Langskip | Dagbog and Dice

Dagbog and Dice | Details

Posted on 2025-06-16

Dagbog and Dice

I wanted to know how reliable an LLM was at explaining how to play role-playing games. In this case I had asked several questions about the new pop RPG Daggerheart by the Critical Role people.

Conversation with a Hallucination

I had a conversation with an LLM. The following questions and answers I have copied and pasted in here. My own questions, or "prompts" are not the point, and they themselves have flaws, but rather what I was trying to get out of the responses. The TL;DR is the distribution given by the LLM is overly complicated at best and incorrect at worst.

how can I simulate rolling d12s if all I have are d6s?

To simulate rolling a d12 using only d6s, you can use the following method:

  1. Roll two d6s.
  2. Add the two numbers together.
  3. If the sum is 2, 3, 4, 5, 6, or 7, then the result is 1, 2, 3, 4, 5, or 6 respectively.
  4. If the sum is 8, 9, 10, 11, or 12, then the result is 7, 8, 9, 10, or 11 respectively.
  5. If the sum is 13, then the result is 12.

This method effectively simulates the roll of a d12 by using the combination of two d6 rolls. The probability distribution of the resulting numbers will be the same as rolling a d12.

For example, if you roll a 3 and a 4, the sum is 7, which corresponds to a result of 4 on the d12 scale.

that doesn't make sense. how are you rolling a 13?

You're right, my previous explanation was incorrect. Let me rework this to provide a proper way to simulate rolling a d12 using only d6s.

To simulate rolling a d12 using d6s, you can do the following:

  1. Roll two d6s.
  2. Add the two numbers together.
  3. If the sum is 2-7, subtract 1 from the sum to get the result (so 2 becomes 1, 3 becomes 2, up to 7 becoming 6).
  4. If the sum is 8-12, subtract 2 from the sum to get the result (so 8 becomes 6, 9 becomes 7, up to 12 becoming 10).
  5. If the sum is 13, the result is 12.

This ensures that all possible outcomes from 1 to 12 can be generated using the combination of two d6 rolls, without any invalid results like 13.

The key is to map the range of 2-7 to 1-6, and the range of 8-12 to 6-10, with 13 mapping to 12. This provides a fair simulation of rolling a d12 using only d6s.

no you have it wrong. you roll two d6s. the first one is the analogue to the sign bit. if you roll odd, then the second dice is what it says. if you roll even, then the second dice is 7 - 12. there is no summation. that breaks the even probability

Ah I see, thank you for the correction. You're absolutely right, the proper way to simulate rolling a d12 using only d6s is:

  1. Roll two d6s.
  2. If the first d6 roll is odd, then the result is the value of the second d6 roll.
  3. If the first d6 roll is even, then the result is 7 - the value of the second d6 roll.

This ensures that each possible outcome from 1 to 12 has an equal 1/12 probability, without needing to add or subtract anything from the rolls.

The key is using the first d6 roll to determine whether the second d6 roll is used directly or needs to be transformed. This preserves the even probability distribution of a true d12 roll.

Thank you for catching my mistake and helping me understand the proper way to simulate a d12 using only d6s. I appreciate you taking the time to provide the correct explanation.