# Conditioning changes the question

**Part G · L27 · Prerequisite: L26 · One modelling idea**

## Situation, question, and prediction

Compare “reroll a 1 once,” “keep rolling until not 1,” and “a 1 scores zero.” They sound similar. Predict which can still end at 1, which removes it, and which adds a zero outcome.

## Build

`.remove(1)` removes the face and renormalises the rest: the conditional distribution given not 1. `.keep([2,3,4,5,6])` is equivalent. `.ignore(1)` keeps the probability mass but changes its score to zero. Neither is the once-only policy from L26.

```dice
def once(faces):
    if faces[0] == 1:
        return faces[1]
    return faces[0]
output("Reroll once", pool_map(dice_pool(2, 6), once))
output("Conditioned on not one", d(6).remove(1))
output("One scores zero", d(6).ignore(1))
```

## Run, read, and check

The respective means are 47/12, 4, and 10/3. Conditioning leaves 2–6 each at 1/5. Scoring zero keeps outcomes 0,2,3,4,5,6 each at 1/6. An unlimited independent reroll-until-not-1 procedure has the same final-face distribution as conditioning, but its number of rolls and resource costs are not represented.

## Change one thing

Ignore 1 and 2 instead: zero has probability 1/3. Removing them instead leaves 3–6 each at 1/4. Keeping no possible faces is an error, not an empty successful model.

## Try it yourself

Choose the operation for “given that the die was at least 5.” Answer: `.keep([5,6])`, making both outcomes 1/2. The chance of that evidence on the original d6 remains 1/3, queried with `.p_ge(5)`; conditioning is not that chance.

## Rules and model notes

Generic probability transformations, not a certified game rule. No random process length, stopping cost, or dependent replacement dice are implied.

## What you now know / where next

Conditioning changes the question is the reusable idea. Follow the generated
previous/next links below, or return to the [course index](index.html).
For a complete self-contained application, see [filter-or-zero](../cookbook/filter-or-zero.html).