well and good, but it's not weighted. A 30% chance of getting 2+ attack is different from 30% chance of 1+ attack. So we have to multiply the entire formula by 'a'. Let c2(a) be the weighted chance of getting 'a' ARB.
c2(a)=a*c1(a)=.25a(2/3)^(a-1).

A	1	2	3	4	5	6	7	8	9	10
c2(a)	.25000	.33333	.33333	.29630	.24691	.19753	.15364	.11706	.08779	.06503

Unfortunately, this still doesn't give us the correct values for the average attack added by ARB, because we're only looking at the chances of getting at least a value of ARB. We need the chances for getting a specific value of ARB, and to do that, we have to assume infinite ARB and go from there (It'll make sense in a sec). Assuming infinite ARB, let c3(a) be the chances of getting only 'a' ARB. So c3(4) is the chance of getting 4 ARB and stopping. c3 is just subracting exponents, so:
c3(a)=[.25(2/3)^(a-1)]-[.25(2/3)^a]

A	1	2	3	4	5	6	7	8	9	10
c3(a)	.08333	.05556	.03704	.02469	.01646	.01097	.00732	.00488	.00325	.00217

Now we weight the above by multiplying c3 by 'a':

A	1	2	3	4	5	6	7	8	9	10
c4(a)	.08333	.11111	.11111	.09877	.08230	.06584	.05121	.03901	.02926	.02168

That's all we need to finally get the average attack added by ARB. I have no idea how to put sigmas in here, so I'll just explain it out. For the power of the ARB (Let that be c5(a)), we need to add (all values being weighted) the chances of getting each ARB prior to that, and the weighted chance of getting that ARB. Obviously we can't have infinite ARB, so the last value must be from c2. Therefore, the chances for, say, 5 ARB would be:
c5(5)=c4(1)+c4(2)+c4(3)+c4(4)+c2(5)

ARB	1	2	3	4	5	6	7	8	9	10
c5(a)	.25000	.41667	.52778	.60185	.65123	.68416	.70610	.72077	.73049	.73699

Don't act like you didn't scroll straight down here. Alright, long story short, that table right up there ^ shows the average attack added by each ARB level. So if you have 2 ARB, it will, on average, add .41667 to every attack you make in battle. As I said at the top, one attack unit is worth .5 average attacks- the average attack of a tank (att=8) is 4.5. This is found by taking the average of all possible attack values, (1,2,3,4,5,6,7,8). from that we can see that average attack = half the attack value plus .5.
So to get the attack units added by each level of ARB, we multiply c5 above by 2:

ARB	1	2	3	4	5	6	7	8	9	10
Attack Added	.50000	.83333	1.05556	1.20370	1.30247	1.36831	1.41220	1.44147	1.46098	1.47399

Crazy enough, Lucky Bastard ARB ADDS 1.5 ATTACK/DEFENCE TO EACH UNIT. LB militia are more powerful than IF militia! Unfortunately, LB only adds about .5 attack to tanks and infantry, because they already have 2 or 3 ARB.

Even outside of LB, ARB is amazingly effective. 3 ARB adds about 1 att/def to a unit. This means that the 4/7 att/def ratio for infantry is now 5/8, enough to give the attacker a sizeable advantage.

However, this information doesn't make most of the Lucky upgrades any more useful. Lucky Infantry adds .2 att/def to a unit, Lucky Marines adds .15. But Lucky Militia, on the other hand, is incredibly useful, as it adds .5 att/def to your militia, turning the tide of battles. I got that upgrade right after realising out how much of a difference it makes.



I hope this allows you to make more sense of ARB. Just goes to show that doing the Math leads to better understanding of the game. If you have any questions, clarifications, corrections, or improvements, comment below. Thanks for reading, cheers!

This makes me want to experience with LB myself.

I feel accomplished for actually reading the math. c: I'm in algebra 2 right now, so it all made sense.
But kudos to you for figuring out all the math in this.

