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I like this question: Flip a coin. If it is heads, flip again and answer Yes for heads and No for tails. If the first coin is tails, answer the question: "Ha

I like this question: Flip a coin. If it is heads, flip again and answer Yes for heads and No for tails. If the first coin is tails, answer the question: "Have you ever cheated on your spouse?" Now, what is the probability the person is a cheater given they answered yes?

deleted 13y ago
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bernardom 13y ago
Wait, is there an interesting answer to this? I'll label the four outcomes 1Y, 1N, 2Y, 2N. Let X = probability person cheated on spouse. If I flip heads first, I get yes or no equally. So the first two outcomes are: 1Y(50%) or 1N (50%). If I flip tails, I answer 2Y (X) or 2N (1-X). So the distinct pdf is: 1Y (25%) 1N (25%) 2Y (0.5X) 2N (1-0.5X) And combined, Y = 0.25+0.5X What am I missing? (Wait, oh, is this a stupid joke about whether they lied about cheating on their spouse? Dammit.)
mildavw 13y ago
Did I read somewhere that this technique is actually used to gather survey data where the subject may have reason to lie? They can answer yes but it remains plausible that it was because of two coin flips coming up heads so they are not individually implicated. When you aggregate the data over many subjects, however, you have a better idea of how many actual cheaters there were than if you asked directly. Anyway, here is the code below: http://jsfiddle.net/nQ3Gu/ http://jsfiddle.net/nQ3Gu/ var runs = 1000, honest_admissions = 0, automatic_admissions = 0; function flip() { return Math.random() < 0.5 ? 'h' : 't'; } for (var i = 0; i < runs; i++) { if (flip() == 't') { honest_admissions++; } else if (flip() == 'h') { automatic_admissions++; } } document.write('The "yes" is honest ' + honest_admissions / (honest_admissions + automatic_admissions) + ' of the time.') 2/3 of yeses are honest and 1/3 didn't have to answer the question!
- mildavw 13y ago
  Oops. That's completely wrong. It's the case where every honest answer is yes! I think this is indeterminate. You have to know the ratio of total yesses to total participants. The real world use-case I was recalling is simpler: flip a coin and answer honestly if it's heads and "yes" if it's tails. Any individual response is non-incriminating, but if N is large enough where you can assume N/2 got heads, you can know the number of "honest cheaters."

deleted 13y ago

[deleted]

bernardom 13y ago

Wait, is there an interesting answer to this? I'll label the four outcomes 1Y, 1N, 2Y, 2N. Let X = probability person cheated on spouse. If I flip heads first, I get yes or no equally. So the first two outcomes are: 1Y(50%) or 1N (50%). If I flip tails, I answer 2Y (X) or 2N (1-X). So the distinct pdf is: 1Y (25%) 1N (25%) 2Y (0.5X) 2N (1-0.5X) And combined, Y = 0.25+0.5X What am I missing? (Wait, oh, is this a stupid joke about whether they lied about cheating on their spouse? Dammit.)

mildavw 13y ago

Did I read somewhere that this technique is actually used to gather survey data where the subject may have reason to lie? They can answer yes but it remains plausible that it was because of two coin flips coming up heads so they are not individually implicated. When you aggregate the data over many subjects, however, you have a better idea of how many actual cheaters there were than if you asked directly. Anyway, here is the code below: http://jsfiddle.net/nQ3Gu/ http://jsfiddle.net/nQ3Gu/ var runs = 1000, honest_admissions = 0, automatic_admissions = 0; function flip() { return Math.random() < 0.5 ? 'h' : 't'; } for (var i = 0; i < runs; i++) { if (flip() == 't') { honest_admissions++; } else if (flip() == 'h') { automatic_admissions++; } } document.write('The "yes" is honest ' + honest_admissions / (honest_admissions + automatic_admissions) + ' of the time.') 2/3 of yeses are honest and 1/3 didn't have to answer the question!

mildavw 13y ago
Oops. That's completely wrong. It's the case where every honest answer is yes! I think this is indeterminate. You have to know the ratio of total yesses to total participants. The real world use-case I was recalling is simpler: flip a coin and answer honestly if it's heads and "yes" if it's tails. Any individual response is non-incriminating, but if N is large enough where you can assume N/2 got heads, you can know the number of "honest cheaters."