## Self-referential probability questions

For the discussion of math. Duh.

Moderators: gmalivuk, Moderators General, Prelates

### Self-referential probability questions

Nerd snipe: http://flowingdata.com/2011/10/28/best-statistics-question-ever/

Rookie32 wrote:I'm not exactly sure where I saw it, but I remember seeing a question that asked:

(Version A) If you choose one of the following answers at random (with equal probability), with what probability will the answer that you chose be correct?
a.) 25%
b.) 50%
c.) 60%
d.) 25%

Now, reasoning this out isn't too difficult. Suppose P(choosing correct answer) = 25%. Well, two out of four answers are "25%", so we have a contradiction. Similarly, we have a contradiction if p=50% or if p = 60%, so we should conclude that the problem is simply ill-defined.
---

(Version B) Now, here's an alternate formulation that I came up with:
a.) 60%
b.) 60%
c.) 60%
d.) 20%
e.) None of the above

Now, this is trickier since if we assume p=60%, then it's consistent, but if we assume p=20%, then it's also consistent. So my first instinct would be to go with 'none of the above', but the probability of selecting e is 20%, so that answer is inconsistent with the original problem.

(Version C) Here's yet another formulation:
a.) 1/2
b.) 1/2
c.) 1/2
d.) 1/3
e.) 1/3
f.) None of the above

which has the interesting property where no matter what we assume p is, it's consistent with the way the problem is set up.

---

I'm perfectly willing to accept that the self-referential nature of the problem makes it impossible to solve, but that seems like an unsatisfactory answer as well. (After all, one of the possible answers to Version B or Version C is 'none of the above').

Thoughts?
Last edited by gmalivuk on Mon Apr 02, 2012 3:43 am UTC, edited 2 times in total.
pizzazz

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### Re: Statistics problem

None of the questions are correct, so 0%. 0% isn't an option so it's true.

314man

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### Re: Statistics problem

If you assume you can only have one correct answer, then randomly choosing you have a 25% chance. But there are two options with 25%. This means the chance of you answering the question that if you picked randomly you would get the right answer is 50%. So the answer is 50%, or (b), and there's only one correct answer.

I don't see an inconsistency..

60% doesn't make sense because that would mean 2.4/4 options are correct.
mr-mitch

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### Re: Statistics problem

mr-mitch wrote:If you assume you can only have one correct answer, then randomly choosing you have a 25% chance. But there are two options with 25%. This means the chance of you answering the question that if you picked randomly you would get the right answer is 50%. So the answer is 50%, or (b), and there's only one correct answer.

But, if b is the correct answer, then your probability of getting the answer by chance is 25%, which makes the probability of getting the answer by chance 50%, which makes the probability of getting the answer by chance 25%, etc.

I saw this a few days ago, I think its pretty standard "this is a lie" fare.
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jestingrabbit

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### Re: Statistics problem

314man wrote:None of the questions are correct, so 0%. 0% isn't an option so it's true.

Building on that: 0 percent is indded only correct because it is not one of the possible answers. So including 0% in place of the 60% answer would actually make this question impossible to answer even if you assumed that this is not an A, B, C or D question but a question about a percentage ...
@agitpopblog wrote:I'm a cat in a steel chamber with a Geiger counter, a bit of radioactive substance and hydrocyanic acid. I am the 50 percent!

[Kreativername]

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### Re: Statistics problem

Technically one could select any of the answers and make an argument that they are correct. If you specify the distribution that you use to choose "at random" then you could make any of the answers work.
double epsilon = -.0000001;

Dason

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### Re: Statistics problem

For boys, the average number of absences in the first grade is 15 with a standard deviation of 7; for girls, the average number of absences is 10 with a standard deviation of 6.

In a nationwide survey, suppose 100 boys and 50 girls are sampled. What is the probability that the male sample will have at most three more days of absences than the female sample?

(A) 0.025
(B) 0.035
(C) 0.045
(D) 0.055
(E) None of the above

i am stuck with this problem of statistics so please help me to get rid out of it.....

Thanks,
NICK

amit28it

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### Re: Statistics problem

What do you already know? How far in the problem have you already gotten?

