Say NO to nerfism against goblinoids!

[Chazar]

Consider a battle of 6 "balanced" orcs vs 6 light Infantry: only 33% chance for the orcs to win!

Ok, now add the dreaded assassin to both sides:
Assassin +8% group ambush: 37%
Assassin +40% group ambush: 40%
Assassin +64% group ambush: 46%
...so the poor green skins still get pelted.

Or how about a paladin in shiny armor added to both sides:
Paladin +4: 42%
Paladin +10: 45%
Paladin +19: 47%
...nope, even the best Palading preaching cannot turn them to the better either.

So lets see how 6 orcs fare against real strong units - after all, that is were ambush helps best. Let's pit both against 2 battle blessed griffins in a city with +15 wall:
6 orcs: 7% (with 61% survival chance for 1st first griffin)
6 lInf: 1% (with 83% survival chance for 1st griffin)
...the advantage is as meager as a nerfed orc.

Adding a level 6 assassin with 6*8%=48% group ambush against those griffins:
6 orcs + Assassin 48%: 98%
6 lInf + Assassin 48%: 95%

Or adding a level 6 paladin with +13 Leadership against those griffins:
6 orcs + Paladin(+13): 36%
6 lInf + Paladin(+13): 27%
So heroes mitigate that slight advantage again.

However you turn it, the orcs are no fearsome warriors after the rebalancing. For 100% more upkeep and 50% more initial cost as compared to light infantry, you get a unit that is clearly inferior, unless it seeks shelter in a damp swamp. Of course, the orcs are a bit better at plundering, but isn't this balanced by the loss of a movement point and the construction ability already, as compared to light infantry or light cavalry that gets good terrain bonuses? Note that the same obvservations about the ambush nerf roughly also applies to the nerfed wolf rider.

So the poor green skins are driven into their secluded swamp reserves by the newcomer assassin, which was (and properly still is) way too powerful. Now that sounds unfair somehow, doesn't it?


PS:
This tongue-in-cheek post is intended as food for though, as a ground for further discussion based on actual numbers. Note that an orc would actually need an impressive ambush value of 25% to reach a 50% success chance against light infantry. I found 25% ambush to balance out 2 meager strength points quite surprsing - did I get the numbers wrong? (Note that the value applies to groups of 6 orc vs 6 light Inf; while 16% ambush is enough for 50% success for 1 orc vs. 1 light Inf.)

I actually like the diversity introduced by non-10 stremgth 1-turners. For flavour, I would prefer the orc to be throw-away shock troops, that kill en masse, and that are also killed in masses. A blood bath unit. Say, how about strength 14, 14% ambush and just 1 hit point? Wouldn't that be fitting?

[KGB]

Chazar,

I could not agree more. The Orc and Wolfrider suffered because of the Assassin hero. The only change even remotely justified was the increased upkeep for the Wolfrider and even that was iffy at best.

I found 25% ambush to balance out 2 meager strength points quite surprsing - did I get the numbers wrong? (Note that the value applies to groups of 6 orc vs 6 light Inf; while 16% ambush is enough for 50% success for 1 orc vs. 1 light Inf.)

Nope, you have it right although it's better to look at 1 unit vs 1 unit rather than 6v6 or 4v4 etc. Your math is slightly off though. I realize you calculated out the winning % as 58-42 for the Lt Inf and came up with 16% (difference between the 2). But it's actually only 14% that's needed because what you want is 50% outcome. To get that you take 50% (what you want) divide by 58% (what you have) to get .86 or 86. The remainder is the ambush: 14%. Looking at that from the other way, subtract the ambush (14%) from 100% to get 86% left that they really fight over. Then from that 86%, 58% of that is 49.88% Lt Inf win chance.

The reason is not because it's 2 strength points, but as I pointed out in another thread it's 20% strength difference between the 2 units. So that's a 16 vs 20 or a 24 vs 30 or a 32 vs 40 equivalent change at higher base 10 strength values. That's why adjusting low strength units is always a bigger change compared to say just making a 35 strength unit 33 strength - a 6% difference.

I actually like the diversity introduced by non-10 stremgth 1-turners. For flavour, I would prefer the orc to be throw-away shock troops, that kill en masse, and that are also killed in masses. A blood bath unit. Say, how about strength 14, 14% ambush and just 1 hit point? Wouldn't that be fitting?

This would definitely make them 1 dimensional. As in only useful against high strength units because they would be wasted trying to kill Lt or Hv Infantry since you'd need a 2-1 or 3-1 unit ratio. So if they are going to be that 1 dimensional they may as well just be 10 strength with 14% ambush. It doesn't matter to me either way as long as the current Orc is fixed from his current useless state.

KGB

[LPhillips]

Just to weigh in, orc infantry indisputably useless now. Before, they were a very viable 1-turn unit. I don't really see what is accomplished by this aspect of the Assassin nerf.

