Balance question

Help
1

So, hi! I’m back! I’m in Colorado[1], so I need help with balancing a fighting system. The way it works is like so: two people each pick a type of die[2], and roll them. The dice are then compared from highest to lowest[3] and each win that a player accumulates takes 1 hp off of the other player. When either person makes it to 0 hp, the other player wins. My question is this: would this make a fair fight for all dice types? Each type of dice needs to be able to not crushingly lose to any other type. Thanks in advance!

  1. and over a thousand miles from my computer ↩︎

  2. d4 x3, d6 x2, or d8 x1. ↩︎

  3. I rolled a 2 (used the d8) and @VoidFluffy rolled a 1, 2, and 4 (used d4 x3). The largest die on her side is compared to the largest die on my side, and the larger number wins. This is repeated with the second largest and third largest numbers. ↩︎

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3

Nonono, that’s a legit question! It’s actually three different tetrahedrons that you roll at once. It’s easier to visualize for me than rolling the same die thrice.

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So, lemme get this straight, the 3 in d4x3 is 3 tetrahedrons that you roll, but then what does d4 mean? Is that the numbers available? 1, 2, 3, 4?

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Yep! The 4 is just the max number you can get!

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And d6x2 is basically two dices that have 6 numbers, highest 6, lowest 1?

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Sorry, I don’t think I can help with this, but [1]

  1. I hope I won the game… lol ↩︎

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I have another topic that I need help with, don’t worry![1]

  1. You won. The mist in my beak and chest is affecting my RNG. ↩︎

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9

Sorry, I don’t think I can help with this either. But as far as I can tell, you did balance it out.

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Alright. I’m going offline for now though, so bye. [1]

  1. uhh that wasn’t me uh heheheheheheh ↩︎

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Thank you! I’ll leave this open just in case, but you’ll get the solution when I close it.

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12

Yes, maybe getrithekd could help?

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So i used chat gpt to make a small html demo [1] and it seems that the system is not balanced at all. I’ve put this on codepen since it works best for mobile

https://codepen.io/mandalorianwins/pen/rNEeybN

If you click the button in the top right or go to javascript and make changes you can tweak it

I made a comment in the javascript about where you can change what dice and how many times it rolls to try to balance it. Hope this helps!

  1. cause I’m unfortunately unable to code on mobile ↩︎

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Heres a simple ai that should decipher which combination is best (its insanely slow)

https://codepen.io/mandalorianwins/pen/bGPpWEO

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15

I’m sorry, but I cannot see the code at all.

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Not even if you click on the link?

Let me try preventing oneboxing

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17

It isn’t fair. I thought of two possible interperetations, and each one is imbalanced.
Interpretation 1: For the d8 and 2d6, the roll (or one of the two rolls for the 2d6) is used as the third roll. This would disadvantage the 3d4, as a 5-8 result in an automatic win.
Interpretation 2: For the d8 and 2d6, a 0 is used for the extra number(s). (e.g. A d8 and 3d4, the d8 roll would count as the highest number amd the other two would be 0.) This would give the 3d4 an advantage, as you would have a guaranteed point against the 2d6, and a guaranteed win against the d8.

My suggestion: Have 3d8, 4d6, and 6d4 (feel free to tweak the numbers.) Whoever rolls the highest total deals some amount of damage (maybe the difference between the two rolls).

If I interpereted the system wrong, please let me know.

The Codepen seems to show a balance though, so maybe I got it wrong.

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18

Another thing to note is that the d8 has a 50% chance of landing above the max of the d4, and even if it landed 4 or below, it still has a chance of winning, so d8 vs any amount of d4s would not be fair

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19

You got it mostly right, but the way I was going to have it work was to discard any extra dice rolls from lowest to heighest. (So a 3d4 versus a d8 would really be a d8 versus the heighest d4 roll.)

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Yea, I don’t really have a good solution for that. If I just remove the d4s, that removes a bit of choice that the player has.

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You could try using d5s maybe? One possible solution for close chances is 1 d8 vs 2 d6s vs 2 d5s 1 d6. If I’m not wrong, you could use the average of score of every possible combination to find how fair it is. Using this method, the first and last roll an average of 4.5, while the 2 d6s roll an average of 4.47ish.

Although, if you want to only use the same dice types for each group, you can use 4 d5s which averages to 4.43ish on a roll, but this is slightly more unfair.

Edit: This doesn’t account for 2 d6s vs 4 d5s, which might be more unfair. Another solution to balance the dice that does not require as much precision is by changing the damage that the dice deal, making some dice more risk-reward.

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