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You want to find out how many students on campus smoke marijuana. However, you don't want to just go up and ask them if they smoke pot, as they might be embarrassed or lie for some other reason. Therefore, you decide to fill a bag with copies of two essentially the same question:

Question 1: Do you smoke marijuana?
Question 2: Do you not smoke marijuana?

Now, you know the % of Question 1s in your bag (as well as the % of Question 2s in the bag). You go out and ask a bunch of students to grab a question out of the bag, answer it on a separate piece of paper, and put their answers in the answer box.

At the end of the day, you have a full box of answers, and you also know the ratio of Q1s to Q2s that were answered.

Hints: Bayes theorem and set up the experiment such that the ratio of Q1 to Q2 is not 1:1.

The probability of a student on campus being a smoker is P(studentIsSmoker) = something involving P(student said yes)

The question is, what is the "something involving P(student said yes)" formula, and how would you set up the experiment?

Additionally, apparently this could turn out the wrong results, but "probabilistically it will be right"

Seems impossible to me, but there's probably some math magic I don't know about.

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I suppose you're expected to assume without argument that (a) you can convince the respondents that the protocol of the experiment is what you claim it is, (b) that the respondents convince themselves that this protocol makes it impossible to trace their answer back to them, more than an ordinary secret-ballot style setup would, and (c) that the respondents therefore don't lie. Right? – Henning Makholm Sep 6 '12 at 23:43
and (d) that they agree with you on what a "yes" or "no" response to a "do you not ..." question means. – Robert Israel Sep 7 '12 at 0:54
Hahha, exactly, unlikely right? But those are the assumptions we are supposed to make – user13327 Sep 7 '12 at 0:57
Another, similar, protocol might ameliorate @Henning's (a) and (b) problems (and Robert's (d) problem as well). If a respondent does smoke, they should answer "yes". If they don't, they should flip a coin and answer "yes" if the coin comes up heads and "no" otherwise. They should have little difficulty convincing themselves that even if the Dean of Students manages to match names and answers, a "yes" answer would not be an admission of guilt (though Dean Wormser might not require evidence beyond a shadow of a doubt, I suppose). – Rick Decker Sep 7 '12 at 1:59
@RickDecker: Even then, answering "no" will release definite information about the student's smoker status. Better to: (1) Decide silently on two numbers between 1 and 6, (2) Roll a die once. (3) Reveal the result of "I smoke, XOR the number that came up was one of the two I picked initially". – Henning Makholm Sep 7 '12 at 14:19
up vote 1 down vote accepted

Let $f$ be the fraction of questions that are Question $1$, and let $p$ be the probability that a student is a smoker. If there are $n$ responses altogether, we expect about $fpn$ yes answers from students drawing Question $1$ (why?) and about $(1-f)(1-p)n$ yes answers from students drawing Question $2$. (We expect $(1-f)n$ students to draw Question $2$, and each of those has probability $1-p$ of being a non-smoker and therefore answering yes.) Thus, we expect about $$\Big(fp+(1-f)(1-p)\Big)n=(1-f-p+2fp)n$$ yes answers altogether. In other words, we expect that the fraction of yes answers will be about $1-f-p+2fp$.

Let $y$ be the actual fraction of yes answers in the sample; $y$ is the observed probability of a student’s saying yes. Then $y$ should be approximately ... what? If you write down the right expression, you can easily solve it for $p$ in terms of the known numbers $f$ and $y$.

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Doesn't (fp + (1 - f)(1 - p))n = (1 - f - p + 2fp)n instead of (2 - f - p + 2fp)n? – user13327 Sep 11 '12 at 21:51
@Silver: Yep. Looks like I typoed it in the display line and then copied it mechanically. – Brian M. Scott Sep 12 '12 at 1:19

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