Saturday, December 3, 2011

Weekend Egnorance: Just in case you didn't already have a headache today...

A math problem.

My answer would be the same one Joe Carter endorses.

My head still hurts, though.


  1. My answer is 0%.

    Let P(X) the probability of randomly drawing the answer X from the list of answers. Let X* be a correct answer, i.e. P(X*)=X*

    P(25%)=50%, so X* does not equal 25%
    P(50%)=25%, so X* does not equal 50%
    P(60%)=25%, so X* does not equal 60%

    But clearly P(0%)=0%, therefore X* equals 0%.


  2. @troy
    But clearly P(0%)=0%, therefore X* equals 0%

    So you answer by giving no answer and giving no answer is an answer?

    Help! I'm in a loop...


  3. To put it simply - 25% is wrong because it shows up twice, 50% is wrong because it shows up once.


  4. 0 is the right answer. It is obviously self-consistent. If the answers must be picked randomly from those on the board, no one will chose the missing answer 0, so nobody will be correct.

  5. If one of the questions is correct and there is 4 answers, the answer would be 25% (1/4).
    The problem is that there is TWO slots that hold the answer 25%, and so there is only THREE real answers. On one hand the indication is that the reason for TWO slots being allocated to 25% is that it is a hint, on the other hand it reduces the odds to 33.3% (1/3) if we are to assume repetition is a NOT hint.
    I will assume it is not.
    once we disqualify the 25% slots then we are left with two answers and the odds become 50%/50%. That is answer 'B'.
    So... Seeing as only ONE answer is set to 50% and that seems the most tactically sound scenario - that is the answer that most fits the circumstances:
    B)50% is the answer.

    Is this some sort of philosophical test maybe? hmmm, Mike? Or just a mind bender...

  6. Oleg & KW,
    Zero is not an option.
    You have a choice between A, B, C, D
    None of those values are zero.

  7. Read the question again, crus. The 0 answer is perfectly valid.

  8. Oleg,

    It is exactly because I examined the wording that I came to the answer that I did.
    Consider: As with all riddles, we must make several assumptions that are embedded in the question that it's author intended for us to eke out: IE The solution.
    First we must assume the answer is evident in the choices presented (A,B,C,D). This assumption is the gift of the inference that is made in the expression 'choose an answer'. That wording is an explicit instruction/clue - to choose from the available responses A,B,C, or D.
    Also in this wording is the key to the next step. 'an answer' - singular. Not 'choose answers' or 'come up with an answer', but rather: 'choose an answer', in the singular.
    So our riddler lets us know two things.
    1. The answer is listed.
    2. It is a singular answer
    As I noted in the previous comment, if all the options had been different, the the obvious response would have been 25%.
    But, our riddler makes it harder than that. He adds a further dimension.
    He REPEATS that VERY value (25% or 1/4) in option A and D. So we are left with a dilemma.
    A real CHOICE is forced on us. We must decide if the clue 'an answer' outweighs the word 'random'.
    How random? A random guess, or a random keystroke?
    For if a random keystroke (ie non intelligent agent) then the odds are STILL 1/4 or 25%, as their is STILL 4 options - even if one is a repeated one.
    But we are talking about 'you' in this question. The word 'you' indicates a person and a mind, and not a simple random number generator. this case 'random' means a guess, is my guess.
    A guess based on the wording (the closest approx of random) leads us to disregard the repeated options as false, or at least as ONE response (1/3).
    That leaves us with the two NON repeated options and ONE repeated.
    ONE guess, and TWO or THREE options. That is 1/2 ( 50%) odds or 1/3 or (33.33%) odds.
    Answer B is 50%. There is no option for 33.3%.
    Therefore, it seems to me the most sensible and strategic answer is B)50%
    The value zero, none, or nil is not represented in the choices, and thus cannot be a valid response. It's a riddle, Oleg. Probability graphs and equations don't come into it.

    Zero is simply not an option, and Troy's little equation (while very pretty) is irrelevant.

  9. Can't think outside the box, crus? No surprise.

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  11. No, Oleg.
    Not about thinking outside the box. Nor is it beyond me to do so. It is quite within my responsibilities to think inside, outside, and even before and after the box. Sometimes even to eliminate the box, or protect the box.
    But your snide remark aside, you have no answer.
    VERY simply put:
    My comments are about solving a riddle.
    They are about working with the options you have, not inventing new ones to avoid a making a choice and/or an intellectual commitment, Oleg.
    You and Troy do exactly that.
    You're just moving the goal posts.
    Funniest part, is YOU don't even seem to know why.
    Show us the words in this question that lead you to believe there is an 'E' option that reads null/none/zero, or that you do not need to 'choose an answer', but rather are free to invent one... please do. Or maybe you could illustrate in the wording where it says that for the odds to be correct ALL answer must be equal the others?
    Just NOT there. But PLEASE do try!
    Do so without relying on invented symbols and pretensions about probabilities and randomness.
    Let's see you 'think outside the box', while ANSWERING THE BLOODY RIDDLE!