If the general nerfs were intended to weaken the Assassin until he can be properly balanced (more significant code changes like a group ambush cap, no main ability stacking, or whatever Piranha and Snotling decide is best for the next version) then I can see the reasoning. However, the practical effect on the orc is awful. He's only got 6% ambush; even a reduction to 9 strength wouldn't really be a balanced move. The dynamic is very different for a unit with "significant" ambush ability, like wolf riders. I'm still looking forward to the time when these units can be restored to their proper strength. I don't see any direct correlation between the orc and the Assassin, given the very low ambush ability. The assassin adds nearly double that to any unit at level 1.

Thanks for your consideration,
LP

[SnotlinG]

We hear you

We do not plan to remove stacking from Hero group ambush, arguments are:
- adding complexity since group ambush for units are stacking so makes no sense for it not to stack for heroes
- Assassin group ambush has been a bit decreased already, so it will take more time to reach high values (also you will sacrifice other herotypes to get 2 assassins etc...)
- We will add a max value on group ambush of 70-75%, so you will only be able to reach higher if you have normal ambush also.
- Leveling of assassins are a bit harder since they now start with UL 7, and also after the bug fixing split xp when having 2+ heroes in the same stack.

As for the Orcs (we also like specialized units), what would you think of
- battle 7, ambush 10?
or
- no ambush but make the orc a relatively weak swarm unit?
btw, the current idea for orcs are not that they should win agains light infantry, but rather that they are a specialized unit which should be used to try to take out high-cost units/superstack, or to fight/lead armies through swamp terrain. So a comparasion with lightinf isnt saying that much

[Chazar]

math is slightly off though.

@KGB: No, I think my math is correct. However, quarreling over 2% is silly.

Unfortunately, I happen to be a mathematician and thus I take both pride and pleasure from silly quarrels over numbers, so here we go in all glorious detail: Everyone else, please skip to the next post. Thanks!

The chance of a unit with strength x to score a hit against a unit with strength y is

X*(1-Y) / (X*(1-Y) + Y*(1-X) )

where X=x/100 and Y=y/100, as explained in the Warpedia and in this post. So for a single hit, the orc with strength 8 has a chance of 43.9024% versus strength 10.

Now let us look at the whole battle, which include both units having two hit points each. There are 4 possible outcomes for the 1vs1 battle, in which the orc is winning:
1. Orc makes a successful ambush.
2. Orc fails to ambush; Orc scores a hit, Orc scores a hit.
3. Orc fails to ambush; Inf scores a hit, Orc scores a hit, Orc scores a hit.
4. Orc fails to ambush; Orc scores a hit, Inf scores a hit, Orc scores a hit.

The probability for each path is obtained by multiplying along the entire path. Let K denote the ambush probability, i.e. ambush score divided by 100, then we have:

1. K
2. (1-K)*(0.439)*(0.439)
3. (1-K)*(1-0.439)*(0.439)*(0.439)
4. (1-K)*(0.439)*(1-0.439)*(0.439)

The overall successchance is then determined by summing up over all these exclusive outcome paths:

K + (1-K)*(0.439)(0.439)*(1+(1-0.439)+(1-0.439))

Now we are looking for a value k such that the above formula equates to 0.5.

For K=0.16 we get 0.50355 and for K=0.15 we get 0.49764, so I'd say an ambush value of 16% is closer to the balanced outcome of a 50% chance for each combatant.

Of course, for safety, let us check the inverse. The chance for the infantry winning is similarly computed by the formula

(1-K)*(1-0.439)*(1-0.439)*(1+0.439+0.439)

For K=0.16 we obtain 0.49644. Since 0.50355+0.49644 = 1 (i.e. there is a 100% chance that the orc or the infantry wins), my calculations seems to be quite on track.

So again, where exatcly do you suppose that I go wrong?

[Chazar]

the current idea for orcs are not that they should win agains light infantry, but rather that they are a specialized unit which should be used to try to take out high-cost units/superstack,

Yes, that is understood, but my point was that throwing 6 orcs against strong units is just as good as using 6 light infantries. The kill chances are almost the same, but the light infantry is at least usable otherwise.

Let's pit 8 orcs and 8 infantries against 2 spiders in a 10-wall city, a more reasonable battle to happen:
8 nerfed orcs attack: 36.6% and 78.2% survival chances for the two spiders
8 lInf attack: 48.2% and 88.5% survival chances for the two spiders

I think that is too small a change compare to the difference in cost and the overall utility. When I buy production, I don't yet know what I will need. If I buy a specialist, then it better be good at is role. So lets try your suggestion:

8 units with Str=7, Amb=10%: 28.6% and 69.2% survival chances for the two spiders.

