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Any and none are complementary questions

Part D · L14 · Prerequisite: L13 · One modelling idea

Situation, question, and prediction

The Pool uses any natural 1 as its success trigger. Given an already assembled pool of two d6, how often does at least one 1 appear? Predict why adding a die helps less when the pool is already large.

Build

.p_any(1) asks whether one or more faces match. .p_none(1) asks whether no face matches. Both return probabilities, not counts. Pool assembly, gambling, rewards, and narration decisions are outside this trigger component.

pool = dice_pool(2, 6)
output("Any natural one", pool.p_any(1))
output("No natural ones", pool.p_none(1))
rows = []
for count in range(1, 6):
    rows.append((str(count) + " dice", dice_pool(count, 6).p_any(1)))
output("Success trigger by supplied pool size", prob_table(rows))

Any natural one · Prob

No natural ones · Prob

Success trigger by supplied pool size · Table

Run, read, and check

No ones on two dice means 5 × 5 of 36 pairs: 25/36. Its complement is 11/36. For n dice, any one is 1 − (5/6)^n. Each added die rescues only some previously unsuccessful rolls, so the gains diminish.

Change one thing

Use one die: any one is 1/6 and none is 5/6. No probability can grow above one.

Try it yourself

Ask for any six instead. Answer: the probability is unchanged for fair d6, but the event is different. For a biased custom die you must check face weights. “Any” does not tell you how many matches occurred.

Rules and model notes

The Pool trigger component; author-rules source review remains pending. The fully specified mathematical question here is at least one 1 on n independent fair d6, not a certified complete game procedure.

What you now know / where next

Any and none are complementary questions is the reusable idea. Follow the generated previous/next links below, or return to the course index. For a complete self-contained application, see the-pool.

Content ID: L14 · Review status: source pending

Prerequisites: L13