  12. crus,

    I am using, in fact the standard probability theory. Nothing fancy.

    Let's start by reading the question: If you choose an answer to this question at random, what is the chance that you will be correct?

    How do we evaluate the probability ("the chance") of a particular outcome in a random process? You perform the process many times and count the fraction of the trials in which the outcome is achieved. For example, if you throw a fare coin many times then the outcome will be heads in half of the trials and tails in the other half. The chance of getting heads is thus 50%.

    In our riddle, we select one of the four answers, A, B, C, or D, at random many times. After many trials, each of those outcomes will be achieved 25% of the time. Thus the answer "25%" will be given 50% of the time, the answer "50%" will be given 25% of the time, and the answer "60%" will be given 25% of the time.

    None of these answers is correct, which means that you will be correct in 0 trials out of however many. In other words, the probability that you the question correctly is 0.

  13. Oleg,
    Thank you for explicitly outlining your understanding of that theory in application to a question like this one. It actually makes some sort of sense the way you explain it. But it does not work for this question, with this wording.
    it is broken in two ways.
    1. It requires 'wiggle room' for external theoretical application. Riddles afford no such wiggle room.
    2. Your answer is VERY well thought out, and thus not a guess or random in any way.

    Here is the question you have created with your application of theory - (at least as I understand it):
    "What is the probability of a correct answer being randomly chosen with many trials in a 4 option multiple choice (A,B,C,D) which correlates with an answer presented in those choices, if the they are a)25% b)50% c)60% d)25%."
    With seeming authority, you have projected a calculation that with these sets of probabilities the chances of the answers correlating with the correct odds (A,B,C,D) is impossible, or at least very near impossible in the predictable, immaterial world of numbers.

    But, that is NOT the question. Nor are the numbers any more important than the words that frame them. They RELY on them.

    For our purposes this question is 'out of the blue', and we are expected to answer it.
    It is not framed as an exercise in probability, philosophy, mathematics. All those would be an avoidance...or as I now begin to see your position - a failed attempt at cutting of a Gordian knot.
    Perhaps it should have been framed as a probability question, or even originally was?
    Of that I have no real interest. The task at hand is of interest.

    The solution?
    They keys are in the language. Here is the ACTUAL question with my own emphasis on key words.

    "If you choose an answer this question randomly, what is the chance you will be correct."

    The answer lays in the balance of the language, assuming the author crafted the wording with care.
    Here broken down with instructions.

    Choose: Choice.
    Pick from the options.
    AN Answer:AN is Singular. Not the multiple runs required for a scientific or statistical approach.Not multiple choices (IE A&D)
    You may choose one answer.
    This question: As in THIS question. Not a series of questions, nor a multiple choice question LIKE this... but rather implicitly THIS QUESTION (above).
    You may choose one answer from THIS Question
    Random: Make your choice in a 'random' fashion. In relation to reality (ie humans, life etc) as noted in the 'you' found at both ends of the riddle - 'random' translates to 'guess', in plain English.
    YOU may GUESS at ONE choice from THIS question. are the clear instructions.
    Chance: Answer format will be numerical. 'What is the chance?' is the brunt of the question. It is the key to the riddle. It asks for the odds of picking the correct answer to the riddle. The odds and the answer are the same.
    You (both):A person. An intelligent agent. In this context it is, again, singular. NOT an algorythm or computer simulation or mathematical model. YOU.

    So finally we come to a rough approximation:
    What are the odds you will guess the unique/single correct answer in a single try?
    The answer to this question, and the one on the blackboard are the same: b)50%
    The option for zero is not present to choose from, and 25% is not unique, and if accounted for changes the odds to 33% (also not present). We are left with B)50% and C)60%. I am not arrogant or optimistic enough for I go with B.
    A simple 1 in 2, if I (you) guess (choose/random) from (an) unique answer.

  14. crus,

    I understand perfectly well what the puzzle is asking. My calculation above determines the chance—the probability that the answer is correct. That's how probabilities are calculated. You specify outcomes (answers A, B, C, D) and their probability distribution (uniform), then you calculate the odds of whatever functions of outcomes (A+D, B, and C in this case). The respective probabilities do not match, so the chance to correctly answer the question is zero.

    This shouldn't be hard to understand.

  15. No, not hard to understand. Easy. Too easy.
    You cut the knot and avoid the disentanglement of the rope.
    You answer the probability factor perfectly. No argument there.
    The argument is that you have NOT answered the Riddle properly. Only the disembodied question is addressed in your equation. The question is only part.
    The framing is the foundation.