Strange "90%" rule occurence

[Seraad]

gameid = 71023
turn 56
5th battle of the turn - lone lc attacking Paris full of units.

Propability to win: (79%)
Attacker: 0%
Defender: 100%
Battle outcome: 78.7%

My lc was killed by the rule (cube indicating), but Paris had not lost any defenders - so the battle outcome was the best oucome for the defender. - but still it was slightly lower then "probability to win".
Why the lc was killed by the rule then?

The battle has no importance - I just want to make sure I understand the rule and there is no bug in it.

Regards!

[KGB]

Seraad,

'Probability to win' is a misleading title. This number is really the 90% value for this battle. Not sure if know that the 90% value is not absolute. It starts at 90% and gets reduced downwards based on the total strength of all the units in the battle (the more the units, the lower the value). So when you see 79% it means the rule is 79-21 instead of 90-10. This evens out large battles even more toward what the average outcome for that battle is.

The battle outcome (78.7) is the % chance that the battle ended exactly as it did (your Lt Calv died and exactly 0 defenders died).

It looks OK to me because pretty much the only valid outcome is no defenders dying (remove 21% on both sides leaves the middle 68% range for the battle and 78.7% on 1 single outcome eats that entire 68% up).

KGB

[Seraad]

KGB

I'm fully aware that "probability to win" is misleading in the game. That's why I've put it in quotation.

You are mistaken about the second part though. It is math - there is no "look ok" in the science. It is either "ok" or "not ok".
In this case you don't "remove" the 21% from both sides (by the way - it would be 58%, not 68%).
It is true that the outcome should be within 58% middle range, but the 78,7% IS within the range! as it "started from 0% probability (there was no "better" battle outcome for the defender). A pity I can't draw here to show it...

The rule is to prevent both sides from extremly unfortunate loses.
In this case:
78,7% - no loses
21,3% - one or more unit lost

So the 21,3 % is the worst what could happen nto the defender. But as it is still more then 21% I should be able to kill one defender before the rule make sure my lc dies.

Seraad

[KGB]

Seraad,

Yeah 58%, not 68. Late at night after a night of beers

But 'look OK' in this case = 'OK'.

Let me show you with a simple 3 unit example.

A = attacker
D = defender

Code: Select all

A  |  D     D
X  | 78.7   Y


Where X and Y are what ever you want them to be as long as they total 21.3.

Lets take X = 21 and Y = .3. You then get 21 | 78.7 .3. The rule says remove 21% on each side. So on the left that consumes ALL of A and on the right that consumes .3 so the only valid outcome is 78.7.

In order for A to have a chance to kill the numbers must be 21.1 | 78.7 .2. (or 21.2 and .1). Now given the city was full of units do you think the A chance was 21.1 or greater or was the sum of all the other units in the city at least .3? My guess is the sum of all the units in the city was >.3 so that means that A was entirely invalid and so was auto-killed.

KGB

[Seraad]

KGB,
I can't agree with you.

Let's put everything into a scheme:


I----------------------------------A------------------------------------I------B------I--C--I
I--------------------------------------------------------------------------------------------I
I--------X--------I-------------------------Z-----------------------------I--------Y--------I

where (probabilities):
A - 0 defenders die
B - 1 defender dies
C - 2 or more defenders die
X = Y - represents our "90% rule"
A+B+C = 100%
X+Z+Y = 100%

CASE 1 (my case, roughly showed in the "diagram" above)):
A = 78.7%
X = Y = 21%
Z = 58%
B = 16.3% (rough assumption we can agree with?)
C = 5%

As I understand your reasoning - in my case B is ruled out because A+B > X+Z ?
Well, I can't agree with this.
B should not be ruled out because B+C > Y


To better show the problem let't put our scheme in different case:

CASE 2:
Suppose my 10str lc is attacking 0 walls city full of units with no other bonuses
Suppose my lc first fights another str10 lc, then a barbarian. Behind the barb lots of other units.
so, according to my scheme:
A - defender lc survives
B - defender lc dies, barb survives
C - barb dies

A = 50% (pure math)
B = 45% (rough assummption for the case)
C = 5%
Let's assume X stays the same:
X = Y = 21%
Z = 58%

1) Do you still belive B should be ruled out because A+B > X+Z?
2) If not, what is wrong with my scheme?

Regards!
Seraad

[KGB]

Seraad,

CASE 2:
Suppose my 10str lc is attacking 0 walls city full of units with no other bonuses
Suppose my lc first fights another str10 lc, then a barbarian. Behind the barb lots of other units.
so, according to my scheme:
A - defender lc survives
B - defender lc dies, barb survives
C - barb dies

A = 50% (pure math)
B = 45% (rough assummption for the case)
C = 5%
Let's assume X stays the same:
X = Y = 21%
Z = 58%

I want to first make sure you aren't confusing the % you wrote in your example above (50/45/5) with the one being reported in the battle (78.7). The ones you wrote above are the chance of beating a single unit in battle (eg 50% is lc vs lc). The one being listed in the battle (78.7) is the % chance the battle ENDS exactly with that unit (in other words that's not the same as beating the unit in a heads up battle).

So based on your example above the game would list the % as 5% (chance your lc beats both units), 50% (chance the defending lc wins), 45% (chance the barb wins the battle after the lc dies). This is what the 90% rule then works on (5 | 45 50). In this case it would be valid for the attacking lc to kill 1 defending lc.

Now lets flip the fight order so its lc | Barb lc. Now the % for the battle are going to be 5% (attacking lc chance to win the battle is still the same), 90, 5 (Barb doesn't consume the other 95% since there is a chance lc can kill it but its at least 90% so only 5% or less goes to the defending lc). Now the numbers are 5 | 5 90 for the 90% rule. There is only 1 valid outcome now which is the Barb auto kills the attacking lc because cutting 10% on the left cuts out the 2 5's).