It is so tedious to create tables in BBcode... Anyway, the above is all well and good, but it's not weighted. A 30% chance of getting 2+ attack is different from 30% chance of 1+ attack. So we have to multiply the entire formula by 'a'. Let c2(a) be the weighted chance of getting 'a' ARB.
c2(a)=a*c1(a)=.25a(2/3)^(a-1).

c2(2)=2*c1(2)=0,25*2*(2/3)^(2-1)=(1/2)(2/3)=3/6=0,5

c2(3)=2*c1(3)=0,25*3*(2/3)^(3-1)=(3/4)(4/9)=12/36=0,33 recurring

c2 value is an overlook I guess, same for c4(1,2), off by 0,00001

Though I don't understand why you would multiply by a in c2, if you explain this it's likely I have more questions and your conclusion of adding 1,5 att/def only happens when you're extremely lucky. I think I can't get that weighing you do, the only thing I recall that has weighing in mathematics is weighted average or something similar, not sure about english terminology.

btw Σ, either character map or control panel localization greek keyboard layout

Skrevet av Aristosseur, 13.09.2012 at 10:14

your conclusion of adding 1,5 att/def only happens when you're extremely lucky.

Actually If a person were lucky the att/def bonus would be even higher. Mathdino calculated an average so, statistically that 1.5 bonus will always be there, lucky or not.

Skrevet av Guest, 15.09.2012 at 17:38

totaly agreed!!!

Skrevet av Aristosseur, 13.09.2012 at 10:14

c2(a)=a*c1(a)=.25a(2/3)^(a-1).

c2(2)=2*c1(2)=0,25*2*(2/3)^(2-1)=(1/2)(2/3)=3/6=0,5

c2(3)=2*c1(3)=0,25*3*(2/3)^(3-1)=(3/4)(4/9)=12/36=0,33 recurring

c2 value is an overlook I guess, same for c4(1,2), off by 0,00001

Though I don't understand why you would multiply by a in c2, if you explain this it's likely I have more questions and your conclusion of adding 1,5 att/def only happens when you're extremely lucky. I think I can't get that weighing you do, the only thing I recall that has weighing in mathematics is weighted average or something similar, not sure about english terminology.

btw Σ, either character map or control panel localization greek keyboard layout

Been busy lately, sorry for not responding. For c2(2), you made an error near the end: (1/2)(2/3) is not 3/6, it's 2/6, making it 1/3 or .33 recurring.

I'm not sure/can't remember if there's such a term as a weighted probablity, and if not, I coined it . I'm weighing them because, in the long run, (using an arbitrary example here) a 10% chance for 1+ attack is not equivalent to a 10% chance for 2+ attack; it's more equivalent to a 5% chance for 2+ attack.

Why do I still feel like the ARB is useless =/

Skrevet av Dr Lecter, 17.09.2012 at 22:58

Why do I still feel like the ARB is useless =/

Because it doesn't give you concrete results; in smaller battles, ARB has little to no effect. In larger battles, ARB is the difference between capping your opponent, and losing your giant general stack. You just don't notice and attribute your wins and losses to it, much like how charismatic people don't notice themselves being charismatic, and intelligent people don't notice when they're being smart.b

Ahhh,ARB,not just that it's hard to understand,it's the only thing that keeps 1 **ING MILLITA,WHO RUINS ALL OF YOUR 1VS1,3VS3 OR ANY **ING THING IN THE GAME AT ALL

AFTER AFTERWIND CIVIL WAR: ARB IS ABOLISHED!

Skrevet av Mathdino, 13.09.2012 at 00:27

Yo yo yo waddup, peeps. That sounded painful. So I've seen a lot of people doing the Math for the battle mechanics, but almost none factor in ARB. Lucky Bastard is seen as the worst strat, because of ARB. Fact is, it's one of the most hated mechanics in the game, not because it's unfair, takes away from skill, or is too hard to understand, but because people think it's plain useless. I set out to prove them wrong.

The mistake that so many people make when thinking of ARB is that it's less useful because you need that original 25% to get its effects. However, consider a battle of 16 infantry vs. 8 infantry. Assuming average attack/defence (average attack is half the attack value plus .5- remember this), that's going to have around 30 battles, and that means ARB will kick in about 7 times. Get it out of your minds that ARB has no power- once you have that first 25%, it's extremely strong.

The point of this is to find out the average attack added by values of ARB, and to show it's more powerful than most people think. For the purpose of simplicity, I'm using average attack to mean average attack/defence.

WARNING. LOTS OF TEXT AHEAD.

My Calcululations

If you guys hate, are bored by, or suck at Math, just skip this section and go to results please; don't bother going "tl,dr hurr durr". Otherwise, you may find this interesting. This is pretty much all algebra, anyway, so it shouldn't be that bad.

