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The city of Roboville has been invaded by robots, monkeys, and dancers, who will dance if the beat's funky. Some of the invaders can be two of them (e.g. both monkeys and robots) and some can be all three. But they all look like normal humans, and you have absolutely no clue how the invaders are and their types.

But you know that among all the residents (invaders and non-invaders), 1 in 2 are robots, 1 in 3 are monkeys, and 1 in 4 are dancers (all independent).

It is known that everytime they say three sentences, robots lie on their first sentence, monkeys on their second, dancers on their third, and non-invaders always tell the truth. So robots who are also monkeys will lie twice then tell the truth.

The Roboville government has set up a reCaptcha test to get rid of all the invaders. Only non-invaders (those who are not robots, monkeys or dancers) will pass the test.

Questions:

  1. Suppose someone in Roboville says, 'I'm not a monkey. I will not dance even if the beat's funky. I'm not a robot', what is the chance of them passing the reCaptcha?

  2. Suppose the current population of Roboville is 12,000,000 (including invaders), which is divided into six equal groups,
    in which everyone from each group says a different combination of the sentences 'I'm not a robot, I'm not a monkey, I will not dance even if the beat's funky'
    (e.g. One group says 'not a robot, not a monkey, not dance', another says 'not a robot, not dance, not a monkey' and so on),
    what is the expected number of people who will pass the reCaptcha? (if not an integer, round off to three significant figures)

This puzzle is heavily based on Linkin Park's song When They Come For Me.

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  • $\begingroup$ Is it only dancers who will dance if the beat is funky, or all types of invaders (the sentence is a little ambiguous)? Also, for "every time they say three sentences" do previous sentences matter? For the first question could "I'm not a monkey" be an invader's 2nd sentence? $\endgroup$
    – Rob Watts
    Jun 8, 2022 at 20:47
  • $\begingroup$ I had an answer that was deleted for some reason, but do we know anything else about the distribution of robot-monkey-dancers in the population? There could be none of them, or they could make up as much as a quarter of the population. $\endgroup$
    – SQLnoob
    Jun 24, 2022 at 13:43

2 Answers 2

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Question 1

If statement 1 is true, then they're neither a robot nor a monkey. If it's false, then they're both a robot and a monkey.

If statement 2 is true, then they're neither a dancer nor a monkey. If it's false, then they're both a dancer and a monkey.

If statement 3 is true, then they're neither a robot nor a dancer. If it's false, then they're both a robot and a dancer.

Thus, if they're any one type of invader, then they're all three types of invader.

Among all the residents, (1/2)(2/3)(3/4) = 6/24 are non-invaders, and (1/2)(1/3)(1/4) = 1/24 are all three types of invader.

The speaker is in one of those groups, so the probability that they're in the first group are 6/7.

Question 2

Each of the six groups cntains 2,000,000 people.

"I'm not a robot, I'm not a monkey, I'm not a dancer": This doesn't eliminate anything. 1/4 of 2,000,000 = 500,000 are non-invaders.

"I'm not a monkey, I'm not a dancer, I'm not a robot": Per question 1, 6/7 of 2,000,000 = abut 1714285.714 are non-invaders.

"I'm not a dancer, I'm not a robot, I'm not a monkey": Similar to question 1, about 1714285.714 are non-invaders.

For each of the other groups, if they're an X then they're a Y and vice versa, but that's all we know.

One has proportions of 6 non-invaders to 3 two-type invaders to 1 three-type invader, so 6/10 = 1,200,000 are non-invaders.

Another has proportions of 6 to 2 to 1, so 6/9 = about 1,333,333.333 are non-invaders.

The third has proportions of 6 to 1 to 1, so 6/8 = 1,500,000 are non-invaders.

Total number of non-invaders is about 7,961,904.761, so round up to 7,961,905. (As it happens, if we round each group and then add, we get the same total.)

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  • $\begingroup$ Actually, the answer to question 2 rot13(znl fvzcyl or guerr zvyyvba, onfrq ba bar-sbhegu bs gjryir zvyyvba orvat aba-vainqref. Gur fvk tebhcf zreryl erdhver gung nyy bar-glcr vainqref ner vapyhqrq va rvgure gur "V'z abg n ebobg, V'z abg n zbaxrl, V'z abg n qnapre" tebhc, be bar bs gur "vs K gura L naq ivpr irefn ohg fnlf abguvat nobhg M" tebhcf, jurer M vf gurve bar glcr). $\endgroup$
    – Ed Murphy
    Jun 24, 2022 at 8:32
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Question 1

7/24. Those who are robots must be monkeys (due to their answer to the first question) and must therefore also be dancers (due to their answer to the second question). Those who are not robots cannot be monkeys (due to the answer to the first question) and so cannot be dancers (due to their answer to the second question). The proportion is then (1/2)(1/3)(1/4)+(1/2)(2/3)(3/4)=7/24

Question 2

6,330,000. Call a statement where the status claimed is one which could lie at that point redundant (e.g. if the first statement is "Are you a robot?" it is redundant as everyone would say this). There are two groups where no statement is redundant and by a similar analysis to the above 7/24 of respondents pass. There is one group where all statements are redundant and 24/24 of the respondents pass. There are three groups where one statement is redundant and can only be pass if both of the other statements are both true or both lies the proportions in these groups are 1/2+1/12 (redundant robots) 1/3+1/6 (redundant dancers) and 3/8+1/8 (redundant monkeys). As there are 2,0000,000 people in each group we expect the number passing to be (14/24+24/24+14/24+12/24+12/24)*2000000=19/3 million.

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