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# Too much fidling?

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questccg
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Joined: 04/16/2011
I don't have such a problem...

X3M wrote:
In my proto-type game and my "public" game. I had accuracy. But also a roll for damage.

My card game only has a roll for accuracy.

Which means that a maximum accuracy always does a fixed set of damage.

In my other 2 games. It was still a random number for everything.

The question is, should I think of something to make every weapon "random". Even if the accuracy is 100%?

How to do that?

Because 8/8 weapons always do 100% damage.

Don't you also have a QUANTITY (as in "Number of Units") too??? That also is variable too... And all my "units" do a RANDOM (1D8) accuracy that ranges between 20% and 100%. Basically you MERGE three (3) "Squadrons" into ONE (1) PLATOON... And roll 3D8s for accuracy that you choose how to assign... So in my version it's rarely 100%... The odds a 1:8 x 3 dice.

Sure there is a chance... But only much more rare to roll 100%. So I don't have this problem and I have also "custom" D6s for certain accuracy rolls too.

These dice also vary in probabilities also... Again so I don't have this issue either (whether it be standard rolls or custom ones).

Note #1: I'm not sure how to compute the probability of rolling a "1" on any three D8s. I THINK it's 1/8 + 1/8 + 1/8 = 3/8 = 37.5% and near to 40% given 3 dice... But I could be wrong... (Not sure)

Note #2: For ME there is no "maximum accuracy" it is always RANDOM using a D8 (as I explained above 3D8s, you choose how to distribute the dice roll values among your three squadrons). There is both a DMG (Damage) and QTY (Unit Count)... But Accuracy is ALWAYS random. That's just how I simplified it in my version...

X3M
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Joined: 10/28/2013
3d6, 3d8

That is what the 3d6 and 3d8 where for. Having only 3 dice. Making a big variety of results with all the cards.

However, the 100% accuracy cards would always hit 100%.

Now, I have been thinking. Should I have the exact same roll do something for the damage as well?
No, that would make the game imbalanced.

A 100% accuracy would have 1 to 8 damage.
A 12,5% accuracy would have 1 damage.

The average damage factor between the lowest and highest, would not be 8/1=8.
But 4.5/0.125=36. When comparing the 100% accuracy to the 12.5% accuracy.
The true weight factor would be:
100% of 4.5 = 4.5

Maybe... I could do this. And calculate for each accuracy, the new factors. Thus a chance to hit. And the hit has a damage value. The damage is on average for the whole range of the accuracy.

d8:
1) 12.5% of 1.0 = 0.125
2) 25.0% of 1.5 = 0.375
3) 37.5% of 2.0 = 0.750
4) 50.0% of 2.5 = 1.250
5) 62.5% of 3.0 = 1.875
6) 75.0% of 3.5 = 2.625
7) 87.5% of 4.0 = 3.500
8) 100.% of 4.5 = 4.500

d6
1) 16.7% of 1.0 = 0.167
2) 33.3% of 1.5 = 0.500 (This is suitable for the game)
3) 50.0% of 2.0 = 1.000
4) 66.7% of 2.5 = 1.667
5) 83.3% of 3.0 = 2.500
6) 100.% of 3.5 = 3.500

***

Another approach is a third set of dice, to roll. One that would be a factor to any die that was a hit. Just like in my proto-type game. Where accuracy could be 100%. But the damage roll would be d6-2, d6-1 or d6.

I don't know what else I could do.
d6 where the result is actually a factor on the hits?
If 3d6 or 3d8 are rolled. Each accuracy would have 0, 1, 2 or 3 hits.
Maybe roll 3 coloured dice. And all accuracy with 1 hit are influence by the white die.
All accuracy with 2 hits are influenced by the white and grey die.
And all accuracy with 3 hits are influenced by the white, grey and black die.

The result would be a factor of 0, 1 or 2 for each die.

The 100% accuracy still have a chance now on doing 0 damage. But if the white die has a hit, and the grey and black die have a miss. Then the 100% accuracy would deal relatively less damage than a lower accuracy that had only 1 hit.

So, now we have the player roll 3d6(red) and 3d8(blue) for the cards. Then another 3d6(white+grey+black)

What do you think?
Then the player rolls

X3M
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post note

There is a true difference between 3d6 and 3d8 now. For the 50% and 100% accuracy. The other accuracies already where different. But the 2 that where the same. With the roll being used as a double result. 50% and 100% accuracy are different for the d6 and d8.

4/8) 50.0% of 2.5 = 1.250
3/6) 50.0% of 2.0 = 1.000

8/8) 100.% of 4.5 = 4.500
6/6) 100.% of 3.5 = 3.500

Also, the
7/8) 87.5% of 4.0 = 3.500
has the same value as the
6/6) 100.% of 3.5 = 3.500

But the distribution of possible damage is different.
The maximum damage roll for 7/8 is 7 and for 6/6 is 6.
A bit more gamble to get a bit of a higher damage.

questccg
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Hmm... I did something "similar" with my "exploding" dice...

But rather than start with 100% (1/1), I went to the direction of 50% (1/2). What this ended up in doing was "simplifying" the math since this limits the HIGHEST possible roll to be 300% (3/1). So on MY D6 you get:

1> 1/2 = 0.5 = 50%
2> 1/1 = 1.0 = 100%
3> 3/2 = 1.5 = 150%
4> 2/1 = 2.0 = 200%
5> 5/2 = 2.5 = 250%
6> 3/1 = 3.0 = 300%

This greatly "improves" the MATH and making it easier to work with. But this flavor of dice was used to simulate "exploding" weapons. Again for accuracy but depending highly on the "nature" of the weapon.

That "simplification" makes the dice more flexible and easier to work with the MATH considering that the roll needs BOTH to make sense and be easy to compute with... That's my first comment.

The other comment that I have is regarding the dice rolling.

So you roll EITHER 3D6 or 3D8, right? Depending on the unit...?! That's my first question concerning the dice. And then what do you DO with these dice?? Set them aside, select units using them or is there another purpose to simply react with the ADDITIONAL 3D6 (White + Gray + Black)???

I'm not sure. How do these new dice INTERACT with the PREVIOUS ones?!

It could be 2-Factors: one Accuracy and one Perc-Hit (giving you two random variables when computing an attack). The question I ask you is: "Do you NEED both sets of dice???" Because I am unsure how they "interact" with the initial dice... I need some explanation on HOW the dice will work with each other...

Before I can comment further.

X3M
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Joined: 10/28/2013
The first method is. A player

The first method is.

A player rolls the dice required by the cards that attack.
3d6 or 3d8 or both.

Then each die is compared to the accuracy of that attacking card. If the die roll is equal or less than the number of that card. 1 projectile hits.

***

Method 1.
Each hit will also do damage corresponding to the roll.
For example, if 3d8 rolls 3, 6 and 8. And the accuracy was 8. Then all 3 dice give a hit. The damage is 3+6+8.

I don't like this method. It gives too much of a high numbers.

Since I use the same dice. I can't turn it into a yes/no situation.

Method 2.
Another set of dice are rolled.
3d6
Where we now use the 3 different colours.
Again we had the 3, 6 and 8 for the accuracy 8 card.
3 hits, all 3 dice from the second set are used.
Roll 1 or 2, the hit is a mis. 3 or 4, the hit remains. 5 or 6, the hit is multiplied by 2.

If thenoriginal roll had only 2 hits. Then a player would use only the white and grey die.

Maybe I can do this with 1 extra die. But I need to think on that one.

questccg
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Comparing both methods and the use of custom dice

X3M wrote:
Method 1.
Each hit will also do damage corresponding to the roll. For example, if 3d8 rolls 3, 6 and 8. And the accuracy was 8. Then all 3 dice give a hit. The damage is 3+6+8.