This is what happened in your battle. There was no way your lc could kill the first unit because it's % chance to win was more than the entire middle section of the battle (58%) so your lc couldn't reach any other units. If the fight order in the city was flipped to your example (a leading weak unit) then yes your lc could have killed someone.

KGB

[Seraad]

KGB
I'm not confusing anything. The only thing that confuses me is how stubborn you are

1) In my scheme "A" is not the probability of beating a single unit, but the probability that defender does not lose any unit! In CASE 2 it happens to be 50%! (50% my lc dies, 50% my lc fights further). I thought I wrote it quite clear.

2) As I undersand you think that:
A>Z => A is the only possible outcome.
What I understand is:
A>X+Z => A is the only possible outcome.

The rule "cuts from both sides" - yes! But it cuts from one side for the defender OR cuts from the other side for the attacker!!! So only MIDDLE is valid.
In my cases "A" IS in the middle range as A< X+Z
Also, thats why initially the rule was called 90%, not 80%. Despite the fact that it cut 10% from both sides.

Let's make

CASE 3:
my 2hp10str lc attacks first 2hp13str eagle, then 3hp30str barb and then lots of other units.
According to my scheme the exact math numbers should be:
A = 60.95% (probability that the eagle kills the lc)
B = 38.33% (probability that eagle dies and barb kills the lc)
C = 0.72% (probability that the lc kills both the eagle and the barb)
X=Y stays as it was = 21%

According to your reasoning the lc could not kill the eagle as A>58% That would mean that the rule does not cut 21% of te worst outcome for the attacker, but almost 40%!!! (B+C)


And what if the rule would "cut" 30% from both sides? According to you that would mean that any battle outcome with more then 40% would be granted. Welll. what if 2 even units would fight - probability 50%/50%. Both would be granted the win as they have more than 40%?

Please sleep well
Seraad

[KGB]

Seraad,

It's not me who is stubborn. I'm telling you that the rule is working and what you wrote from the initial battle seems perfectly valid to me and is not a bug. You are insisting that it is.

And what if the rule would "cut" 30% from both sides? According to you that would mean that any battle outcome with more then 40% would be granted. Welll. what if 2 even units would fight - probability 50%/50%. Both would be granted the win as they have more than 40%?

This tells me you still don't quite understand what's happening or that you understand it but wrote something nonsensical by accident. In a 50/50 battle the numbers will be obviously:

50 | 50

Now if the rule is cutting 30% from both sides, then it cuts NOTHING because 50 > 30. That means either outcome is valid as you would expect.

Now to your example #3:

.72 | 38 60 (number on the right indicates defender wins with no losses. Ie Eagle wins vs Lt Calv).

Now cut 21% from both sides. You cut out .72 on the left and NOTHING on the right because 60 > 21. So the valid outcomes are (Eagle + Barb) survive OR (Barb) survives. This is exactly what you would expect. What do you think is wrong here?

Try re-running the above battle with Barb in front. I'm guessing the Barb vs Eagle will be >90% (based on Barb chance being 38.33 out of (1-.6095) 39.05 = 98%). In that case the number would be something like:

.72 | 1.28 98

Now cut 21% from both sides. You cut out .72 and 1.28 on the left and NOTHING on the right. So the valid outcome is (Barb + Eagle survive) because the Barb auto kills the Eagle.

KGB

[Seraad]

KGB
You are expleining, yes. But your reasoning is not coherent. That's the problem.
Please focus now and tell me which of the following quotes is true:
1)

This is what happened in your battle. There was no way your lc could kill the first unit because it's % chance to win was more than the entire middle section of the battle (58%) so your lc couldn't reach any other units.

or
2)

Now if the rule is cutting 30% from both sides, then it cuts NOTHING because 50 > 30. That means either outcome is valid as you would expect.

In (1) you explain that if any outcome probability is geater then 58% then the win is granted, yes? or not?
The 58% is my Z = 100% - 2x21% where 21% is X in my scheme.
So please tell me - do you think any battle outcome greater then Z is granted or not?


In (2) X=30% so Z = 100-2x30% = 40%

In (1) A is granted because it is greater then Z (according to you)
in (2) Z= 40% so ANY battle outcome with bigger probability then 40% should be granted according to what you said in quote (1)

I can't believe we still have this conversation.
Let's do it step by step:
Can you look at my scheme and agree or not agree with the following:

As I undersand you think that:
IF [A>Z] THEN [A] is the only possible outcome.
What I understand is:
IF [A>(X+Z)] THEN [A] is the only possible outcome.


As for explaining - well, I would welcome ANY comment from SnotlinG or Piranha

[Seraad]

BTW. I think I know the reason for the "bug"
My guess is the system counts all the probabilities from one side to sum up to 21%. In my case:
P0 - my lc kills 32 and wins whole battle
P1 - my lc kills 31 units
P2 - my lc kills 30 units
......
P30 - my lc kills 2 units
P31 - my lc kills 1 unit
P32 - my lc kills 0 units


So the system sums P0 + P1 +...+Pn to the point it reaches 21% and rules out all the probabilities before reaching 21%
The problem in my case is that probabilities from P0 to P27 are extrelmy low and near 0.
So my guess is that the system rounds the probabilities to 0 - this way sum of all possible probabilities is lower then 100%.
So P32= 78,7% bu sum of P0 to P31 was lower then 21% due to the roundings.

SnotlinG, Piranha?Could you confirm?

Seraad