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Division by 0

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X3M
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TLDR post ahead. I have made [[[skip]]] blocks so you know you can skip the rest of a block until the next set of ***.

So first a summary:

A Risk like game, but with different rolls. You only roll to hit a number of times.
1d6 currently rolls miss-miss-hit-hit-2hits-2hits.
I want to change this into hit-hit-2hits-2hits-3hits-3hits.

The miss gives chance to infinite battles. (It is easy to roll 3 miss in a row) I want to make each roll with a certainty in hitting.
Since the average roll doubles, so do the health points of all my soldiers.

This post is my train of thoughts about this matter.

***

[[[You might skip this part of reasons]]]

I want to get rid of that 0. For 3 reasons. Don't read them if you want to safe time.

1
While the game is balanced, there are situations of a few (1 to 3 soldiers) against a few. And when both sides roll only misses, notching happens. Which is annoying. This happens mostly at choke points where only a few 1d6 are rolled. Those little skirmishes often take ages compared to the big battles.

2
Another annoying thing about misses is that it makes my probability test tree's infinite in size. Sure I can keep overview for some simple battle's. But the really complex ones that are only tested by play tests. Well, I want hard numbers. Even chess is better to predict than my game at this moment.
I want my game to be easier in prediction (also for the math loving players). And I want my game to keep rolling.

3
Some pieces on the board already have an accuracy effect by terrain, by movement choices, by event cards, by deflecting rolls and by choice of weapon rolls. So plenty of 0 chance there. But those are the gamble units with extreme rewards.

***

You need 3 average rolls (3 hits) for 1 kill (3 health). This is very balanced. And I need to keep the 1 to 3 ratio.

[[[The rest of this block can be skipped]]]

The mechanic is currently like this:

1d6 provides:
1=miss
2=miss
3=1 hit
4=1 hit
5=2 hits
6=2 hits

An average roll provides 1 hit. The "steepness" of the roll is 2.

You guessed it right, those misses have to go.
Also important to know is that each unit has to be hit 3 times. A double hit can be placed on 2 units. There is no such thing as over kill in bigger battles. But there still is when only 2 units fight.

I need to keep the balance. So for every 3 health, an average of 1 hit.

***

I want to change this dice roll into certainty. Yet maintaining the 1 to 3 ratio. I think I need help on this one. Or opinions are also welcome.

Further, I don't want it to be escalating in even higher numbers. So a some sort of minimum requirement is what I seek.

I still want to use only 1d6.

***

[[[skip if you like]]]

It will be a huge change to the game, since you roll, well, almost every second.

Personally I was planning on an average of 2 hits, with 6 health (awww yesss!!!, the 6 health is back).

Thus giving me this:
1=1 hit
2=1 hit
3=2 hits
4=2 hits
5=3 hits
6=3 hits

The "steepness" is also important here. It is now 1,5.
While 3/2 means overkill. 6/1,5 is a round number.
The number of overkill is actually reduced while the game is speeded up.
(And when I noticed this, I whacked my head several times)

***

Ok, here it comes, the problem:
My friends don't like this...6, while they agree on changing the roll into certainty.
They say it has to do with the beauty of the set of numbers for health for the units. Urrrg...

My first solution:
3,12,27,48,75 etc.
becomes
6,24,54,96,150 etc.

While they rather see numbers like
5,20,45,80,125 etc.
or
10,40,90,160,250 etc.

But that would give me, bad balance and unwanted overkill.

Further, they liked the fact that some units still had a chance in winning. Even if it was just 1/1000. The new certainty roll takes this away.

What now?
Tell them to take it anyway?
Or find a way with "rounder" numbers?

I too like the sequence starting with 5. But I don't know a good 1d6 roll going with it. Suggestions?

I too like the fact that a lonely soldier can hold of 3 tanks, just a bit longer than expected. Could I keep this in another way? A new rule might do the trick, so shoot if you have idea's for me.