So Accuracy "8" means all three (3) projectiles would hit (8/8)? Right?! And this was your initial concern that an 8/8 unit will ALWAYS hit three (3) times no matter what the dice roll results give. Correct?

X3M wrote:
I don't like this method. It gives too much of a high numbers.

Indeed for powerful units (8/8), it will mean that they ALWAYS hit. To reflect upon this... I don't have this issue. WHY? Because you ALWAYS need to ROLL 3D8s and ASSIGN the dice to the given squadron.

For example: If I roll 1/2, 1/5 and 1/1, I have "1" 100%, "1" 50% and "1" 20%. And so for my platoon, you need to "choose" which die is assigned to which unit. The opponent does the same when he rolls too.

X3M wrote:
Method 2.
Another set of dice are rolled (3d6). Where we now use the 3 different colors.
Again we had the 3, 6 and 8 for the accuracy 8 card.
3 hits, all 3 dice from the second set are used.
Roll 1 or 2, the hit is a miss. 3 or 4, the hit remains. 5 or 6, the hit is multiplied by 2.

If the original roll had only 2 hits. Then a player would use only the white and grey die.

Maybe I can do this with 1 extra die. But I need to think on that one.

So you are saying it's sort of a "binary" result (kinda). You roll 3D6 and the results is MISS (1 & 2), HIT (3 & 4), and DOUBLE HIT (5 & 6) for EACH DIE, right? If I was you then, I would use a "custom" D6 with the values:

MISS (2x), HIT (2x) and DOUBLE (2x).

That's why I am using fractional dice for my attacking. I don't want the players to MEMORIZE that "1 = this" and "2 = that", etc. A custom die would make it much more obvious the result of the "binary" attack.

Just my thoughts on this 2nd method...

Note #1: Also note that 33.3% of the time the result no matter what the Accuracy is ... is a MISS. Maybe a HYBRID solution would be BETTER. And I think you alluded to this. If your "Accuracy" is based on values of "8" ... Well then I would use ONE (1) D8 change the probabilities just a bit:

1> 1/4 = 25%
2> 1/2 = 50%
3> 3/4 = 75%
4> 1/1 = 100%
5> 5/4 = 125%
6> 3/2 = 150%
7> 7/4 = 175% (the only computation which is "harder")
8> 2/1 = 200%

And this varies between LOW (25%) and HI (200%), sort of like you HIT/MISS dice with Fraction you can use to compute the damage done.

Like I stated, only the "7/4" is a bit tricky... Multiplying by "7" is NOT obvious for many players... And then dividing that total by "4" is ALSO NOT obvious too... But it happens 1:8 times per roll of 1D8.

ANOTHER possibility is to "abstract" the FRACTIONS and make EVERYTHING "divide by 4" and then you get:

1> 1 (/4) = 25%
2> 2 (/4) = 50%
3> 3 (/4) = 75%
4> 4 (/4) = 100%
5> 5 (/4) = 125%
6> 6 (/4) = 150%
7> 7 (/4) = 175%
8> 8 (/4) = 200%

AND HOLY CRAP!!! That makes it a STANDARD D8!!! So you could use normal D8s and simply divide the result of the roll by "4". Obviously you would make the multiplication of the Accuracy, Damage and Quantity (of units) FIRST and then divide by "4" to get the RESULT of the amount of damage caused!

That could be some real MATH GENIUS because they would be regular D8s and they can offer "fractional" outcomes that are more "acceptable" when rolling dice for something like "Hit-Percentages" and "Modifying Damage" (as a couple of examples)...

X3M
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Joined: 10/28/2013
Editing, so here is only the conclusion

I just was editing my last post. Trying to explain a bit better. But you got the gist of it.

It's all gone, lol.

***

I have a new option. Which is more or less the number of hits that you need on an enemy card.

Everytime when you roll the dice. You have a number of hits. But you also need to hit each card or chip, 3 times in order to remove it.

An 8/8 and 6/6 card would always have 3 hits. And therefor always remove an enemy card.
If you have less hits per card, you simply need more cards.

So, I thought of turning this around with the second set of dice.
You roll 1 die. And it shows you the number of hits you need on your opponent. In other words. How many times do you need to exceed the armor.

The balancing is rather complicated.
But I think that the next d8 will proof very usefull.

2 2 3 3 3 4 4 6

These are the number of hits that you need to make on an opponent card.
So if you make 2 hits per card. And you roll this die. And you roll a 2. Then each card of yours will defeat one opponent.

How did I balanced this die? I looked at how much "damage" each number of hits represents. A 3 is the default and thus 100%. If you need only 2 hits, then each hit is 150% "damage".
The 2 is worth 150%.
The 3 is the standard 100%.
The 4 is 75% and the 6 is 50%.
If you add up all the percentages. You reach 800%. Which gives 100% average again.

Rolling a 6 means your cards are only half effective.
Your round will be only half effective.
Which is not a great loss.

The biggest difference is 3 to 1. The chance of that happening is 1/32.

questccg
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Joined: 04/16/2011
Very clever!

How do you compute 1/32 = biggest difference between 3 and 1??? I understood pretty much everything except for the probability you computed at the end...

2, 2, 3, 3, 3, 4, 4, 6 = 800% = 100% average!

Very, very clever indeed! So you do "3" DIVIDED by the value obtained by the roll... Okay it works... But it's a bit more challenging to do the multiplication by "3" and then dividing by the die roll.

I'm not saying that it's NOT in the spirit of the standard D8 which does a division by "4", that feels more natural.

But it does work and the mean is 100% which is good too. The largest difference is only a 3 to 1 difference.

Note #1: What you are effectively doing is this:

3/2, 3/2, 3/3, 3/3, 3/3, 3/4, 3/4, and 3/6.

Where as I was proposing:

1/4, 2/4, 3/4, 4/4, 5/4, 6/4, 7/4, and 8/4.

The principle is very similar and both distributions have their own "flaws". Like I said the "7/4" is hard to compute. While the "3/6" dividing by "6" is going to be a challenge. Even if fundamentally this could be expressed as "1/2"... It's just something the players will have to figure out and get used to I think...!

Cheers @X3M!

questccg
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Joined: 04/16/2011
Why I think it's EVEN MORE CLEVEL!

Quote:
2, 2, 3, 3, 3, 4, 4, 6 = 800% = 100% average!

Another way of looking at this DIE roll... Is to completely IGNORE the "%"s or the "Fractional values" and JUST USE THE VALUE!

So if you ROLL a "2" ... It means you need "2x" "2" to defeat THREE (1) UNITS. If it was a "3" ... That means that you need "3x" "3" to again defeat THREE (3) UNITS.

Same goes for "4" and "6"... "4" means "4x" and "6" means "6x"...

You don't even need to THINK about the fractional "meaning" behind the values. You just USE the values AS-IS. And knowing what the value means: "4" I need to hit this unit "4x"...

The interesting part is when you must use TWO (2) or more values to compute how effective a combination of attacks are. I think that's where the challenge lies. Like I said, it's CLEVER and SIMPLE if you use the SAME value... Where it becomes more of a challenge is when you open it up to say a "2" + "4" what does that mean???

Fundamentally it's 150% + 75% = 225% ~= 200% = 2 units.

Not overly complicated... But not simple either. The percentages work great in this scenario... Let's see another:

"3" + "6" + "6" what does this mean???

Again this is 100% + 50% + 50% = 200% = 2 units.

This TBH is simpler to work with. From what I have seen... And this is VERY STRANGE for me to SAY: "It would be better off with the 'percentile' than just numbers, IMHO."

So your dice would look like this:

150, 150, 100, 100, 100, 75, 75, 50.

These would all be "percentiles" and you would NOT need the "%". I think that this would be more "SIMPLER" to use even though it is percentiles. I just showed you with a couple examples that it's easier to work with PERCENTILES and then figure out how many hits (100%)...

