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Dice 2x6D Action / Attack / Run

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lindseth
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Joined: 09/08/2010

I'm working on a game where I need a good solution for how to balance the relations between the attack, damage, or to escape. You do not need more than 2 dice. The goal is to find a solution that allows you to vary if something is hard to break, be damaged or that it is easy to slip from. Every roll of the dice can be manipulated before they are rolled, bye both player and opponent. What I now need to know is what kind of numbers I should have on the stats to get a good flow of the game.

Casamyr
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Joined: 07/28/2008
You could always have a

You could always have a standard Difficulty Value (DV) based on the difficulty of the action which players need to beat in order to succeed.

Jumping to grab a ledge might have a DV of 5, but if that same ledge was coated in ice, the DV could become a 9 for instance.

lindseth
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Joined: 09/08/2010
Yes I have triad that, but I

Yes I have triad that, but I need to find a way to make 3 different things happen depending on what you get on your dices.

InvisibleJon
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Joined: 07/27/2008
Figure out how likely you want each event, then set targets.

lindseth wrote:
Yes I have tried that, but I need to find a way to make 3 different things happen depending on what you get on your dice.

I could get really detailed here, but I'm going to try to keep this brief – for the sake of everyone involved...

On 2d6, you have 11 different sums (2 to 12) and 36 discrete die roll combinations (1,1; 1,2; 1,3; ... , 6,5; 6,6).

Knowing this, I suggest that you start by figuring out how often you want each of your three results to occur in a given situation.

Example: I want A to happen 50% of the time, B to happen 25% of the time, and C to happen 25% of the time.

Note that if you want to ensure that *something* will happen, the sum of your probabilities needs to add up to 100%.

Once you have odds for your three events, match those odds to the result odds on 2d6.

Example, continued: The odds of rolling a five or less (or a nine or greater) on 2d6 are 28%. That's close enough to 25% for me. This means that B will happen if the result is two, three, four, or five; C will happen if the result is nine, ten, 11, or 12. Any other result (six, seven, eight) will cause A to happen.

Sidetrack: You might be wondering, "How do you know what the odds are?" It's all about probability, and would take me a bit of time to type and explain. However, I Googled for "the odds of rolling a given number on 2d6" and found this article on a D&D forum: http://forums.relicnews.com/showthread.php?t=151889 – It should be helpful.

Note that I'm assuming that you never want a "no result" result *and* that the results are mutually exclusive (You can't have A and B; B and C; C and A, or A, B, and C happen at the same time.). It's easy enough to do either of these things, though.

Example: I want A to happen 50% of the time, B to happen 40% of the time, and C to happen 25% of the time. 15% of the time, I want A and B to happen.

The odds of rolling a six or less on 2d6 is 42%, which is close enough to 40% for me. From the previous example, we know that the odds of rolling a six, seven, or eight come close to 50% (It's ~44%). Those ranges overlap at six, which has a 14% chance of occurring – darn close to 15%. We already know that the odds of rolling a nine or greater on 2d6 are 28%. That's close enough to 25% for me.

That gives us A on a six, seven, or eight; B on a six or less; and C on a nine or greater. Note that a six gives us both A and B.

One more note: You don't have to make your "result zones" contiguous. You could say that a two, three, seven, or 11 causes an A; a four, five, nine, or ten causes a B; and a six, eight, or 12 causes a C. (FYI: That's a 31% chance of A, 39% chance of B, and 30% chance of C – all mutually exclusive.)

If this is hard to understand, I strongly recommend spending some time learning about (and playing with) the odds of any given result on 2d6.

I wish you the best of luck in your design. I hope it comes out the way you want it to!

hulken
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Joined: 04/18/2009
An easy way also of including

An easy way also of including a certan action would be to have it ocur on ither odd or even numbers. this way you have 50/50 if you have one action for odd and one for even. So using the posted A and B you can easaly ad on a C or D to that using the odd or even as a decider...

Good luck to you And keep us posted on youre progress...

lindseth
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Joined: 09/08/2010
Thank you. You gave the info

Thank you. You gave the info I was looking for. The game can get the final design before Alpha relies now. Sometimes you can't see the solution even when your holding it in your hand.

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