In advance, thanks.

Masacroso
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I dont know if this is good

I dont know if this is good for your system or not but you can still maintain a miss chance, only one, and lesser the power to lesser the overkill with this distribution of 1d6: miss, 1, 1, 1, 2, 3 or something similar.

X3M
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Nah, the point is to

Nah, the point is to completely remove all the misses in the basic roll. Yet to keep randomness.
A basic battle with basic units, will always lead to an end.

We thought of something for the sequence of 5:
1 hit
1 hit
1 hit
2 hits
2 hits
3 hits

The average is 1 2/3th. And this times 3 is 5.
So we have the 1 to 3 ratio, while the health counters are 5 to begin with. The battle ends rather fast. And the initial test tree is now very limited.

However, the distribution is not really linear. Seems a bit exponential. But lets say the roll 6 is just special now.

With 2 units, there is an overkill if both roll a 6. But the chance on that is 1/36th.
Other combinations give a rather pleasent result. But I have not found time yet to draw out the complete combat of a 2 vs 2. Which is what will show me the results. I will compare it with the 2 vs 2 with the average of 2 roll.

Masacroso
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More than exponential

More than exponential it is logarithmic. Anyway it is non linear, ofc. But it never is linear when you fight more than a 1vs1.

When you increase the number of fighters, by example 3vs3 or 5vs5, etc..., the distribution is transforming in a normal shape... This is one of the more important laws on mathematics: the central limit theorem. Please see here if you want to understand in an easy way.

And this means too that is easier to predict a result when there are more actors/actions that when there are fewer. In mathematical words this is that the probable outcomes of 1vs1 have a bigger standard deviation than when it is 3vs3 or 4vs4. See here.

A normal distribution is the more general and present distribution of probabilities. And we copy it in many games, it is mathematical form of "extraordinary and rare events and stronger/worse than normal events", in role games this is generally put as critical strikes and misses.

The distribution 111223 have less "power" (logarithmic distribution or similar) than 112233 (linear distribution) so the first cannot snowball more than the second.

Logarithmic shape "counters" exponentiality. By example the senses of life beings (the 5 classical senses of humans, by examples) transform an exponential input in a linear one, this is, the relation of stimulus-sensitivity is logarithmic... this is why volume is measured in decibels, but the same happen with all the other senses not only hearing.

If you want X3M I can show to you in numbers or graphics how the probabilistic distribution evolve when you are increasing number of fighters for any kind of die distribution you want. It is not hard using CPU power to do all computations, it is just a "classical" generating function.

X3M
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I am not good with formula's

I am not good with formula's regarding distribution etc. But I know my stuff by using Excel. And I have seen all these graphs too. :)
These graphs showed me that I might as well use the absolute average numbers for big battles.

Besides, having a battle story on paper is handy. You see all the play tests in one glance for a given situation. Having the 0 damage removed, means no infinite basic battles.
Even if one of the 2 players is using risky units, the other one will end the battle either way.

For a quick average calculation, I always had 1 die is 1 hit. Since 1 is the average. With a lot of units, this is only logical since the result is centred. With 6 projectiles having an accuracy of 5, I simply counted 5/6 damage instead of 6/6 damage.

A problem that I had half a year ago was the unit with 9 and 16 damage compared to that of 25 damage. The only way to do that was also calculating in Excel. It gave me great insights.

---> However, if I want to calculate something difficult for my standards, can I ask your help?

***

Another worry that I tackled last evening.

I was comparing the number of hits needed for 5 and 6 health. The less overkill, the better. This is not by dice, but by how much 1 hit is worth.
An example:
Tier 3 unit hit by a tier 2 weapon gives
45health /4damage =11,25hits needed.
You need a 12th shot for killing, which is 0,25 overkill.

My intuition told me that 6 health would be better balanced. (6/2=3, 6/3=2, while 5/3=1,67, 5/2=2,5) But it turns out it is equally in overkill compared to the 5 health.
This for the first 6 tiers in the game.