X3M
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Joined: 10/28/2013
1/32

That chance was rather easy for... me. Yeah, I know that a lot of people have trouble with probabilities and such.

So let me explain a bit.
Let's start with the 3 to 1.
If you need 2 hits for removing your opponents card. And the opponent needs 6 hits in order to remove your card. The ratio of card removal is 3 to 1.
The chance on rolling a 2 with that custom d8 is 25%. Or 1/4th.
The chance on rolling a 6 with that custom d8 is 12.5%. Or 1/8th.

1/4 * 1/8 = 1/32.

***

questccg wrote:
Note #1: What you are effectively doing is this:

3/2, 3/2, 3/3, 3/3, 3/3, 3/4, 3/4, and 3/6.

That would be a result for the cards that have 3 hits, yes.

Of course, you simply add up all the hits. Then divide by whatever you rolled with that last d8.

So it could be possible for a player to roll like roughly 91 hits and then roll the 4. In that case 91/4 is the number of cards to remove... Well, I guess it is chips at this point in the game...

Smarter players will probably do 40 hits, 40 hits, and thus remove 10 cards, 10 cards. Then 11 remains, which is a bit easier to handle. I guess 2 more cards can be removed now. And the remaining hits are not sufficient.

***

Now, for the medics and mechanics and engineers joining the battle.

If the custom d8 is playtested. I could see if these "shield" cards can modify the number of hits needed per card/chip. A +1 would make the
22333446 into 33444557. And that is a bit weird to balance. So I need to think of that one. Maybe not do it and simply have them function entirely different.

X3M
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Joined: 10/28/2013
Ah, i see you are posting too

Ehm, the divider dice will be only 1 die.

So you use the first 3d6 and 3d8 for deciding the number of hits that your side makes.
Then the custom d8 decides how many hits the opponent cards need to take.

The only thing missing here is that each hit might have a different damage value.

What players really do is decide with the custom d8, how many times the threshold (armor) of a card has to be reached.

So, if a player has a number of hits with different damages.
Let's say 6 times a 1 and 4 times a 3.
And the target cards are those of 9.

Then rolling a 2 with the custom die means. You need to reach the threshold 2 times. You use 3x3=9 for the first threshold. Then 6x1 + 3=9 for the second one. And the other card is beaten.

questccg
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Understood 80% of what you posted...!

X3M wrote:
Ehm, the divider dice will be only 1 die.

Okay... Not sure of the implications. I need to think about it some more...

X3M wrote:
So, if a player has a number of hits with different damages.
Let's say 6 times a 1 and 4 times a 3. And the target cards are those of 9.

Then rolling a 2 with the custom die means. You need to reach the threshold 2 times. You use 3x3=9 for the first threshold. Then 6x1 + 3=9 for the second one. And the other card is beaten.

So are you saying 6x "1" (one type of unit) and 4x "3" (another type of unit)???

And the target card(s) is "9" defense. I'm I understanding so far correctly??

questccg wrote:
And I'm definitely not getting "threshold 2 times".

If you roll a "2" that means 150%, so it means you need to hit it TWICE to defeat "3 units" (300%). Does this mean a "dynamic" value instead of "3 units" as per my earlier understanding???

Now where I need some explanation (better) is with these statements:

Quote:
"3x3=9 for the first threshold" and then "6x1 + 3=9 for the second threshold"... And then the "card is beaten"?!?!

I really can't make sense of this until the previous assumptions that I have made are corrected. Please let me know IF I have understood before explaining how to HANDLE multiple damages...

Note #1: If this "threshold' you are trying to have a DYNAMIC value instead of just "3", it can adapt to different values... Well that will totally blow my mind! (LOL) If this is correct, you'll need to provide like 2 or 3 examples where I can see the difference and understand properly. Cheers!

X3M
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Joined: 10/28/2013
Ok, cardgame examples

First examples will only contain my riflemen.

Costs 2
Armor 1
Damage 1
d8 Accuracy 4

Quote:
Most factions use the 3d8 for the accuracy roll. However, the original had 3d6 in mind. But d6 simply had less options for the first tier.
d6 Accuracy 3 would have been the original
Trivia: In my proto-type game, the accuracy is a d6 roll for 5. However, each rifleman there gets its own die to roll.

Both players have 6 riflemen.

A attacks B or vice versa. And all cards fight.
The 3d8 is rolled. But also the divider die, d'8.
For now, I want the attacker to also roll the divider die itself. Because then this can go in 1 go by the attacker. I want to actually cut down the downtime for the other player to get the divider die and make confusion.
Also, if a defender rolls the divider die for itself. Chances are that in a 3 or more player game. Everyone will target the player rolling a 2.
So, the divider die is the double edged sword for the attacker.

Quote:
With only the 3d8, the chance for a card killing itself was only 12.5%. Now with the divider die, this is increased to 17.2%. This due to expanding the need for 3 hits, with the chance of 2 hits in combination with the divider of 2.

Round 1
A rolls 238, 6
B rolls 288, 2

The 3d8 are compared to the accuracy.
2 and 3 are lower than the 4, these are hits.
8 is higher than the 4, these are miss.

A has 2 hits per card it owns, thus 12 in total.
B has 1 hit per card it owns, thus 6 in total.

However, A rolled a 6 with the divider die. Thus each target needs to be hit 6 times. Only 2 cards of B will die that round.
B rolled a 2 with the divider die. Each target needs to be hit only 2 times. 3 cards of A will die that round.

A has 3 cards left.
B has 4 cards left.

Round 2
A rolls 358, 4
B rolls 667, 3

A has 1 hit per card it owns, thus 3 in total.
B has no hits...

However...A rolled a 4 with the divider die. Thus each target needs to be hit 4 times. A had only 3 cards left, and each did hit only 1 time. 3 hits is not enough this time. A too, will not remove any card.

A has 3 cards left.
B has 4 cards left.

Round 3
A rolls 118, 3
B rolls 556, 4

A has 2 hits per card it owns, thus 6 in total.
B has no hits...

A rolled a 3 with the divider die. Thus each target needs to be hit 3 times. This means 2 cards can now be removed from B. B is down to 2.

A has 3 cards left.
B has 2 cards left.

Round 4
A rolls 167, 2
B rolls 455, 3

A has 1 hit per card it owns, thus 2 in total.
B has 1 hit per card it owns, thus 2 in total.

A rolled a 2 with the divider, and thus managed to remove yet another card of B.
While B rolled a 3 with the divider, and thus managed to remove no cards of A. Simply because B had only 2 hits.

A has 3 cards left.
B has 1 cards left.

Round 5
A rolls 123, 4
B rolls 477, 2

A has 3 hits per card it owns, thus 9 hits in total.
B has 1 hit per card it owns, thus flipping the table, barely missing A, since B already knows it lost.

Still, A wants to know how many cards would have been removed from B. With 9 hits and a divider of 4. 2 cards could be removed. With 1 hit to spare.

B says: "Good Game, A(hole)"
A raises its bottle of beer and says: "Good Game, A(cohole)" And drinks from the bottle.

Need another example? But with more different cards?
Then I should do one with the riflemen and some light tanks.

questccg
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Joined: 04/16/2011
Please give me time to review this post

There's a lot of EXAMPLE to digest... So let me go through it for protocol-sake in order to see if I can comprehend your method. Honestly, as long as it works for you it doesn't matter. But IF I can help you with simplification or showing what may be "too complex" or making it smoother by making some minor adjustment(s) then my duty to you is done!

Besides I hope that your game benefits from some "suggestions" and it seems like you are working through most of your design "details".

Whenever you get into the dirtier MATH ... That's when I get the impression of pulling away. If it's not understandable by ME... Most likely MOST people will not understand it.