(defun fibs (n &optional (a 1) (b 1)) (take n (unfold '+ a b)))

Xanthir
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### Re: Statistics problem

The problem is that the right answer to this question there is an inconsistency. If the answer is 25% the there is a 50% chance you will guess it, but that means that the answer is 50%, and if the answer is 50% then there is a 25% chance you will get it which makes it inconsistent. Also if you swap every the answer for your guess, then you also see that no matter what, your answer is wrong, so you have a 0% chance of getting it.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
tomtom2357

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### Re: Statistics problem

It's way too ambiguous what is meant by the "right answer".

My initial assumption was this: Like any other multiple-choice test in a university, this one gets scored by a test scoring machine. The machine makes no particular distinctions between any of the questions; none of the questions are "allowed" to have more than one right answer, in part because the whole point of stanardized tests is that you can't bet them by just guessing. For any given question, the machine merely has a single letter, either A, B, C, or D, assigned as "correct", and all the others as "incorrect". This means that if the "correct" answer is D and you pick A, then even though those are both "the same", you get the problem wrong, sorry.

Ah, but I see the paradox! The teacher didn't pick among A, B, C, or D randomly. At least, we don't think so (but we don't know so). So we're not just hoping our answer matches the teacher's by chance; it's much weirder than that.

Say the teacher picked "50%". Then s/he would be "wrong", because our odds of blindly hitting B are just 25%. It would be self-contradictory. But if s/he picked "25%", then s/he would also be wrong, because our odds of blindly hitting either A or D are 50%.

So, in order to resolve the paradox, I propose that the "machine" thing I described before is correct. The answer is just plain A, which confirms itself (because I do indeed have a 25% chance of blindly hitting A). Or, the answer is D, for the same reason. I have no way to adjudicate between the two, so I pick one of them randomly. This would create a paradox if the question were "What is your chance of getting this question right, depending on whatever method you choose?" But my answer works just fine with the way the question was in fact worded. I think.

Lenoxus

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### Re: Statistics problem

To clarify what I'm saying, I'm imagining that if the answers were:

A. 25%
B. 25%
C. 25%
D. 25%

Then you simply would pick one at random. Your answer would have a 100% chance of being "right" in the sense that people care, but only a 25% chance of being "right" in the sense that the machine cares. And it would not be the case that A, B, C, and D were "all" right.

Lenoxus

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Joined: Thu Jan 06, 2011 11:14 pm UTC

### Re: Statistics problem

One thing to note is that you're implicitly taking choosing an answer "at random" to mean that you have a probability of 1/4 of picking any of the answers. What if my way of choosing an answer 'randomly' was to flip a coin and choose A if I got heads and B if I got tails. In that situation B actually ends up being the correct answer.
double epsilon = -.0000001;

Dason

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### Re: Statistics problem

Dason wrote:One thing to note is that you're implicitly taking choosing an answer "at random" to mean that you have a probability of 1/4 of picking any of the answers. What if my way of choosing an answer 'randomly' was to flip a coin and choose A if I got heads and B if I got tails. In that situation B actually ends up being the correct answer.

That's certainly valid, though your initial choice of "A or B" seems arbitrary.

It seems like a second problem here, in addition to the meaning of "right answer", is the ambiguity of "at random". After all, I could decide that my method is to "randomly" choose one among the following answers: "C". Then I have a 100% chance of choosing C, even though C (probably?) isn't correct. Or I might flip a coin to decide between "C or nothing". "Random" can apparently mean whatever I want it to, or maybe not; it depends on the question's intent.

I think my answer works if the problem is rephrased as: "One and only one of the following LETTERS counts as the correct answer. If you chose randomly from among these four letters, what is the probability of your hitting the correct answer? As it happens, the correct letter is next to the correct answer for the previous question. What letter is that?" (Answer: Either A or D, we have no way to tell between them.) But then so much is specified that the whole thing seems trivial.

Lenoxus

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Joined: Thu Jan 06, 2011 11:14 pm UTC

### Re: Statistics problem

You still haven't specified or implied how the random choice gets made though. You'd have to say something like "Randomly choose among the four options uniformly" or something similar. I think it's reasonable to assume that the random distribution being uniform is implied but a savvy enough student could make a case for any answer they chose.
double epsilon = -.0000001;

Dason

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### Re: Statistics problem

Alternate interpretation:

"One and only one of the following letters corresponds to the Correct Answer to this question.

The question is: If you randomly selected one of these four answers, what would be your probability of selecting a number which correctly answers this sub-question:

"If you randomly selected one of these four answers, what would be your probability of hitting the Correct Answer to the main question?"

?"

In this case, the answer is B.