Doesn't matter much. 1 Worry less.

***

I am gathering votes at the moment. The question is not which list is better, that would require play testing. But.
Which list looks better to you?
Which one would be easier for you to remember?
Which one gives the impression that it is better balanced?

List of 5 (1,1,1,2,2,3)
5, 20, 45, 80, 125, 180, 245, 320, 405, 500

or

List of 6 (1,1,2,2,3,3)
6, 24, 54, 96, 150, 216, 294, 384, 486, 600

The old one:

List of 3 (0,0,1,1,2,2)
3, 12, 27, 48, 75, 108, 147, 192, 243, 300

Masacroso
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X3M wrote:---> However, if I

X3M wrote:
---> However, if I want to calculate something difficult for my standards, can I ask your help?

Ofc, if I can help you. My knowledge is very limited but maybe in some topic I can help.

Quote:
Another worry that I tackled last evening.

I was comparing the number of hits needed for 5 and 6 health. The less overkill, the better. This is not by dice, but by how much 1 hit is worth. An example: Tier 3 unit hit by a tier 2 weapon gives 45health /4damage =11,25hits needed. You need a 12th shot for killing, which is 0,25 overkill.

My intuition told me that 6 health would be better balanced. (6/2=3, 6/3=2, while 5/3=1,67, 5/2=2,5) But it turns out it is equally in overkill compared to the 5 health. This for the first 6 tiers in the game.

Doesn't matter much. 1 Worry less.

I think the difference of exact number to approximate one is not very important globally but the cases when the number of rounds is low. By example there is a big difference killing an unit in 2 or 3 rounds... way bigger difference of kill an unit in 11 or 12 rounds.

Quote:
I am gathering votes at the moment. The question is not which list is better, that would require play testing. But. Which list looks better to you? Which one would be easier for you to remember? Which one gives the impression that it is better balanced?

List of 5 (1,1,1,2,2,3) 5, 20, 45, 80, 125, 180, 245, 320, 405, 500

or

List of 6 (1,1,2,2,3,3) 6, 24, 54, 96, 150, 216, 294, 384, 486, 600

The old one:

List of 3 (0,0,1,1,2,2) 3, 12, 27, 48, 75, 108, 147, 192, 243, 300

To remember I think that all lists are equal, maybe the first a bit easier to remember, so I will focus on balance. But to understand the balance that you want I need to understand better the complete game.

I cant see clearly the difference in the impact in the game of the three lists. But if you throw multiple dice the first choice seems to me with less probability for snowball because higher values have lower probability... and the snowball is strong if you add higher values than lowers.

But anyway you must test to see it, I think the best you can do is test shortly in test battles the three lists with your friends and see what feels better.

Instead to play complete games you can play different scenarios (playing the same scenario multiple times to see how can evolve one kind of battle).

X3M
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Yeah. This sunday, mission 3

Yeah. This sunday, mission 3 is going to be tested again.

In this mission your first battle tank is introduced.
All that you can support with is rifle infantry.

It is this mission where you learn about damage and armor types.

Infantry armor is 1.
The rifle does 1 damage.
1 damage on any armor remains 1 damage

The tank has 36 armor. And remains 36 damage on tanks.
However, the damage of the cannon on the infantry is only 1.

It is up to the player to decide which combination of units is the best. Further, will there be formations to consider during the battle. The counters are hard to begin with. This so a player can learn a good basic feeling of what is better.

With the division by zero, this one was a pain in the @$$. But back then i managed. But also stopped going deeper with the missions afterwards.

With the new possible changes, i hope to get the answer faster. Not just for me, but the players as well.
5 and 6 are both going to be tested.

I hope it gives the same results. But then with a certain end in time.

***

I think you like it if I say that 1 tank costs 6 infantry. With the numbers given, you might notice that they are sort of balanced. It is something that always made me proud.

