Say NO to nerfism against goblinoids!

[KGB]

Chazar,

@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:

True. Writing 3-chose-2 wasn't the right way to express what was occurring there. As you say in your longer explanation we just want all the cases where the Orc wins with taking 1 hit. That of course has to work out mathematically to the same number for each distinct case since the odds of getting 1-4 or 4-1 with 2 6-sided dice is the same.

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)

But you do agree this works out to 14.75% (round down to 14) as I showed above right? Your original calculation of 16% must have just been an error when you expanded everything.

Effectively the formula is just:

k + (1-k) * Orc Win % = .50

The reason I use the simple formula is so it can be determined by simulation instead of expanding lots of terms that get bigger and bigger if you start dealing with units that have 3 hits (imagine all the terms for 2 3-hit units).

KGB

[Chazar]

But you do agree this works out to 14.75% (round down to 14) as I showed above right? Your original calculation of 16% must have just been an error when you expanded everything.

No, I stick with my original 16%.

However, my last post contained indeed a bad typo in the formulas at the end, which I have now corrected. I apologise for that. Let me repeat the correct formula in a better way.

There are 4 winning paths for the orc, whose probabilites we have to add up to compute the overall winning probability: (case1) + (case2) + (case3) + (case4), since case 3 and case 4 have the same probability,
this is equal to (case1) + (case2) + 2*(case3). Putting in the numbers we agreed upon earlier, we get the formula:

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

For k=0.16 I thus compute:
0.16 + (0.84*0.439*0.439) + 2*(0.84*0.561*0.439*0.439)
= 0.16 + 0.1619 + 2*(0.1619*0.561)
= 0.16 + 0.1619 + 0.1816
= 0.5035

Effectively the formula is just:

k + (1-k) * Orc Win % = .50

That formula is also correct, if you put in the correct OrcWin%. However, since both units have two hit points, this is not 0.439, but 0.40899 instead. 0.439 is the chance for a strength 8 unit to score one hit against a strength 10 unit. 0.40899 is the chance for a strength 8, hitpoints 2 unit to win an entire battle against a strength 10, hitpoint 2 unit.

Similarly we compute:
0.16 + 0.84 * 0.40899 = 0.50355

and for 14% ambush we get
0.14 + 0.86 * 0.40899 = 0.49173

So even with your formula, the poor orc need 16% ambush to be on an equal footing with a light infantry in 1vs1. Of course, due to having 2 hitpoints, things get worse for the orc in large numbers.

[Jeremy]

Well, it's obviously they enjoy math the way I enjoy puzzles. And turn-based strategy games.

The point is, the Orc is too weak at 8 strength. It's strength should be 9 or 10.

[KGB]

If you put in the correct OrcWin%. However, since both units have two hit points, this is not 0.439, but 0.40899 instead. 0.439 is the chance for a strength 8 unit to score one hit against a strength 10 unit. 0.40899 is the chance for a strength 8, hitpoints 2 unit to win an entire battle against a strength 10, hitpoint 2 unit.

Correct. I was never using .439. I was using what I my simulation computed which was about .4143.

Similarly we compute:
0.16 + 0.84 * 0.40899 = 0.50355

and for 14% ambush we get
0.14 + 0.86 * 0.40899 = 0.49173

So even with your formula, the poor orc need 16% ambush to be on an equal footing with a light infantry in 1vs1. Of course, due to having 2 hitpoints, things get worse for the orc in large numbers.

I tend to round down, as in under 50% when you can't hit exactly 50% (which is almost impossible on the accuracy Warbarons uses). The reason for that is because the Ambush chance always leaves the victor with max hp (2 in this case) so what you are really doing is heavily weighting the 2 HP success chance (.1619) so that with 14% ambush it's now .3019 when the Orc wins, it has 2 HP left. The Lt Inf will have 2 hp left .561*.561=.3147. With 15% ambush the Orc will have 2 hp left .3119.

So in reality 15% ambush is probably the closest thing possible to a 50/50 battle.

KGB

[Chazar]

... The reason for that is because the Ambush chance always leaves the victor with max hp (2 in this case) ...

I don't see how that changes anything, as all surviving units start the very next battle with full hitpoints anyway, even if it is in the same round?! (Unless you want to write a general battle calculator by chaining single battles, but this would be a bad way to design it.)

So in reality 15% ambush is probably the closest thing possible to a 50/50 battle.

Well, let us agree on 15% then. I don't mind rounding either way, as long as we agree on the math.

So, back to the important point of this thread. Is anyone else still here?

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.

Totally agreed! There ought to be more greenskins. I love the suggested Goblin.
Also, I think a 1 hp Snotling with ambush would be totally cool.

[Moonknight]

Also, I think a 1 hp Snotling with ambush would be totally cool.

I approve this as well, along with the 1 hp Piranhas with swarm ability

[KGB]

Chazar,

I don't see how that changes anything, as all surviving units start the very next battle with full hitpoints anyway, even if it is in the same round?! (Unless you want to write a general battle calculator by chaining single battles, but this would be a bad way to design it.)