So 10% ambush seems about to be ok, but I think you can still keep the strength at 8, for otherwise they would still suck too much: just 21.3% of 8 of these against 8 light Inf seems pretty bad. The survival chances for the infantries are: 0.1%, 1.1%, 5.1%, 14.7%, 29.9%, 47.9%, 65.0%, 78.7% So one can expect about 3 Infantries to survive the carnage. Meh.

Ambush for the Orc and the Orcish Wolf Rider is pretty thematic, so please keep that!

One Hitpoint and high strength & ambush might work out towards their role for killing high-strength units, but being somewhat useless against masses. Here are the numbers with 14 strength, 12% ambush and just 1 hitpoint per Orc:
8 of these against 8 lInf: 31.1% chance for the orcs to win.
8 against 2 spider in 10 wall city: 22.8% and 61.3% survival for the spiders
8 against 2 blessed griffins in 15 wall city: 30.2% and 70.0% survival for the spiders.

[KGB]

Chazar,

Unfortunately, I happen to be a mathematician

Me too! I have an actual BS in Math, although it's over 20 years since I graduated and I only use math now as it relates to programming robots and motion control systems. I took an online calculus course recently at Stanford as part of a Machine Learning course and was surprised at how rusty my advanced calculus was

The chance of a unit with strength x to score a hit against a unit with strength y is
X*(1-Y) / (X*(1-Y) + Y*(1-X) )

Now let us look at the whole battle, which include both units having two hit points each. There are 4 possible outcomes for the 1vs1 battle, in which the orc is winning:
1. Orc makes a successful ambush.
2. Orc fails to ambush; Orc scores a hit, Orc scores a hit.
3. Orc fails to ambush; Inf scores a hit, Orc scores a hit, Orc scores a hit.
4. Orc fails to ambush; Orc scores a hit, Inf scores a hit, Orc scores a hit.

The probability for each path is obtained by multiplying along the entire path. Let K denote the ambush probability, i.e. ambush score divided by 100, then we have:

1. K
2. (1-K)*(0.439)*(0.439)
3. (1-K)*(1-0.439)*(0.439)*(0.439)
4. (1-K)*(0.439)*(1-0.439)*(0.439)

Absolutely correct.

The overall successchance is then determined by summing up over all these exclusive outcome paths:

K + (1-K)*(0.439)(0.439)*(1+(1-0.439)+(1-0.439))

You have a small mistake here which I bolded. Not sure if its in your math or just a typo in what you entered or both. Cases 3 and 4 are one and the same so you just need 2*case 3 (probability wise you'd write this as 3-chose-2 where the 2 are 2 hits on the Lt Inf). So the formula should be:

K+(1-K)*(0.439)*(0.439) + 2*((1-K)*(0.439)*(1-0.439)*(0.439)) = .50

factoring like terms gives:

k + (1-k)*[(0.439)*(0.439) + 2*(0.439)*(1-0.439)*(0.439)] = .50
k + (1-k)*[.4134] = .50 // Note: .4134 = Orc win % vs Lt Infantry without ambush
k + .4134 - .4134k = .50
.5866k = .0866

multiply both sides by 1.704 to get:

k = 14.75


I think mine (below) is much easier to do:

I did my math for the Orc by monte-carlo simulation. I ran 10000 combats of 1 Orc vs 1 Lt Inf. The Orc win 41.5% of the time (virtually identical to the actual .4134). So the K value (ambush) needs to reduce the Lt Inf winning value so it reaches 50% instead of 58.5. This is where you use the formula I gave:

50/58.5*100 = 85.4. So you need a K value of 14.6 or 14%.

You can see this is right by imagining the Orc fighting a unit that was 90% likely to beat the Orc. How much ambush would the Orc then need to be 50/50 against such a unit?

50/90*100= 55.5. So you need a K value of 45.5 or 45% ambush when the Orc is only 10% like to win.

KGB

[KGB]

SnotlinG,

I'd just prefer the Orc cost increase to 175 and retain it's 10 strength and increase to 8 ambush. That way it can still be semi-useful against 1 turn units while at the same time having a chance to kill a super strong unit with it's ambush.

If you look at the example Chazar gave but instead change it to defending instead of attacking and imagine having 20 strength Hv Inf (no wall bonus) against 2 spiders then the Orc ambush is totally underpowered. It really only has any value against stacks with strength of 65+ where even Hv Inf on defense are useless. That's just too rare an occurrence to justify ever having /making Orcs. It's even worse finding them in neutrals because you have to pillage them out because they can't be used for expansion due to their low strength.