This test is just a check to see if 5 or 6 still give the same desired balance as 3.

X3M
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Well, I tested mission 3.

Well, I tested mission 3. Somehow I feel that things have gone in the wrong direction???

Checking with tools...
So, I have a bit better understanding of anydice.com. But this doesn't yet get me where I want. Yet it did provide me with an insight on things. A scary "I have screwed myself" kind of insight.
Instead of looking at when an unit dies. It is better to know how much unit there is still alive!!!

I lost track of how a medium balanced battle unfolds.
In summary: Infinite small units vs an infinite big unit, are only worth 50% in damage output.

This effect is normally countered with division by 0, taking cover and increasing durability to all.

The third one is applied in this topic. However, with more durability, division by 0 is removed. This means that with certainty in the rolls for eventual killing, the bigger units have gained an advantage.

So just throwing the next thing out on this forum.

***

I would like to see the impact of my decision in changing the roll.

The given situation (which is a basic for mission 3):
6 infantry vs 1 tank.

If an infantry unit is hit, it takes cover. Meaning another infantry unit is going to be the next target. In other words, the infantry unit with the highest health will always be targeted.
Once dead, the unit cannot fight any more.

3 versions that are need in a test. Just to see the percentage of how many times a side wins. In overall, the tank is just a little bit better.

Old game; version 3
one infantry has 3 hp
the tank has 108 hp
a dice roll has 0,0, 1,1, 2,2

New game A; version 5
one infantry has 5 hp
the tank has 180 hp
a dice roll has 1,1,1, 2,2, 3

New game B; version 6
one infantry has 6 hp
the tank has 216 hp
a dice roll has 1,1, 2,2, 3,3

Masacroso
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X3M wrote:3 versions that are

X3M wrote:
3 versions that are need in a test. Just to see the percentage of how many times a side wins. In overall, the tank is just a little bit better.

Old game; version 3 one infantry has 3 hp the tank has 108 hp a dice roll has 0,0, 1,1, 2,2

New game A; version 5 one infantry has 5 hp the tank has 180 hp a dice roll has 1,1,1, 2,2, 3

New game B; version 6 one infantry has 6 hp the tank has 216 hp a dice roll has 1,1, 2,2, 3,3

Ok, I can see it but I need to know how the game works: alternate turns throwing dice? It is the same die for tank and infantry?

X3M
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Apparantly it is confusing.

Apparantly it is confusing. No matter. I simply tell what is going on with version 3.

The tank is bad against infantry. And these cheap infantry have the worst weapon against a tank.

The infantry and the tank, both only do 1 damage per hit against each other.

When all roll an average. The battle goes as following:
6 infantry do 6 × 1 = 6 damage against the tank.
1 tank does 1 × 1 = 1 damage against the infantry.

The balance is shown when looking at the number of hits needed.
There are 6 infantry units with 3 health each. This would mean 18 health or rounds in total. The tank has 108 health. With 6 damage in total against it, thus means again 18 rounds.

Of course 1 tank has a randomness in the number of hits. With the dice, it is 0, 1 or 2 hits per round. This means that there is a chance that an injured soldier dies too soon. This only can happen when all healthy soldiers have suffered injury once.
The infantry too can do 0 to 12 damage in one round.

Round 1
Tank rolls a 4. 4 is 1 hit.
Infantry rolls 2,6,2,4,5,1; miss,2hits,miss,hit,2hits,miss; 5 hits in total.
Tank suffers 5 damage and has 103 health remaining.
Infantry suffers 1 damage and the health are 3.3.3.3.3.2

After around 6 rounds we might have 72 health remaining for the tank and 3.3.2.2.1.1 for the infantry.
After 9 rounds there is 54 health remaining for the tank and 2.2.2.1.1.1 for the infantry.
Now there are chances that infantry die and will do no more damage.

