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Throwing dice to overcome a defense threshold rock paper scissors-style

4 replies [Last post]
Thu, 11/13/2014 - 10:45
Garwyx
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Not for the game I'm currently working on, I just thought of this idea because of the 4th version of Dungeons and Dragons. I just want to get your ideas. The combat system is as follows:

Every player/hero and opponent/monster does damage in one of three schools. To make it easy, let's call them melee, ranged and magic. Melee has advantage over magic, magic has advantage over ranged and ranged has advantage over melee (let's keep the argument of whether this makes sense lore-wise out).
On encountering an opponent, the player rolls two dice. If he has the advantage (e.g. player is melee and opponent is magic), the player chooses the highest roll.
The opponent has a threshold. If the threshold is exceeded the player attacks, if the roll and the threshold are equal none attack, and if the roll is lower the player receives damage.

I still need to think what would happen when equal combat schools meet. It would be possible to throw with 3 dice at all times, or just throw a single die in this case.

Like I said I have no intentions (yet) to use this myself, so if you feel inspired please feel free to use (parts of) this idea. I am curious to your guys' feedback :).

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Thu, 11/13/2014 - 15:15
#1
X3M
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Joined: 10/28/2013
this, me like

I like this idea.
It can be implemented in so many ways, yet keeping the RPS system as well.

The right target?
2 Dice, highest counts
This can be upgraded to
3 Dice, highest counts

Not the right target? But not the wrong either? (Melee vs melee)
3 Dice, middle counts

Thus 3 dice would be best then, right?

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Thu, 11/13/2014 - 16:05
#2
Garwyx
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Joined: 10/28/2014
I like your suggestion to

I like your suggestion to have an upgradable system. If you could increase the number of dice used, care must be taken what to do when one has a disadvantage, because obviously the chance for a low roll increases as well as the chance for a high one. Another option would be to change the die-type, as I do in my current game (e.g. from D4 to D6).

I also though 3 dice would make a lot of sense because then you have low, middle and high. However, the amount of dice in a system such as this has to be tweaked very well, because the chance for an extreme (high or low roll) increases with an increasing dice pool: imagine using 100D6, then every roll will get at least one 1 and one 6. Then you might as well not use dice and say "I have a disadvantage vs I have an advantage".
Therefore my initial thought is that players that encounter an opponent with the same combat school use a single die.

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Fri, 11/14/2014 - 02:27
#3
X3M
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Well, with 100D6 you expand

Well, with 100D6 you expand the game in an extreme way. The rule for this is, simply not doing this :).

3 Dice would allow for example rock vs rock to be balanced. Rock vs scissor is of course a +, and rock vs paper is a -.
When using only 2 Dice, you have a +, but twice a -. However, - vs - is also equal.

This issue arises when you have a rock with a scissor weapon against another rock that has a rock weapon. Logical speaking, the difference should be less then when facing a rock with paper weapon.

Therefor, this technique works best with 3 Dice. That is my opinion.

***

Other ways of using this technique.

How about first the 3 dice. Then using the number that you get (low, medium or high). And roll that number of dice to determine damage.
This way, you work with higher numbers.

***

I am curious how the chances are divided when using just 2, or 3 dice.
Of course I can calculate 4 or higher. This takes time though. I'll come back later with results. Might be monday ;)

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Fri, 11/14/2014 - 14:53
#4
X3M
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Probabilities

1 Die; lowest or highest, out of 6 possibilities
Each roll is 1 out of 6

2 Dice; lowest roll, out of 36 possibilities
1: 11 = 30,556%
2: 9
3: 7
4: 5
5: 3
6: 1 = 2,778%
highest roll is upside down.

3 Dice; lowest roll, out of 216 possibilities
1: 91 = 42,130%
2: 61
3: 37
4: 19
5: 7
6: 1 = 0,463%
highest roll is upside down.

4 Dice; lowest roll, out of 1296 possibilities
1: 671 = 51,775%
2: 369
3: 175
4: 65
5: 15
6: 1 = 0,077%
highest roll is, you guessed it, upside down.

I could continue with this list. But I think with 2 or 3 dice, the chances are already extreme for targeting right or wrong. When a rock with scissor weapon attacks a paper with also a scissor weapon. Thus the right weapon and right defence.

The winning chances when targeting right and the draw chance.
1 Die, out of 36
15 win = 41,667%
6 draw = 16,667%
15 loss = 41,667%

2 Dice, out of 1296
995 win; 76,775%
146 draw; 11,265%
155 loss; 11,960%

3 Dice, out of 46656
42771 win; 91,673%
2442 draw; 5,234%
1443 loss; 3,093%
And thus, allowing both players to use 3 dice would be too extreme here. With 10 out of 11 chances to win.

Going to neglect 4 dice already.

Of course it is possible to allow the defender to use less dice then the attacker. Thus effectively increasing its chances again. But it doesn't help much though.

When using this mechanic, perhaps allowing the defender to roll 1 die would be best. Then there is at least 1/6th of a chance of survival.

The chances now are:

1 Die, out of 36
15 win = 41,667%
6 draw = 16,667%
15 loss = 41,667%

2 Dice, out of 216
125 win; 57,870%
36 draw; 16,667%
55 loss; 25,463%

3 Dice, out of 1296
855 win; 65,972%
216 draw; 16,667%
225 loss; 17,361%

4 Dice, out of 7776
5501 win; 70,743%
1296 draw; 16,667%
979 loss; 12,590%

That's it for now.
Any comments?

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