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Unfortunately, for large dice, the number of possible multisets can grow quite quickly---in this case, \$\Theta \left( n^{12} \right)\$ where \$n\$ is the number of scores. … If you would like to know more, you can read my paper on the subject: @inproceedings{liu2022icepool, title={Icepool: Efficient Computation of Dice Pool Probabilities}, author={Albert Julius Liu …

The sum of dice also converges to a normal distribution. Therefore, we can try using the margin of success as the number of hits. …

d1000 In a brief search I found the following: Here's a forum post including images of a Games Workshop publication using the d1000. The d1000 MUTATION TABLE! is simultaneously everything that was ri …

Goodman Games' Dungeon Crawl Classics dice. Long dice that are unstable on their ends. Teetotums are similar. …

Approaches of this type have been proposed on StackExchange, with examples including: Formula for dropping dice (non-brute force) Generating Function for sum of N dice [or other multinomial distribution … Examples include: Roll and Keep in Anydice How to calculate the probabilities for eliminative dice pools (dice cancelling mechanic) in Neon City Overdrive? …

following identity: $$ \left\lceil \frac{n}{d} \right\rceil = \left\lfloor \frac{n + d - 1}{d} \right\rfloor $$ In this case the numerator \$n\$ is the sum rolled on the 4d4 and \$d\$ is the number of dice … As for "scalar" they probably mean a result that is a single number (e.g. the sum of some dice), as opposed to a list of individual die rolls. …

Each of those dice has a 9/20 chance of rolling below a 10. Do a little binomial theorem and we get an 84.7% chance of at most 4 of those dice rolling below a 10. … The most likely case is that exactly 10 out of 17 dice rolled 15+, in which case none of the remaining dice rolled 15+, effectively reducing the d20 to a d14 for those remaining dice. …

According to this MathExchange answer, the 90-degree configuration of a pentagonal trapezohedron has the additional property that the angle between a face and the opposite edge on a vertex is also 90 …

set "maximum function depth" to 10 function: cascading X:n MAX:n { if MAX <= 4 | X < MAX { result: X } else { result: X + [cascading d(MAX-2) MAX-2] } } function: cascading D:d { result: [c …

@ die + max(n - 12, 0) @ extra result = raw.map(score) print(result) for triggers in [4, 3, 2, 1, 0]: area(result.map(lambda s, t, b: s if t >= triggers and not b else -1), f'{n} dice … : +{triggers}') output(result.map(lambda s, t, b: s if b else 100), f'{n} dice: clean') limit(0, 5) degenesis(5) The dice have vector values, represented by tuples (4+s, 6s, 1s). …

Adding the middle dice together If you just want to add the middle dice together, you can use the @ operator to select individual sorted positions and add them together: output {2..4}@5d12 Or, perhaps … , rather than 3 values for the individual dice. …

If we sort the dice from highest to lowest, and we observed that the player's \$i\$th roll (starting from \$i = 1\$) rolled \$x\$, then we know that all \$i\$ rolls up to that point rolled at least \$x … (Presumably the player automatically fails if they have fewer dice than the obstacle.) …

Mixing different dice also makes things much more difficult. … Fortunately, it turns out that, as long as you can express the dice mechanic in terms of an incremental calculation where you're told how many dice rolled each number in each pool for one number at a time …

# action_set: The set of surviving action dice. # leftover_stress: The number of surviving stress dice that rolled # this number or higher. … : 6, 3, 3") but the number of possible such sets grows rapidly with number of action dice (e.g. this example results in 252 rows), so I won't reproduce the table here. …

Compared to an ideal Laplace distribution: Opposed step dice In this type of system, instead of adding a modifier to their roll, the two sides use larger and larger dice as they advance. … Proof Suppose two players roll step dice against each other, with respective sizes \$A\$ and \$B\$, with ties being broken by coin flip. …