dnd-5e – Analyzing Homebrew Casting Mechanism

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Background

I'm a DM and setting builder working on creating a new homebrew set of classes/subclasses for my world.

The basic in-universe mechanic is that the caster is channeling their energy through runes that he or she has carved into a focus (usually a staff). These runes are unstable (for various setting reasons), and so have a tendency to "burn out" and need to be reworked.

Runecasting is mostly elemental and/or buffing.

Casting Mechanic: Runecasting
As a runecaster you have only one spell slot of each level of spells you can cast, but can attempt to cast without spending a slot by making a special check. If the check fails, no more spells of that level can be cast until you finish a long rest (fictionally the runes have failed and need to be reworked). Only spells learned via the Runecasting feature can be cast this way.

The first time you attempt to cast a spell without using a slot, roll a d20. If the result is higher than twice the spell level, you succeed and the spell takes effect as normal. If you fail, the spell does not take effect and you cannot attempt the check again for a spell of that level until you finish a long rest. For each additional attempt after the first for that same level of spells, the die rolled decreases by one size (d20 → d12 → d10 → d8 → d6 → d4), to a minimum of a d4. The die resets to a d20 after you finish a long rest.

Question
What is the expected value for the number of times a hypothetical full-casting (spell levels 1-9, so level 17+) rune caster can cast each level of spells? Is there a simple formula for this?

Best Answer

There isn't a simple formula which covers all the cases, but they can be calculated

Expected number of successes before failure for spell levels 1-9:

  1. \$4.3 =\frac{43}{10} = 1 + \frac{18}{20} + \left(\frac{18}{20} \times \frac{10}{12}\right) + \left(\frac{18}{20} \times \frac{10}{12} \times \frac{8}{10} \right)+ \left(\frac{18}{20} \times \frac{10}{12} \times \frac{8}{10} \times \frac{6}{8}\right) +\left( \frac{18}{20} \times \frac{10}{12} \times \frac{8}{10} \times \frac{6}{8} \times \frac{4}{6}\right) + \left(\frac{18}{20} \times \frac{10}{12} \times \frac{8}{10} \times \frac{6}{8} \times \frac{4}{6} \times \frac{2}{4}\right) + ...\$
  2. \$2.8667 \approx \frac{43}{15} = 1 + \frac{16}{20} + \left(\frac{16}{20} \times \frac{8}{12}\right) + \left(\frac{16}{20} \times \frac{8}{12} \times \frac{6}{10}\right) + \left(\frac{16}{20} \times \frac{8}{12} \times \frac{6}{10} \times \frac{4}{8}\right) + \left(\frac{16}{20} \times \frac{8}{12} \times \frac{6}{10} \times \frac{4}{8} \times \frac{2}{6}\right)\$
  3. \$2.225 = \frac{89}{40} = 1 + \frac{14}{20} +\left( \frac{14}{20} \times \frac{6}{12} \right)+\left( \frac{14}{20} \times \frac{6}{12} \times \frac{4}{10}\right) +\left( \frac{14}{20} \times \frac{6}{12} \times \frac{4}{10} \times \frac{2}{8}\right)\$
  4. \$1.84 = \frac{46}{25} = 1 + \frac{12}{20} + \left(\frac{12}{20} \times \frac{4}{12}\right) + \left(\frac{12}{20} \times \frac{4}{12} \times \frac{2}{10}\right)\$
  5. \$1.5833 \approx \frac{19}{12} = 1 + \frac{10}{20} + \left(\frac{10}{20} \times \frac{2}{12}\right)\$
  6. \$1.4 = \frac{7}{5} = 1 + \frac{8}{20}\$
  7. \$1.3 = \frac{13}{10} = 1 + \frac{6}{20}\$
  8. \$1.2 = \frac{6}{5} = 1 + \frac{4}{20}\$
  9. \$1.1 =\frac{11}{10} = 1 + \frac{2}{20}\$

1st level spells is the only case which could potentially go infinite (but the expected total number of successes converges to 4.3), but everything else automatically fails on the d4 because you can't roll higher than twice the spell level.

6th level spells and higher are easy to calculate because they will never succeed on anything other than the d20 throw.

The nth summand in each calculation is the probability of succeeding n times; since each success contributes 1 more to the total, we can sum all these probabilities to obtain the expected number of total successes. (Noting that for level 1 spells this is an infinite sum).

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