Damage is determined first. Then the casualities. Thus an infantry unit that dies, still does damage that round.
When 2 hits occur by the tank. And the infantry has only 1 health, the remaining damage goes to the next infantry unit.

I hope this has clarified a lot. If not, please ask.

Masacroso
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One question, the tank damage

One question, the tank damage is cycling between the soldiers?

By example we have the six health of soldiers like this: 3,3,3,3,3,2; and tank is attacking again then:

-if tank roll 2-hit damage I suppose is taken from 5th soldier and the healths change to 3,3,3,3,1,2

-if tank roll 1-hit damage I suppose is taken from 5th soldier and the healths become 3,3,3,3,2,2

But, what happen if tank roll 0-hit damage? Then the next round, damage is taken form the 5th soldier or the 4th.

In general: the damage is taken in a fixed order or it is just taken from where it cant kill (if it is possible)? In your example it seems to be the last case, just to be completely sure.

X3M
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Just where it can't kill.

Just where it can't kill.

Masacroso
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Let me show to you the basics

Let me show to you the basics to evaluate probabilities, in some days (maybe) I will try to do it with full information using Maple or a similar program.

In mathematics exist something called generating function. To simplify a die is represented as

f(x)=p0 x^0 + p1 x^1 + ... + pn x^n,

where p0 is the probability for 0 on dice, the p1 is the probability for 1 on dice and so on.

Then to know the probabilities of some value throwing k dice we just can do

g(x)=f(x)^k, where the coefficients in the expression are the probabilities of the exponent for each x. An example:

For case 1 we have a dice that is 001122, so the probabilities are p0=1/3, p1=1/3 and p2=1/3. Then

f(x)= 1/3 + 1/3x + 1/3x^2

And if we throw six dice then the different probabilities for the values of the throw is

g(x)=f(x)^6= (1/3 + 1/3x + 1/3x^2)^6,

and here we can use WolframAlpha to expand the expression and see the coefficients and exponents using some expression as

expand N[ 100(1/3 + 1/3x + 1/3*x^2)^6 ]

(we multiply by 100 to put coefficients as % probabilities). The result is

0.137174 x^12+0.823045 x^11+2.88066 x^10+6.85871 x^9+12.3457 x^8+17.284 x^7+19.3416 x^6+17.284 x^5+12.3457 x^4+6.85871 x^3+2.88066 x^2+0.823045 x+0.137174 (13 terms)

What this means? This means that, by example, the probability for a sum of 12 throwing 6 of these dice is only 0.13%, the probability for a sum of 7 is 17.28% and the probability for a sum of 0 is 0.13%

For case 2 and 3 you only need to change the atomic probabilities, by example for the case using the dice 111223 we have that the probabilities and sums throwing 6 dice are

expand N[ 100(1/3x + 1/4x^2 + 1/6x^3)^6 ], i.e. the expanded form is

0.00214335 x^18+0.0192901 x^17+0.0980581 x^16+0.337577 x^15+0.870065 x^14+1.73732 x^13+2.7545 x^12+3.47463 x^11+3.48026 x^10+2.70062 x^9+1.56893 x^8+0.617284 x^7+0.137174 x^6 (13 terms)

Then, by example, we want to know what is the probability for some amount of damage on T turns. Then we can simply see probabilities and amounts doing

M(x)=g(x)^T

By example, suppose that we are on case 1 (dice 001122), and we want to know the probabilities for different amount of damage over 4 turns. Then

M(x)=g(x)^4= (0.00137174 x^12+0.00823045 x^11+0.0288066 x^10+0.0685871 x^9+0.123457 x^8+0.17284 x^7+0.193416 x^6+0.17284 x^5+0.123457 x^4+0.0685871 x^3+0.0288066 x^2+0.00823045 x+0.00137174)^4

The result is (multiplied by 100 to express in %):

expand N[ 100((1/3 + 1/3x + 1/3*x^2)^6)^4 ]

that is a lot of numbers (you can see it in the link). Ofc all these things is better to see using graphs more than in pure numbers. In this example the maximum probability is for 24 damage that is about 9.89%.