Plus Orcs normally have a lot of brute strength but are dumb. So they should not be weaker than other 1 turn units in straight up fights. Goblins on the other hand would be smaller/weaker but that would require another unit

KGB

[LPhillips]

The 10 strength is really a requirement for basic expansion infantry. If you want a unit with high ambush, you probably ought to be pretty radical with it (6-7 strength, 15% ambush) for around 200 gold. That would be a good "Goblin". Orcs with 10 strength 6-8% ambush for $150 would be good. Or even Orcs with 175 cost, 10 strength, 10% ambush if you want an in between unit. Again, if you had Goblins, you could have orcs with 12 strength, 8% ambush, $225-250. Right now they're more like an 80-90 gold unit.

[Chazar]

If you look at the example Chazar gave but instead change it to defending instead of attacking and imagine having 20 strength Hv Inf (no wall bonus) against 2 spiders then the Orc ambush is totally underpowered. It really only has any value against stacks with strength of 65+ where even Hv Inf on defense are useless. That's just too rare an occurrence to justify ever having /making Orcs. It's even worse finding them in neutrals because you have to pillage them out because they can't be used for expansion due to their low strength.

Yes, that is precisely the main point here! I totally agree!

@KGB: in short, your calculation is off, since you are accounting for a battle that goes "inf takes hit, inf takes hit, orc takes hit", which is not a possible battle in warbarons; long answer:

Chazar,

The overall successchance is then determined by summing up over all these exclusive outcome paths:

K + (1-K)*(0.439)(0.439)*(1+(1-0.439)+(1-0.439))

You have a small mistake here which I bolded. Not sure if its in your math or just a typo in what you entered or both. Cases 3 and 4 are one and the same so you just need 2*case 3 (probability wise you'd write this as 3-chose-2 where the 2 are 2 hits on the Lt Inf).

No, cases 3 and 4 are distinct cases, even though the outcome (winning with 1 hit point left) are exactly the same. (Just like cases 1 and 2, which also have the same outcome "winning with 2 hit points left" albeit cases 1 and 2 have different probabilities.)

To see this, consider throwing two equal six-sided dice and compute the probability to roll a sum of 4. This is achieved by 3 distinct outcomes (1,3),(2,2),(3,1). Each has the probability 1/6*1/6 so we get 3*(1/6*1/6) = 3/36 as the overall probability for rolling a sum of 4. Do this for all outcomes 2-12 and add up the numbers, since there is 100% chance to roll a sum between 2 and 12 with two six-sided dice, this should then add up to one, which it does: Each outcome for two specific numbers has a chance of 1/36 to occur, and there are 6 * 6 = 36 distinct outcomes, so 36*1/36=1. If you would treat (1,3) and (3,1) as the same, and (1,4) and (4,1), and so on, then you would count only 21 distinct outcomes, leading to 21/36. So the order of events does matter if one is looking at probabilities.

In order to simplify computations, one sometimes tries identify outcomes with equal probabilites early on, and this is then where the n-choose-k binomial coefficient sometimes comes in handy, but then the whole computation needs to be changed: we have 2 outcomes for the orc winning, i.e. "orc winning with 2 hitpoints left" and "orc winning with 1 hitpoint left". Note that the two paths leading to outcome "orc winning with 2 hitpoints left" have a distinct probability, so this is a first sign that it is not worth the trouble computing the probability by outcomes rather than paths. Furthermore, note that there are only 2 possible paths to achieve the outcome "orc winning with 1 hitpoint left" and not 3 as your application of 3-choose-2 would tell you. This is because the path "inf takes hit, inf takes hit, orc takes hit" is not possible in warbarons, so you need to adjust for that as well. Overall, it is much simpler to consider all possible paths here, rather than all distinct outcomes with all their special cases.

However, for our poor orc, you are right that the probabilities for case 3 and 4 to happen are exactly the same, so yes, we can just add in the probability for one of these two cases twice. By looking at all paths, we've already accounted for the 2 hitpoints of each unit, so we don't need 3-choose-2 here at all. (3-choose-2 is not a probability, it is just the number of distinct ways to choose 2 items out of 3 distinct ones. However, 3-choose-2 ignores the order in which things are choosen (i.e. 3-choose-2 = 3 and not 6), which is again not what we want here, since we are considering all paths, not all outcomes.)

So we have K for case 1,
(1-K)*(0.439)*(0.439) for case 2, and
(1-K)*(0.439)*(0.439)*(1-0.439) for case 3 and also for case 4, which we have to add twice, leading to the formula I originally had:

K + (1-K)*(0.439)*(0.439) + 2 * (1-K)*(0.439)*(0.439)*(1-0.439)

EDIT: there was a typo in the above formulas within this post, which I have now corrected. The formula it refers to in my previous posts was ok though. Although it is oddly arranged.
Orginially I had written above:

So we have K for case 1,
(1-K)*(0.439)*(0.439) for case 2, and
(1-K)*(1-0.439)*(1-0.439) for case 3 and also for case 4, which we have to add twice, leading to the formula I originally had:

K + (1-K)*(0.439)*(0.439) + 2 * (1-K)*(1-0.439)*(1-0.439)