As you know, I go to GREAT lengths to "over-simplify" matters that are more complex and TRY to make them EASIER to work with with various simplifications. I don't really care about ALL the MATH... As long as it WORKS and is FUN... Those are my two (2) criteria. It doesn't need perfect BALANCE and the proof to this is my "Exploding" Dice: instead of using formulas, I just went and made minor ESTIMATES for each side of the die... Such that it is CLOSE to the REAL DEAL/Ultimately Balanced die but with approximations to make the dice more user friendly and having some simplification.

Give me a chance to re-read over that message and ask where it may be unclear BEFORE you need to post a 2nd comment with different units than the Riflemen Example.

BTW thank you for explaining. It helps me understand how you are designing your game ... It's far from my own approach which is mainly approximations and dice-fudging to whatever FEELS like enough...

Cheers @X3M!

questccg
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Okay ... I understood the 1st Round. BUT...

X3M wrote:
First examples will only contain my riflemen.

Costs 2
Armor 1
Damage 1
d8 Accuracy 4
...
Both players have 6 riflemen.
...

Round 1
A rolls 238, 6
B rolls 288, 2

The 3d8 are compared to the accuracy.
2 and 3 are lower than the 4, these are hits.
8 is higher than the 4, these are miss.

A has 2 hits per card it owns, thus 12 in total.
B has 1 hit per card it owns, thus 6 in total.

However, A rolled a 6 with the divider die. Thus each target needs to be hit 6 times. Only 2 cards of B will die that round.
B rolled a 2 with the divider die. Each target needs to be hit only 2 times. 3 cards of A will die that round.

A has 3 cards left.
B has 4 cards left...

I think this "works" and is fine since as long as all the computation will be kept "under-the-hood".

If it's good for you ... Well then I'm sure it will be good for most players... People who like clever computation and enjoy things that are not as a simple as rolling a dice and doing a multiplication (for example).

Note #1: Let me help you out a bit more... Okay so each card has an "Accuracy" (a D8 value). That indicates how many HITS. That makes 100% sense and is pretty easy to comprehend and compute. Where it's a bit confusing is that you need to roll LESS THAN or EQUAL to the "Accuracy". That's a bit counter-intuitive. If you would FLIP this around and say that "Accuracy 1" means all rolls HIT and "Accuracy 8" means that most rolls FAIL.

So in the case of the Riflemen the "Accuracy 4" means you need to roll a "4" or HIGHER on the D8. That's a bit more INTUITIVE and makes more sense from a dice rolling perspective (and what people are used to). Your way is the OPPOSITE of the general understanding about how DICE ROLLS work in "gambling" and in general (most games).

That's something that you can FIX and there is little impact. (+1)

Note #2: It would be better to express the Riflemen as a FORMULA rather than explaining. Something simple like:

QTY x DMG x HITS

6 x 1 x 2 = 12 DMG
6 x 1 x 1 = 6 DMG

Now all the REST which is a LEVEL HIGHER in terms math analysis and comparison needs to be simplified.

But if it can BE a simple "division" then so be it:

QTY x DMG x HITS / DIVIDER = Total Damage

6 x 1 x 2 = 12 / 6 = 2 Total Damage
6 x 1 x 1 = 6 / 2 = 3 Total Damage

AND THAT WORKS (I think!)... Well then maybe it's NOT TOO BAD. Just make sure the FORMULATION is EASY. If all you NEED to do is DIVIDE the "divider D8 die" well then I think this is good.

BUT I have ONE (1) COMMENT. The "divider" die should be a D6 not a D8. Why? Dividing by "7" and "8" is well... difficult. It it was a D6, dividing by "6" is difficult ENOUGH. Less than "6" is reasonable integer math.

So please clarify if I have understood correctly. Sorry it took some time and I wanted to be sure I understood ... Since this seems like for you a very good method for your card game!

Note #3: In my case, it comes out to something VERY similar:

QTY x DMG x ACCURACY = Total Damage

Where ACCURACY is a custom D8 die roll from three (3) dice (in a pool).

I like your method. It's good. But my version allows me to use a smaller dice pool and allows me to determine ALL three (3) of my Squadrons in my Platoon their "accuracy" rolls. This means ONE (1) DICE roll of 3D8 (Custom).

Where in your case you need to roll 3D8 + 1D6 PER Squadron. So that's 9D8 + 3D6 in total. In my version this is ONLY 3D8 in TOTAL. It's a simplification and does more or less something similar with the requirement of only requiring rolling 3D8s (custom). Basically no matter what, in my version you ONLY roll three (3) dice. There could be 6-sided dice depending on the unit... But still it's only a question of rolling 3 DICE.

But your method is good... Just took the long-way to get a similar result.

Cheers @X3M...

X3M
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Joined: 10/28/2013
questccg wrote:I think this

questccg wrote:

I think this "works" and is fine since as long as all the computation will be kept "under-the-hood".

Yes, players are supposed to compare dice results with the cards.
Then see if a card will have 0, 1, 2 or 3 hits per projectile.(Which can become very complicated by design, I playtested a very weird sniper)
Once the number of hits are determined for each damage value, which they can track by placing chips as damage resources. Players will start distributing the damage over the enemy cards.
This part will be simple, since players are more or less forced to attack what has been determined before a roll. Unless the enemy had a big front. But in that case, optimal targets or tactical targets, are solved first.

questccg wrote:

If it's good for you ... Well then I'm sure it will be good for most players... People who like clever computation and enjoy things that are not as a simple as rolling a dice and doing a multiplication (for example).
The calculations are:
Comparison, Addition and Multiplying (the multiple projectiles times multiple hits times multiple cards)

questccg wrote:

Note #1: Let me help you out a bit more... Okay so each card has an "Accuracy" (a D8 value). That indicates how many HITS. That makes 100% sense and is pretty easy to comprehend and compute. Where it's a bit confusing is that you need to roll LESS THAN or EQUAL to the "Accuracy". That's a bit counter-intuitive. If you would FLIP this around and say that "Accuracy 1" means all rolls HIT and "Accuracy 8" means that most rolls FAIL.

Ah yes, in the testing phase and when we had our own battle's in the proto-type version. We always went for an accuracy roll like that. It isn't much of a change to turn this 180 degree's.

questccg wrote:
So in the case of the Riflemen the "Accuracy 4" means you need to roll a "4" or HIGHER on the D8. That's a bit more INTUITIVE and makes more sense from a dice rolling perspective (and what people are used to). Your way is the OPPOSITE of the general understanding about how DICE ROLLS work in "gambling" and in general (most games).

That's something that you can FIX and there is little impact. (+1)

Noted.
Accuracy 8 will be, you have to roll an 8 in order to hit.
Accuracy 5 will be, you have to roll a 5 or more, in order to hit.

The balance accuracy 4, would be 5 in the game.
The balance accuracy 1, would be 8 in the game.
The balance accuracy 8, would be 1 in the game.

questccg wrote:

Note #2: It would be better to express the Riflemen as a FORMULA rather than explaining. Something simple like:

QTY x DMG x HITS

6 x 1 x 2 = 12 DMG
6 x 1 x 1 = 6 DMG

Oh, ok. Thought that was a bit too hard for some.

It is going to be:
QTY x Projectiles x HITS = Damage Resource

Why like this? The damage resource is simply the number of projectiles that the player now has.
When distributing the damage. The player will most likely do this most optimal or is forced.
In either case. Each damage resource is then multiplied by the damage. Making this a 2 step.
First the player decides in the total projectiles that hit.
Then where they hit with the help of the damage assigned to those projectiles.

questccg wrote:

Now all the REST which is a LEVEL HIGHER in terms math analysis and comparison needs to be simplified.

But if it can BE a simple "division" then so be it:

QTY x DMG x HITS / DIVIDER = Total Damage

6 x 1 x 2 = 12 / 6 = 2 Total Damage
6 x 1 x 1 = 6 / 2 = 3 Total Damage

AND THAT WORKS (I think!)... Well then maybe it's NOT TOO BAD. Just make sure the FORMULATION is EASY. If all you NEED to do is DIVIDE the "divider D8 die" well then I think this is good.

