---
title: "Homework: Bootstrap"
format:
  html:
    theme: [default, "class.scss"]
    callout-appearance: simple
    callout-icon: false
tlda-answer-baseline: "homework-bootstrap.qmd.support/baseline.txt"
filters:
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  - "homework-bootstrap.qmd.support/answer-placement-warning.lua"
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<!-- STUB (2026-09-16): bootstrap problems live inside later homeworks; lift from
  these when filling this in —
  - homework/homework-markov-chebyshev.qmd:305 (bootstrap-vs-normal-vs-Markov
    interval comparison on GA turnout data)
  - homework/homework-sample-size-coverage.qmd:177 (bootstrap-vs-normal coverage)
-->

## What Changes If You Change This

The derivation had two stages, and each one needed something to be true. Stage 1 needed every roll sequence to be equally likely, so that adding them up was counting them. Stage 2 needed every response sequence with the same sum to carry the same probability, so that adding those was counting too. Change the setup and you can ask which of those two survives.

::: {.callout-exercise}
#### The Only Die You Have Is A d12
Ten voters, and no ten-sided die in the drawer. So you roll a twelve-sided one and agree that an $11$ or a $12$ counts as a $10$. Every call is still made the same way, independently of the ones before it.

Which of the two stages still works, and what happens to the answer?
:::

::: {#ans-q1 .callout-answer .callout-note title="Template"}

*(your answer here)*

:::

::: {.callout-exercise}
#### A Third Answer
Now suppose people can say Yes, No, or Undecided, and you're still after the number of Yeses.

Which of the two stages still works?
:::

::: {#ans-q2 .callout-answer .callout-note title="Template"}

*(your answer here)*

:::

## Appendix: Throwing Rolls Away

The d12 has a fix, and it's the obvious one. Instead of counting an $11$ or a $12$ as a ten, roll again. Keep rolling until you get a number you can use.

Every roll you keep is equally likely to be any of the ten, because the rule for throwing one away treats all ten usable faces alike --- an $11$ and a $12$ go in the bin no matter who you were about to call. So stage 1 is counting again, $\theta_1$ is the population frequency again, and the answer is the one we derived.

You've paid for it in rolls rather than in bias: a sixth of them go in the bin. That trade has a name --- *rejection sampling* --- and it's how you draw from a distribution you want using a tool that gives you a different one. Keep what fits, discard what doesn't, and the ones you keep come out right.