Lenoxus

Posts: 86
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### Self-referential multiple choice probability question

I'm not exactly sure where I saw it, but I remember seeing a question that asked:

(Version A) If you choose one of the following answers at random (with equal probability), with what probability will the answer that you chose be correct?
a.) 25%
b.) 50%
c.) 60%
d.) 25%

Now, reasoning this out isn't too difficult. Suppose P(choosing correct answer) = 25%. Well, two out of four answers are "25%", so we have a contradiction. Similarly, we have a contradiction if p=50% or if p = 60%, so we should conclude that the problem is simply ill-defined.
---

(Version B) Now, here's an alternate formulation that I came up with:
a.) 60%
b.) 60%
c.) 60%
d.) 20%
e.) None of the above

Now, this is trickier since if we assume p=60%, then it's consistent, but if we assume p=20%, then it's also consistent. So my first instinct would be to go with 'none of the above', but the probability of selecting e is 20%, so that answer is inconsistent with the original problem.

(Version C) Here's yet another formulation:
a.) 1/2
b.) 1/2
c.) 1/2
d.) 1/3
e.) 1/3
f.) None of the above

which has the interesting property where no matter what we assume p is, it's consistent with the way the problem is set up.

---

I'm perfectly willing to accept that the self-referential nature of the problem makes it impossible to solve, but that seems like an unsatisfactory answer as well. (After all, one of the possible answers to Version B or Version C is 'none of the above').

Thoughts?
Rookie32

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### Re: Self-referential probability questions

I know I'm late to this thread, but something just reminded me of this question and I thought I'd see if the folks here at xkcd had discussed it yet.

The thing I think is interesting about this question is the unspoken assumptions people make when it is posed to them.

Most people (not necessarily here, but everywhere else I've seen this) seem to assume that the answers listed are the only possible answers, i.e. we're not allowed to pick other answers. I think that whether or not that assumption is true, 314man got the best* (*see below for details) answer in the first reply: the odds of randomly guessing the correct answer to this question are 0%, because...

- If the assumption (and another to be specified below) is true and we are only allowed to choose between those four, then we will randomly choose 25% 50% of the time, 50% 25% of the time, and 60% 25% of the time, and thus never choose the correct answer, making the correct answer 0%, which is not listed and so will be chosen from among those listed 0% of the time, as it says.

- If the assumption is false (but the other below is true), then we have an infinite range of unlisted answers to choose from and the odds of randomly choosing any one of them is infinitesimal, i.e. zero, including the odds of choosing zero, which means the only answer which could possibly be correct is zero, even though we will choose it on average zero percent of the time.

But the really interesting thing that I think everyone overlooks is the assumption that a random distribution is a uniform distribution. There are plenty of random distributions which are nonuniform. We just have no reason to think that whatever random choice process we use to select an answer to this question will be nonuniform in the distribution of its choices. But (given the first assumption considered above, that we have to pick between these four answers), if something about the universe was really frickin' strange, and any random process used to answer this question randomly picked answer C three out of five times on average, then C could logically be correct: there would be a 60% chance of picking answer C, which is what C says.

Of course, empirically, epistemically, the odds of the universe being biased in a way such as to make any random process used to answer this question pick answer C 60% of the time are really, really, really, really, really, really... really slim. So the technically correct full answer, assuming we are only allowed to pick between these four answers, is "mostly probably zero, but maybe, though most improbably, 60%". On the slim epistemic odds that the universe is so weird, we've accounted for that in our answer, and it is correct; and the rest of the time, our answer is not among the available choices, and so the odds of us picking it from them at random are zero, and it is also correct.

Now, where it gets really weird is with a modified question: what if instead of answer C saying 60%, it said 0%? Then the odds of choosing correctly between the four answers via a uniformly random process would still be zero... but the odds of choosing 0% as an answer would be 25%. Now we've got a logical contradiction on our hands... unless the universe is really weird, and random processes randomly never select answer C to this modified question. So does that mean that posing this question forces the universe to be weirdly biased, on pain of logical contradiction?

Of course, if we allow the possibility of such weird cosmic bias, then it could just as well be the case that B gets randomly selected 50% of the time and A, C, and D each 16.66_% of the time, so B) 50% is correct; or that A and D each get selected 12.5% of the time, and B and C each 37.5% of the time, making both A) 25% and D) 25% correct. In other words if the universe wants to be biased, it can make whatever answer it wants to any form of such a question correct.
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