The divider die should be renamed though. It is more like a "durability" die for the cards.
You need to remove the armor X times. And X is determined by this durability die. Is this way of explaining to the players, more understandable?

questccg wrote:

BUT I have ONE (1) COMMENT. The "divider" die should be a D6 not a D8. Why? Dividing by "7" and "8" is well... difficult. It it was a D6, dividing by "6" is difficult ENOUGH. Less than "6" is reasonable integer math.
The divider die, or now renamed durability die. Is custom.

The sides are:
22333446

No 7, no 5.
But no division either. The players will simply make groups of whatever is rolled.
Thus if a 6 is rolled. The players make 6 groups of the required damage dice.
If for example a tank with 9 armor is attacked. And the 6 is rolled. The player that attacks this tank will get groups that each do 9 or more damage. It is entirely possible that in the future, inefficient combinations are used. Only in order to get the last group of damage complete.

I had chosen the d8, because on the d6. The distribution would be:
223446

questccg wrote:

So please clarify if I have understood correctly. Sorry it took some time and I wanted to be sure I understood ... Since this seems like for you a very good method for your card game!
We all are busy... I have yet to do gaming today. But it is bedtime tbh.

questccg wrote:
Note #3: In my case, it comes out to something VERY similar:

QTY x DMG x ACCURACY = Total Damage

Where ACCURACY is a custom D8 die roll from three (3) dice (in a pool).

I like your method. It's good. But my version allows me to use a smaller dice pool and allows me to determine ALL three (3) of my Squadrons in my Platoon their "accuracy" rolls. This means ONE (1) DICE roll of 3D8 (Custom).

I have one (1) dice roll of 7 dice in total; 3D8+3D6+1D8(custom).
If the player has no need for the 3D8 or 3D6. The dice roll will be only containing 4 dice.
The 3D6 will have a high chance on not being used in the first decks to start with anyway. And perhaps, I will use them as a faction theme as well. Meaning, the player either needs that 3D8 OR 3D6, for the entire faction deck.

Still, using 3D8 AND 3D6 will bring forth a more balanced roll.

questccg wrote:
Where in your case you need to roll 3D8 + 1D6 PER Squadron. So that's 9D8 + 3D6 in total.
No...

It is mostly 4 dice. And the advanced players use 7 dice.

questccg
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Joined: 04/16/2011
What I am wondering about now...

Is how the Defender handles an attack. For example in TradeWorlds the attacking player decides WHO he will attack. In Magic, the Blockers are declared by the defending player. And so I wonder about each of these methods. What are the PROs and CONs. Since there is a "Tactical Readiness" die in my version, the Platoons are composed of forces given a variable mix. Which means that you don't always get the forces or units that you want TOGETHER into a platoon.

That seems to logically favor the Magic version where the opponent declares which unit or troops will BLOCK an "Attack".

It's not really BLOCKING in this case, it more like DEFENDING FORCES against a specific Squadron's attack.

The other way it would be TOO EASY: you target the weakest opposing force to deal the most possible damage.

Not very convinced that it should be done that way. Thoughts???

Note #1: Other things worthy of note is that since I am using a Dice Pool of "3" custom D8s... I like the idea of ASSIGNING dice like WORKERS. And there would be this "Worker-Placement" mechanic also. So primarily a Deck-Builder with combat featuring Worker-Placement. This will lead to more strategic thinking (in my mind at least). Because it forces you to do the best you can with your roll.

And then like you, the army battle to see WHO comes up on top (wipes out the enemy clear of all their forces). This of course results in the winner selecting a Victory Spot and hopefully draws the White Flag. That's more luck based too ... But there are probabilities that go along with this also. So while you might have a 25% of defeating your opponent, as you move along it goes to 33%, 50% and 100% for the unlucky opponent.

Note #2: I forgot the the "durability" die was CUSTOM:

X3M wrote:
The divider die, or now renamed durability die. Is custom.
The sides are: 22333446

No 7, no 5.
But no division either.

Much better too ... Yes indeed the D8 will be most appropriate in this method of deciding the outcome of battles or skirmishes. Completely forgot that it was going to be "custom".

Note #3: Also you mention NO DIVISION. I think TRYING to THINK like it's in groups or sets of units that are destroyed... is well... COMPLICATED. It would be MUCH easier to say: "Divide by the die result". The divisions are pretty reasonable and integer MATH can be applied. The WAY you explain HOW you get to those groups and sets is ... well... again a bit hard for most people to comprehend.

Simply telling them that it is a DIVISION would greatly simplify the MATH and not cause any Analysis-Paralysis when it comes to CONTEMPLATING HOW or WHAT to do NEXT in the process. But stating that it's purely a DIVISION is much more easier to understand EVEN if in some instances you must divide by "6".

That's just my own impression... Because I'm not too sure about all that COMPARISON analysis may be a tad too complicated. Like saying it's "2 hits" because you roll a "6" on the durability die and therefore meaning that you deal only deal "2" cards that round... Is well not to OBVIOUS thinking.

IF it's JUST a DIVISION: FINE! Works great. That's my take on TRYING to keep it simple. Math aficionados will look deeper to understand the MATH... But overall much simpler to treat this as a STANDARD Integer Division.

Again that's my opinion...

X3M
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Joined: 10/28/2013
I guess...I should slowly expand on the examples

Just to see over complications.
I am literly going to copy a...6?? yeah, 6 year old example :D
But then translated to a card game.

***

A new game

For this example. We imagine that both players have an army, consisting out of only riflemen and combat tanks.

In fact, each player has 2 cards. 1 card represents all riflemen. The other card represents all combat tanks.
Both players have both of them. The number of chips however, indicate how many they have.

The cards
Rifleman
Costs 2
Armor 1
Damage 1
Accuracy 5 (heads up, lets do the roll public wants, roll 5 or higher for a hit)

Combat Tank
Costs 8
Armor 9
Damage 9
Accuracy 5

Please keep in mind, riflemen take only 1 damage from the combat tank per hit. If the durability die rolls a 2. This means you still need 18 hits in total.

How many chips do the players have?

For every Combat Tank, 4 Riflemen.

Seeing as how the cardgame works with a threshold. While the original proto-type board game had health tracking. I have no other choice than make sure the riflemen are at a level where they still matter to the players.

I calculated that the best roll will require 6 Riflemen. But provide a chance of (1/8) x (1/4) = 1/32.
The average roll will provide with 1.5 hits per Rifleman.
With the durability roll being 3 as average. You need 18 Riflemen in order to kill 1 Combat Tanks per round.

Now then, there is a threshold where the riflemen become overkill for the combat tanks. But I am not going to calculate that. Instead, I am simply going to give each player, 36 Riflemen and 9 Combat Tanks.

Quote:
Only 2 rounds! Or else a tldr post would be made

Round 1, A in turn

Player A is in turn and decides to attack. Before the dice are rolled. There are several steps the players undertake.

Player A has 2 cards of equal value. And has several options of formation:
- Riflemen attack. Combat Tanks don't.
- Combat Tanks attack. Riflemen don't.
- Both cards attack separately.
- Both cards attack together, with the riflemen as fodder.
- Both cards attack together, with the Combat Tanks as meat.

When attacking, you always, ALWAYS, select targets for your cards. They can have the same card as target. But it is also possible to choose many different targets.

Player A also has the choice to attack one of the following:
- The enemy Riflemen.
- The enemy Combat Tanks.

There are 10 different options with the 2 cards that player A has. But whatever A does. If a card attacks now, it cannot help defend in the next round.
For fun, I take a random decision here:
Both cards attack together, with the Combat Tanks as meat.
The attack is on the enemy Combat Tanks.