What I meant here is that when you move on to face the next unit in the stack when multiple units are involved. If you win by ambush you are always going to have full HP to face the next unit. In regular combat you can move on to face the next unit with only 1 hp if you get the Orc Hit/Lt Inf Hit/Orc Hit combo cases (3&4). Ideally you want to make the Orc and Lt Inf have as close as possible to the same chance to move on with 2 HP to face the next unit.

So in reality 15% ambush is probably the closest thing possible to a 50/50 battle.
Well, let us agree on 15% then. I don't mind rounding either way, as long as we agree on the math.

We do!

In actual reality of the game even though 15% would make the 2 units equal in 1v1, in an 8v8 attack the Lt Inf wins about 57% of the time (I ran to check it) which you mentioned. But that's OK because the 15% ambush would make the Orc deadly as hell and in truth, over powered.

The real question is what ambush % would make an 8 strength Orc worthwhile. And after checking a couple of combinations, for me it would be 10% ambush. At that level I could overlook the Orc's weakness and be happy with the ambush chance. The reason for 10% is because 8 Orcs then have a chance to kill up to 2 units from ambush:

8 units, 10% ambush:
1) Chance of 0 kills = .9^8 = 43.0%
2) Chance of 1 kill = 8!/7!1!*.10*.9^7=38.3%
3) Chance of 2 kills = 8!/6!2!*.10^2*.9^6=14.9%
4) Chance of 3 kills = 8!/5!3!*.10^3*.9^5=3.3%

Since the 10% rule prevents outcomes <10%, you can't get 3 kills unless the combat part > 6.7%. Against strong units like hero stacks or cities full of Spiders/Gryphons etc it means the only outcomes for 8 Orcs are 0, 1 or 2 kills. Having a 14.9% chance of getting 2 kills with 8 Orcs would make them viable for me to use because killing 2 units in an enemy hero stack probably lets mine move in and finish off the surviving 6.

KGB

[Chazar]

Since the 10% rule prevents outcomes <10%, you can't get 3 kills unless the combat part > 6.7%.

Objection, I don't think it works like that:
From what I have read, the 90% rule means that an entire battle is rerolled if the winning side had a less than 10% chance to win. That means, that a stack of 8 really lucky orcs could always kill all units except for the last one. One can be lucky, as long as the favoured side still wins.

Actually, the game displays the battle outcome%, and here is a battle I just had today

Propability to win:
Attacker: 3.1%
Defender: 96.9%
Battle outcome: 2.3%

So the outcome was well below 10%, but that is ok, because there was still one defender standing at the very end!

Despite this, 10% ambush and 20% for the wolfrider seem ok to me, if the orc is to be kept at strength 8 or at its current price.

[KGB]

Chazar,

From what I have read, the 90% rule means that an entire battle is rerolled if the winning side had a less than 10% chance to win. That means, that a stack of 8 really lucky orcs could always kill all units except for the last one. One can be lucky, as long as the favoured side still wins.

No. That's not how it works. It eliminates 10% of the outcomes on both attacker and defenders sides. This is exactly to prevent what you described: 1 lucky unit killing WAY too many men. For example what it's designed to prevent is when 8 Lt Inf fight 8 Lt Inf you don't want 1 side winning with all 8 men left (0 losses) which can happen since it's non-zero.

So what the rule does is start from the outside edges and eliminate all outcomes until it reaches 10%. I'm making up the numbers here but for 8 vs 8 Lt Inf the numbers look something like for 1000 simulations

1, 5, 14, 30, 60, 110,130, 150 | 150, 130, 110, 60, 30, 14, 5, 1

The bell curve you'd expect with the 150 numbers representing winning with 1 man left, 130 with 2 men left and so on down to 1 chance to win with 8 men left. So the 10% rule says throw out 10% or up to 100 on each side. That eliminates 1+5+14+30 but not 60 since adding 60 more would exceed 100. So the only valid results for this battle would be the winner has 1-4 men left.

So it would be impossible for the Orcs to kill 3 units unless their battle chance + ambush chance exceeded 10%.

Actually, the game displays the battle outcome%, and here is a battle I just had today

Propability to win:
Attacker: 3.1%
Defender: 96.9%
Battle outcome: 2.3%

So the outcome was well below 10%, but that is ok, because there was still one defender standing at the very end!

Yes, it's quite legitimate to have a battle result <10%. It just means THAT particular outcome was small but still in the middle range of the bell curve. You can think about this as 8 units attack a city with 32 defenders. There are 40 outcomes possible. If they were all even, they'd be 2.5% each which is less than 10% but definitely some of them must be in the middle of the curve and thus used (each side would lose 4 outcomes in this example so you'd get a range of 4 attackers to 28 defenders if every outcome was exactly 2.5% likely).

KGB

[Chazar]

No. That's not how it works.

Well, it is really pointless to discuss this unless we have access to the actual programm code, which at least I do not have.

However, I believe piranha has access to the code, and piranha clearly states in this post::

You will see results like 6.2%, but you should not see someone win with 6.2% chance to win.

The battle is run 1000 times to find out what the chances are for each side to win. Then the real battle is run and if its less than 10% chance for one side to win, and they do win the real battle will be run again until the side with 10% to win loses.

Now that could be outdated, but since it matches my observations, it is good enough for me.