As you guys probably know, whether you get to a certain threshold of ARB is determined by (let 'a' be ARB and c1(a) be the chance) c1(a)=.25(2/3)^(a-1). If you have 2 ARB, as infantry do, you'll have a 16.667% chance of getting 2 ARB. Below is a table of A values 1 to 10.

A	1	2	3	4	5	6	7	8	9	10
c1(a)	.25000	.16667	.11111	.07407	.04938	.03292	.02195	.01463	.00975	.00650

It is so tedious to create tables in BBcode... Anyway, the above is all well and good, but it's not weighted. A 30% chance of getting 2+ attack is different from 30% chance of 1+ attack. So we have to multiply the entire formula by 'a'. Let c2(a) be the weighted chance of getting 'a' ARB.
c2(a)=a*c1(a)=.25a(2/3)^(a-1).

A	1	2	3	4	5	6	7	8	9	10
c2(a)	.25000	.33333	.33333	.29630	.24691	.19753	.15364	.11706	.08779	.06503

Unfortunately, this still doesn't give us the correct values for the average attack added by ARB, because we're only looking at the chances of getting at least a value of ARB. We need the chances for getting a specific value of ARB, and to do that, we have to assume infinite ARB and go from there (It'll make sense in a sec). Assuming infinite ARB, let c3(a) be the chances of getting only 'a' ARB. So c3(4) is the chance of getting 4 ARB and stopping. c3 is just subracting exponents, so:
c3(a)=[.25(2/3)^(a-1)]-[.25(2/3)^a]

A	1	2	3	4	5	6	7	8	9	10
c3(a)	.08333	.05556	.03704	.02469	.01646	.01097	.00732	.00488	.00325	.00217

Now we weight the above by multiplying c3 by 'a':

A	1	2	3	4	5	6	7	8	9	10
c4(a)	.08333	.11111	.11111	.09877	.08230	.06584	.05121	.03901	.02926	.02168

That's all we need to finally get the average attack added by ARB. I have no idea how to put sigmas in here, so I'll just explain it out. For the power of the ARB (Let that be c5(a)), we need to add (all values being weighted) the chances of getting each ARB prior to that, and the weighted chance of getting that ARB. Obviously we can't have infinite ARB, so the last value must be from c2. Therefore, the chances for, say, 5 ARB would be:
c5(5)=c4(1)+c4(2)+c4(3)+c4(4)+c2(5)

ARB	1	2	3	4	5	6	7	8	9	10
c5(a)	.25000	.41667	.52778	.60185	.65123	.68416	.70610	.72077	.73049	.73699

My Results

Don't act like you didn't scroll straight down here. Alright, long story short, that table right up there ^ shows the average attack added by each ARB level. So if you have 2 ARB, it will, on average, add .41667 to every attack you make in battle. As I said at the top, one attack unit is worth .5 average attacks- the average attack of a tank (att=8) is 4.5. This is found by taking the average of all possible attack values, (1,2,3,4,5,6,7,8). from that we can see that average attack = half the attack value plus .5.
So to get the attack units added by each level of ARB, we multiply c5 above by 2:

ARB	1	2	3	4	5	6	7	8	9	10
Attack Added	.50000	.83333	1.05556	1.20370	1.30247	1.36831	1.41220	1.44147	1.46098	1.47399

Crazy enough, Lucky Bastard ARB ADDS 1.5 ATTACK/DEFENCE TO EACH UNIT. LB militia are more powerful than IF militia! Unfortunately, LB only adds about .5 attack to tanks and infantry, because they already have 2 or 3 ARB.

Even outside of LB, ARB is amazingly effective. 3 ARB adds about 1 att/def to a unit. This means that the 4/7 att/def ratio for infantry is now 5/8, enough to give the attacker a sizeable advantage.

However, this information doesn't make most of the Lucky upgrades any more useful. Lucky Infantry adds .2 att/def to a unit, Lucky Marines adds .15. But Lucky Militia, on the other hand, is incredibly useful, as it adds .5 att/def to your militia, turning the tide of battles. I got that upgrade right after realising out how much of a difference it makes.

Conclusion

I hope this allows you to make more sense of ARB. Just goes to show that doing the Math leads to better understanding of the game. If you have any questions, clarifications, corrections, or improvements, comment below. Thanks for reading, cheers!

...are you a wizard?