But to do it better, counting dead of soldiers and so on, the easiest way to do is using some kind of software to probability experimentation. I dont know if I had the time to do it on Maple because I never used it, but I will try, ok?

Any questions are welcome.

X3M
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Hey, I finaly understood how

Hey, I finaly understood how that function works. Thanks. You do a good job in teaching.

It is faster then my excel programming.
To bad it doesn't yet get me an end result winning percentage.

I think i need to practice simulating on a small level first. Or is there a way to calculate with losses? Meaning that first 6 is calculated and added up until 5 then 4 etc remain?

The difficult part lies within the damage distribution to the infantry.

Tommorow, i am going to recalculate the way you explained. Thanks again.

Masacroso
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X3M wrote:Hey, I finaly

X3M wrote:
Hey, I finaly understood how that function works. Thanks. You do a good job in teaching.

It is faster then my excel programming.
To bad it doesn't yet get me an end result winning percentage.

I think i need to practice simulating on a small level first. Or is there a way to calculate with losses? Meaning that first 6 is calculated and added up until 5 then 4 etc remain?

The difficult part lies within the damage distribution to the infantry.

Tommorow, i am going to recalculate the way you explained. Thanks again.

I dont understand exactly what you say about "losses".

For damage distribution on infantry we can do different things but the best is just simulate all the game with some mathematical program as Maple.

I will try to see how to do, if I can.

But a lighter approach, maybe ok too, it is just see the probability for die of X numbers of soldiers on a number of turns T and from here manipulate probabilities to approach something... but it maybe a bit complicated. The simulation is the best way, just simulate all the scenario and play it a lot of times to take some graphs and probabilistic measures with all the information, included dead soldiers, distribution and so on.

EDIT: there is a easier way to do simulations... the number of dice rolled is determined by the total hp of soldiers. So you only need the relation between hp and number of dice, for each hp it is determined a number of dice that you roll.

X3M
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Well, here is the thing. Dead

Well, here is the thing. Dead infantry don't shoot. While the living do. Due to my mechanics, the chances on killing a guy with 2 health is 33%. And killing a guy with 1 health is 67%. This means less dice to be rolled at the infantry side.

How strong this effect is, is why i need the simulation. I suspect that the other 2 dice options are more extreme in a negative way for my little infantry.

I can't just simply take the average. Because the average is random as well. 1 die more or less can have a huge impact in the end result.

I am going to experiment this evening with some tables to see if i find a way to determine with a branchtree of chances. (I use tables instead).

Masacroso
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Yes... I know you cant take

Yes... I know you cant take the average. I was thinking the way to make the simulation: there is a probability for some turn for 0 dead soldiers, 1 dead soldiers, etc... This probability is linked to the probability, in any turn, to have some amount of hp.

There are some ranges of remaining hp that reflect an unique number of alive soldiers. Then we can see the probability for this range of remaining hp at some turn.

You will have then branches starting in some turn and ending in other for different number of alive soldiers. In the case of the die 001122 there is not "end turn" cause you can miss forever... anyway we can try to see what happen.

The branches are defined just by a probability for any turn, a number, so we will not get a lot of numbers... at least not as I imagined (maybe my conception of "a lot" is different than yours). For each turn we will have at maximum 6 branches and at minimum 1, ofc each branch will had more branches... but the number of total branches is not too big as I imagined in first place.

And maybe, with a little of luck, these branches can be represented by some formulas not too complex (I know a little about searching closed forms from recursive formulas as it happen in any kind of tree structure. But luck is needed anyway, many times these things cannot be written in closed form.)


I see now that I had a stupid mistake before in a formula. The dice 111223 is not represented by probabilities 1/3, 1/4 and 1/6... the correct probabilities are 1/2, 1/3 and 1/6, so the dice is represented by the generating function

f(x) = 1/2*x + 1/3*x^2 + 1/6*x^3

X3M
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Hehe, i figured that much. I

Hehe, i figured that much. I calculated too with the given formula. On paper.