This means that there is one group that attacks

Once A has made its regrettable choice. B can react as following:
- The Combat Tanks don't return fire.
- The Combat Tanks return fire.
---
- The Riflemen don't help.
- The Riflemen help defend by being fodder.
- The Riflemen help defend by being support.

At this moment, there are 6 different options for B.
If B chooses to take the damage. It will loose Combat Tanks. However, then the Combat Tanks can attack in the next round without being stopped. Seeing as how the Combat Tanks of A are meat. B most optimal decision is to at least return fire with the Combat Tanks.

As weird as it sounds. B doesn't have much limitations here. In fact, the most important decision now is, will B use its Riflemen to help defend or not? And as stupid as it sounds, right now, it is smart to NOT use the Riflemen. Instead, B decides ( I decided for B, B has no free will ) to keep its Riflemen at bay.

If however B did use the Riflemen as well. They could simply take the damage instead, but that is not smart here. Or be support and make it a mirror match.
The attack that a player does is simply forcing the opponent to defend.

Time to roll the dice!!

Player A rolls: 236 / 3
Player B rolls: 568 / 4

We compare the accuracy dice with the accuracy. 5 or higher will be a hit.
A has 1 hit per chip.
B has 3 hits per chip.

These cards only have 1 projectile per chip:
QTY x HITS = Damage Resource

A targets the Combat Tanks. And thus the damage is taken in full.
1 Damage: 36 x 1 = 36
9 Damage: 9 x 1 = 9 (each damage resource here will do 9 damage)

The durability die is 3. Thus in order to defeat a Combat Tank. 3 times 9 damage has to be inflicted.
We have 9 damage resources of 9. 9/3 = 3. We can remove 3 Combat Tanks with this attack.
We also have 36 damage resources of 1. 36/3=12. We can remove 1 more Combat Tank with this attack.
After the attack round. B will have to remove 4 Combat Tanks.
B will then have 5 Combat Tanks remaining.

B targets the Combat Tanks of A. And thus the damage is taken in full. If B gets through, the damage would be overkill.
9 Damage: 9 x 3 = 27

The durability die is 4. Thus in order to defeat a Combat Tank. 4 times 9 damage has to be inflicted.
We have 27 damage resources of 9. 27/4 = 6.75. We can remove 6 Combat Tanks with this attack.
After the attack round. A will have to remove 6 Combat Tanks.
A will then have 3 Combat Tanks remaining.

If you think about it. B did make a gamble here.By not including the Riflemen. And succeeded.
If B rolled a durability of 3 or 2. Then the Combat Tanks of A would be all dead. But vice versa could have happened. And the Riflemen are strong enough that they count as if they are 4 more Combat Tanks, in terms of damage.
Either way, the riflemen stayed out of the combat and would all survive for certain. Defending cards that are not targeted and stay out of combat. Will not be harmed at all. Which is important for the next round.

Table:
A has 36 Riflemen and 3 Combat Tanks
B has 36 Riflemen and 5 Combat Tanks

Round 2, B in turn

The forces of A attacked last round.
And are now exhausted. They cannot defend anymore.

The forces of B are not exhausted anymore.

All B has to do is attack A.

B is very smart. B wants to B like me.

Player B has 2 cards. But they are not of equal value any more. But still has the same options of formation:
- Riflemen attack. Combat Tanks don't.
- Combat Tanks attack. Riflemen don't.
- Both cards attack separately.
- Both cards attack together, with the riflemen as fodder.
- Both cards attack together, with the Combat Tanks as meat.

Seeing as how no cards of A can defend properly. It is going to be one sided. B decides to make the most optimal attack.

All Riflemen are going to attack the Riflemen.
All Combat Tanks are going to attack the Combat Tanks.
Ehm... that would be option 3.

Of course, there is a risk that for a kill on one of the 2 or both cards. Can be avoided by missing just 1 damage.

Player B rolls: 355 / 2

We compare the accuracy dice with the accuracy. 5 or higher will be a hit.
B has 2 hits per chip.

QTY x HITS = Damage Resource

The damage resources:
1 Damage: 36 x 2 = 72
9 Damage: 5 x 2 = 10

The durability die is 2. Thus in order to defeat a Rifleman. 2 times 1 damage has to be inflicted.
We have 72 damage resources of 1. 72/2 = 36. We can remove 36 Riflemen with this attack.
After the attack round. A will have to remove 36 Riflemen.
A will then have 0 Riflemen remaining.

The durability die is 2. Thus in order to defeat a Combat Tank. 2 times 9 damage has to be inflicted.
We have 10 damage resources of 9. 10/2 = 5. We can remove 5 Combat Tanks with this attack.
After the attack round. A will have to remove 5 Combat Tanks.
A will then have 0 Combat Tanks remaining.

Table:
A has 0 Riflemen and 0 Combat Tanks
B has 36 Riflemen and 5 Combat Tanks

Hmmmm... only 2 rounds.
And the RPS wasn't really used yet...

X3M
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Joined: 10/28/2013
Instant kill

Ok, so there is a chance of having an instant kill. And while the chance is 1/32. I wonder how big the chances are on every accuracy. Now that I decided to have that durability die added to the mix.

... after all, in a mirror match, it can be over in 1 attack. Which is something I don't want to happen.

comparing combat mechanics

First I need to think of the combat mechanic again. Because if I look at MtG, one player attacks. And it doesn't matter what the defender does. All what matters is that the defender can defend in the most optimal way. So the attacker will attack only if the forces are a factor 2 to 3 bigger.

In my game. The attacker can target in an optimal way. Is what the attacker might think. Because after that, the defender will put the right units in front again. In antipation of this. The attacker will probably use a mix.

What I neglected to look at is the order of damage distribution.
The damage between 2 mixes is still handled poorly.
Of course, the card in front takes damage first. But if a good RPS match is used. Those projectiles are spend first.

This was less apparent in my proto-type game. Not only was the order of projectiles depending on their range. But the health to damage ratio was 3. The card game has a health to damage ratio of 2.

We know that with an equal range mix, in combination with a mirror match. The damage in the proto-type is also distributed as optimal as possible. But there the cover mechanic is stronger since every "card" thus piece is an object on itself. No chips. And in combination with the H/D ratio of 3, it takes several rounds before the attacker breaks through.

Not only that. But there is also no exhaustion on the attacker side. As long as there are AP avaiable. A squad can also return fire.
The card game is so far, very strict in this.

Combat mechanic improvements?

A possible solution is that I force "frontline" units to fire first. Despite them going in front later on.
Meaning, that if a mirror match happens. And you get this perfect roll. But the RPS matching is worse due to good positioning.
Thus, the tanks and riflemen in front attack the other tank squad. The defender gets riflemen as support instead of a front line. The rps difference would be that the riflemen of A and the tanks of B will spend their damage first on each other. Then the support lines will fire on the frontlines till they die. And remaining damage will reach the front lines.

As example:
All cards have 3 hits with 2 durability.
36 riflemen have 108 damage resources.
9 combat tanks have 27 damage resources with 9 damage each.

A puts the riflemen in front.
B puts the combat tanks in front.

108/(2*9)=6 dead combat tanks of B.
27/2=13.5=13 dead riflemen of A.

26 riflemen remain for A
3 combat tanks remain for B

Now the support lines fire.
27/2=13.5 dead combat tanks of B. There are only 3 left.
27-3*2=21
21/2=10.5=10 dead riflemen of B.

108/2=54 dead riflemen of A. There are only 21 left.
108-2*21=66
66/(2*9)=3.67=3 dead combat tanks of A.

The frontlines of both sides are dead now.
6 combat tanks remain for A.
26 riflemen remain for B.

---

If both sides fight again WITH the best roll there is.
A goes down to 2 combat tanks.
B goes down to 17 riflemen.

A goes down to 0.
B goes down to 14.