I got the dice ready for 1 to 6 soldiers. This only for 111223. With the formula and with my table method. And i have exact numbers, chances out of 46656. These are now blue prints. And can be used for bigger branching.

Next step.
I know that 2 soldiers cannot absolutely win against the tank. I am going to see if i can find a way for easy branching including the death of an unit. On the infantry side, i need 2d6 and 1d6 and the mix of those 2.

With just 2 units, there will be only one moment where it happens. But this moment happens somewhere in all branches. I think that my table is going to get layers. A cube table if you will.

So much work, so little time.

Masacroso
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I was wrong... the analysis

I was wrong... the analysis is more complex. Some hp level not reflect an unique numbers of soldiers... some hp can represent different number of alive soldiers...

This is done in 5 minutes by people that know how to use Maple, in a few lines of "code", and in a perfect way with full information, graphs, etc...

I will try to do it in some moment because is fun but surely it will not be soon, sry.

By now Im taking some fun trying to make the simulation "by hand" through some formulas and analysis :p The distribution of damage and the different number of alive soldiers for the same hp make it very interesting problem, more complex than I thought. My previous post was wrong, it really is not as easy as I thought yesterday.

At math level there are some very interesting counting problems for this simulation.

RyTracer
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Here's an error.

After just skimming this conversation, I see a problem here:

Going from (0,0,1,1,2,2) to (1,1,2,2,3,3) is not a doubling operation, but an adding one operation. So instead of

6, 24, 54, 96, 150, 216, 294, 384, 486, 600

it should be

4, 13, 28, 49, 76, 109, 148, 193, 244, 301

Keeping the numbers closer together will be more like your original mechanic without the misses.

X3M
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The experiment at a smaller size.

RyTracer, please reread your own comment.

And 1 + 1 = 2
That is still a doubling too. That is what the dice roll do.

***

I am currently experimenting with a 2 infantry vs 1 trooper. Just to see how works with the method that I always use.

While it works. It becomes a big hassle when half way. I can fight through. But I think it is better for me to make calculations more automatic by using Excel.

With the results that I have seen so far. I suppose I should not try to calculate 6 infantry vs 1 tank this way.

X3M
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A little update, plus a question

Thisss topic isss resssurfacing to sssee the sssun oncsse more.

---

I have been doing some number crunching with a simulator.
Further, I wanted to add some new rules for balancing as well. And the results have given me new insights in randomness and determinism.

Turns out that determinism unbalances a game greatly. And that the 111223 die does this with a factor of 200%.

I know that I am giving up on speed here. But balance outweighs this.
Am I correct in doing so?

***

Now then, I am still leaving open, the option for 5 health instead of 3 health.
But this time, I need to adjust the die to something more useful.

I made a list of possible dice. And am discarding them one by one. This is what remains right now:

011233
or
012223

I would like to hear from you guys, what kind of distribution you think is the best for the die?

Or should I just completely discard the 5 health system?

X3M
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The choice

{0,1,1,2,3,3} is several times better than {0,1,2,2,2,3}, but why this is? I have no idea.

Maybe a mathematican can tell me.
For now, this case is closed :)

Arcuate
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Simulate instead of calculate?

For problems like this, rather than trying to work out the probability (where it is changing due to deaths etc midway through a battle), I would simulate 1000 battles on a computer, tweak the inputs, try again. If you would like to try this, let me know.

X3M
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After Masacroso's last post.

After Masacroso's last post. I asked my cousin to get a simulator running.
Since then I started changing rules for better balance (3 versions (times 3 possibilities) so far).

Could post the link here, but it goes to his personal computer. And I don't want people to flood it. So if you are interested in the simulation that I use. I will PM the latest to you + explanations.

And the why's?! of some choices.

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