Still, this is 3 rounds. But by circumstances. Every card could fight every time. Not sure if I should remove exhaustion or not.
And...these are all well balanced units.

Maybe I need to increase the H/D ratio?

Where is the average?
The H/D ratio is 2.
But this means that the average hit is 1.5 with the 3 dice. After all, the average durability is 3. Thus 3/1.5=2
But when you roll for the basic cards. A player will either roll 1 hit or 2 hits for all basic cards. This is a factor of 2!!!

So, once again I am looking at the dice. They do not feel right. And a 4d8 or 4d6 might be a better solution. Where the average number of hits is 2. Then I need to re-adjust the durability roll in such way, the average is 4 on that roll.

If 4 is the average. Then 4 is the 100% weight in damage.
2=200%
3=133.3%
4=100%
5=80% (can't really combine this one)
6=66.7%
8=50%

23 44 46 88
200+133+100+100+100+67+50+50=800 ok, so that matches.

Now player roll either 5 or 9 dice.
But with the possibility of 2 hits being the average.
And 3 hits not being a factor 2 but only 1.5. The game has once again reached a bit more balance. And thus the game will be a bit longer.

The chance on the most optimal roll for the basic units, went from 1/32 to 1/64.
Which basicly would result in the same ammount of defeated cards.
Well, I just reduced the chance on it happening by half.
And the rolls around the average are nerved too.
Somehow, I get the feeling that 5d6 and 5d8 where a bad idea. Because then we where dealing with an average roll of 2.5. But the 2 closest to that are 2 and 3. Those too have a factor of 1.5 difference. But you can't get the exact average that way, unless you get a mix of accuracies.

X3M
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Joined: 10/28/2013
H/D ratio

I have been thinking about this one.
Have I not made this too low??

The card game has 2. But if a support unit has 6 projectiles. It means 3 units can be removed at once.

The proto-type game has 3. But 2 effects make this higher.

1. The attack range weight is 50% higher than the movement speed. With a default of 2 for both. The weapon is 20% more expensive than the body. In other words, only 100%/120% effective. Thus the durability is times 1.2

2. The damage roll makes equal or stronger weapons in RPS effects. 20% less effective as well. This just happens to be another 1.2

Thus my board game actually has 3 x 1.2 x 1.2 = 4.32 as durability for the basic units.

I think my card game is simply too fast. I had chosen 2 as H/D ratio because there is also a threshold to overcome.

But with the dice giving factors 2 and 1.5 in a 1/64th chance.... plenty of overkill.

The RPS can't really fight this either.
The first degree RPS is a factor of 1.5.

So....
36 riflemen fighting 18 assault buggies.
The buggies have 3 armor.
3d8 gives 3 hits per riflemen.
The durability die rolls a 2.
36 x 3 = 108 damage resources of 1.
108 / (2 x 3) = 18 assault buggies will die.

This is not what they are supposed to do tbh.

X3M
Offline
Joined: 10/28/2013
The 3 factors, analyses

Factors....
A factor 2 means times 2.
A factor 3 means times 3.
A factor 2 with a factor 3 means 2 times 3 is times 6.
A factor 2 against a factor 3....
Well, in that case we can have either 2 divided by 3 or 3 divided by 2.

There are 3 factors involved to see if overkill is possible:
1. RPS.
The RPS factor here is: "how much difference is there between 2 tiers?" In other words, if the weapon is not right for the target. How much effect is lost?
2. Roll-for-damage.
This factor involves the maximum damage compared to the average damage. Depending on the resolution and number of the dice, the true factor might have to be calculated differently. It also has something to do with a type of overkill.
3. Other factors.
Caused by other game mechanics. For example, accuracy rolls for terrain. Balance calculations on weapon range and movement speed. etc.

There are also different types of overkill:
1. Overkill is that you do too much damage in order to kill a target.
2a. Overkill on an unit, means less damage for the group. This also means that the group will be more durabile.
2b. Overkill on a group, means that some factors are too strong compared to other factors. In this case, the RPS factor is weaker than the roll-for-damage factor.

X3M
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Joined: 10/28/2013
Should I change the RPS factor?

Right now, it is a factor 2 in costs. A factor 3 in armor or weapon.
The true RPS factor is 1.5. (3/2)
Which means, you are supposed to have 50% more troops if your tier difference is 1.

In order to increase this. And perhaps re-modify.
I can still keep the factor 2 in costs. But a factor 4 in armor or weapon. This means that tier 3 will go from 9 to 16.

Now the true RPS factor is 2.

***

I don't know if I make the right decision doing this. For starters. The armor value goes to 2 digits at tier 3.
I also had an imbalance added in order to fight against the cumulative effects. In a card game, these are very strong. But then again, I have much bigger armies now with more units.

I still could go the "public" version route. Where I skip tier 1, have tier 2 to 4. And adjust tier 3 by 1 point.
Then all the armor and weapon value's can be divided by 4.
And we are left with 1, 2 and 4.

Tier 1 and 3 are the new tiers in that example. Because tier 2 would be halfway. And the digits are 1 again. I could easily add tier 4, which would have 8 armor.

Thus, back to the costs of 2, 3, 4 and 6. While the value's are 1, 2, 4 and 8.
The true RPS factor here is also 2. While halfway would be 1.5 again.
Thematic, this is no problem. Still, the tank tier could be costing 8, and we are back to 2 digits in value, 16.... A positive thing would be having tier 2 being troopers, tier 3 again the light vehicles like buggies.

***

I guess, the important question is.
Can players handle adding up, till 16??

questccg
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Joined: 04/16/2011
Hmm... I need a break. Plus I have my own issues to deal with.

I see you posted so much content it may take me a week or two to process it...

In any case, I am considering having a "Chip Limit or Cap". Meaning that you may only be allowed "12" Chips in total. But I am considering having DIFFERENT "levels" for the chips.

For example: Assault/Concussion chips may be in groups of FIVE (5) "Chips". So while each chip COUNTS towards you "12 Cap", each one is WORTH 5 of that unit. So for example:

A Regular Soldier may come in groups of FIVE (5) for each chip used. So this would mean that "3 Chips" = 15 Units (3 x 5). Which would mean that you would have nine (9) more Chips you could configure...

Another reason this is important is to LIMIT the number of chips used by the game. I mean if there are nine (9) tiers and each one has chips, this means 9 x ? (how many chips per tier) and is in a way good, because things can be color-coordinated, but BAD in the way that it requires too much diversification and the availability of more TYPES of chips means you need to control the amount of each chip per tier.

But if there would be FIVE (5) chips groupings from 1 to 5... This means that AT MOST your army could have 60 Units (5 x 12 cap). Is this an "unreasonable" ceiling? Hmm... That's a Company minimum (60 Soldiers minimum). Maybe not!

What I am thinking is the "12" cap is not enough TBH.

What if instead you had "5" squadrons... And you had like "25" cap. Meaning that on average, each Squadron could have FIVE (5) "Chips"! This could mean that some Squadrons could have MORE and other LESS. This could ADD VALUE to "cheaper" (fodder) units as having more "value" per chip.

Like my example of the Regular Soldier which could come in groups of FIVE (5) for each chip used.

This means that with the "25" cap, you could have 5 times (5x) or 125 Soldiers. Which equates a company (50 to 200 soldiers).

Not too BAD! But "25" cap for chips SOUNDS and SEEMS HIGH... "25" x 4 players = 100 chips... It's not working TBH. Ayayayai...!

I need much more thought on this matter ... Because it is FAR from over. And there needs to be more "analysis" when it comes to "Chips".

Sincerely.

Note #1: Makes me think that "Chips" should come in the following values: 1, 2, 3, 4, 5 and 10. And the Regulars should be "10" each with a cap of "15" Chips = 150 Soldiers.

This means 60 chips per player... Hmm... Better. Still more thinking to do.

I'll do some more thought and post my results at a later date. For now, know that I don't have a working solution that will LIMIT the amount of chips needed. And well ... That is very important because "Chips" are EXPENSIVE components to make... They add a HUGE amount of weight to any game box too.

So there are considerations outside "MATH" or "Balancing". There is COST and WEIGHT too... Hmm.

Note #2: The maximum amount of "squadrons" is TEN (10) or "2 rows x 5 slots". So 10 x 10 = 100 Soldier Maximum. I should be thinking in terms of "10" (Ten). And then I need to compute minimums also. Ayayayaii...

More thoughts on this aspect as I figure out what to do...

X3M
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Joined: 10/28/2013
Well, all I can say now is

Well, all I can say now is that you don't need to read those posts.

All you need to know is that the game is too fast. 1 basic card can kill more than its own worth in other basic cards. And the chance on this is also too big.

I need to fix:
- H/D ratio
- Accuracy rolls need to be less basic
- durability roll needs to remain soft enough
- RPS needs to be sharper

questccg
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Joined: 04/16/2011
I think I have solved this issue a while back.

X3M wrote:
All you need to know is that the game is too fast. 1 basic card can kill more than its own worth in other basic cards. And the chance on this is also too big...

This is what I meant when using my Accuracy Dice: AT MOST there is a 1/1 (100%) Accuracy. It figures 12.5% (1/8) probability that an attack is 100%. Otherwise the Accuracy is BELOW 1/1 (100%) and can be as poor as 1/5 (20%). The Player rolls 3D8s from his pool of dice and determines which squadron will use what die.

So YES, you can get 100% Accuracy. Is it impossible to get? No. It has a reasonable probability: 3/8 (37.5%). I think this is the correct probability:
1/8 + 1/8 + 1/8 = 3/8 (37.5%).

In any case, I have moved on to my CHIP dilemma which is figuring out HOW many chips should be allowed and the value of said chips (is it like health?!?!)

Sincerely.

X3M
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Joined: 10/28/2013
Which means

My H/D ratio needs to increase.

And maybe I can combine the accuracy roll with the durability roll.

What was I thinking...
The H/D ratio was only 2 while the RPS factor of the first degree is 1.5.
A total factor of 3.

In my proto-type game it is 3 for the ratio and 2 for the first degree.
A total factor of 6.

The minimum H/D number is 1 with any accuracy roll. Even in my proto-type game. Where 1/6th means a factor 6 with a hit. The card game has 1/8 which becomes 8.
6/6 or 8/3. Tgat is a huge difference between the 2 games.

But the numbers in the prototype game are also treated individually. Not amass. And I constantly felt this would be a problem. Even if I where to use 1 die for accuracy.

questccg
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Joined: 04/16/2011
To me it seems SIMPLE

What you need to do is ensure that whatever dice you plan on rolling that they are all "Fractionary". Meaning that when you ADD them UP together, AT MOST you end up with "1". Anything else needs to be BELOW "1". Why? Let us make the assumption that all your dice results in "2" or HIGHER... This means that the result will lead to "overkill" (as you put it): having HIGHER than "1" means that the unit or troop can EXCEED more than it's own STRENGTH or POWER and inflict more DAMAGE to the opposing units.

While this MAY be desirable under certain conditions (like say a Nuke Bomb is used... Or a Fragmentation Grenade is used...), anything else will result in the problem of "overkill" (being able to kill more than one unit itself).

And if you think about this a little bit, it makes sense. If a Rifleman can ONLY target ONE (1) opposing unit, that should at most be a "1". If however you have a Gunner which can Spread his Machine Gun to more than one (1) target, then values of "2" or higher make sense... But only with this "special" configuration.

So I think in terms of what should concern you most is to ensure that BALANCE exists when factoring all the dice and they should NOT go over 1/1 (100%).

Cheers!

questccg
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Joined: 04/16/2011
Also for you to consider

Working with Friendlier "numbers" is good. Numbers "6" or less are fairly easy to do Integer Math. Yeah "6" is borderline in terms of division... But still DO-ABLE! 2/6 or 4/6 = 1/3 or 2/3 so there is some simplification that CAN be done! Working with "3" is much easier than "6" and since you can simplify some of the "6" terms into "3" or even "2" (3/6 = 1/2) and you are left with 1/6 and 5/6... The only tricky values for Integer Math.

Again I'm pretty sure you are AWARE of all this. Because I know you are SMART and you've worked already with the Fractions and trying to shy away from even the SIMPLER Integer Division...

Best @X3M!

X3M
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Joined: 10/28/2013
questccg wrote:What you need

questccg wrote:
What you need to do is ensure that whatever dice you plan on rolling that they are all "Fractionary". Meaning that when you ADD them UP together, AT MOST you end up with "1". Anything else needs to be BELOW "1". Why? Let us make the assumption that all your dice results in "2" or HIGHER... This means that the result will lead to "overkill" (as you put it): having HIGHER than "1" means that the unit or troop can EXCEED more than it's own STRENGTH or POWER and inflict more DAMAGE to the opposing units.

While this MAY be desirable under certain conditions (like say a Nuke Bomb is used... Or a Fragmentation Grenade is used...), anything else will result in the problem of "overkill" (being able to kill more than one unit itself).

And if you think about this a little bit, it makes sense. If a Rifleman can ONLY target ONE (1) opposing unit, that should at most be a "1". If however you have a Gunner which can Spread his Machine Gun to more than one (1) target, then values of "2" or higher make sense... But only with this "special" configuration.

So I think in terms of what should concern you most is to ensure that BALANCE exists when factoring all the dice and they should NOT go over 1/1 (100%).

Cheers!

Then the durability die was a bad idea.

questccg
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Joined: 04/16/2011
As I understood it... Not how you perceive it.

X3M wrote:
Then the durability die was a bad idea.

Just as a reminder to all who might be reading and following along, the durability die was a "custom" D8 with the following values:

2, 2, 3, 3, 3, 4, 4, 6 = 800% = 100% average!

It may not be necessarily a "bad idea". Remember that 2 = 150%, 3 = 100%, 4 = 75% and 6 = 50%.

And to simplify the logic, it was "3" (the number of HITS) divided by the die value. So if there are "3 hits" you get:

3/2, 3/2, 3/3, 3/3, 3/3, 3/4, 3/4, 3/6.

I think that this doesn't work out that well with "3 HITS". If it was "2", then you would get:

2/2, 2/2, 2/3, 2/3, 2/3, 2/4, 2/4, 2/6

or

1/1, 1/1, 2/3, 2/3, 2/3, 1/2, 1/2, 1/3.

That could be better. So from what I have understood you would need to LIMIT the number of rolled "dice" for your HITS to "2" and not higher.

Which is BAD by design. And so I THINK that the "durability die" is the source of the problem but it's NOT bad. Just needs better odds (less favorable) for it to work effectively.

Maybe something like:

3, 3, 4, 4, 4, 6, 6, 9

And with "3 HITS", you would get:

3/3, 3/3, 3/4, 3/4, 3/4, 3/6, 3/6, 3/9

1/1, 1/1, 3/4, 3/4, 3/4, 1/2, 1/2, 1/3

That means your values would go DOWN from 100% as being the BEST possible roll and going downwards to 33% (on the lowest end).

MAYBE something like that...???

Note #1: This "alternate" D8 is more compatible than the "2 HIT" version above. Okay so it doesn't average out to 100%. If you use less HIT Dice, it works... If you TRY with "4" it will cause it to break again:

4/3 = 125%.

You just need to figure out if it works with your way of thinking: number of successful hits required to defeat x units. Whereas, I prefer to see it as an Integer Division... (For me anyways, it's simpler to understand that needing to reverse think the logic that it's a number of HITS). "2" hits = 150% x 2 = 300% = Too powerful hits... (Overkill as you say)!

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