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Thread: Puzzle: probability to guess one number given a second number...

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    Contributor Speakpigeon's Avatar
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    Puzzle: probability to guess one number given a second number...

    What solution do you give to this puzzle?

    I give three different formulations, if that can help.

    The original one is supposed to be the last one here...

    In this game there are two people, labelled Dealer and Player. The Dealer writes two different numbers down on two slips of paper, and seals them in envelopes. It doesn't matter what the numbers are. They can be 0, 5, 4081922, -382.393193, pi, anything. They just have to be different numbers. The Dealer then hands the two envelopes to the Player in any order they please. The Player then selects one envelope to open. They then must decide whether the number in the other envelope is greater or less than the number in the envelope that was just opened. Obviously it's easy to win 50% of the time. The challenge of the game is to come up with a strategy which wins more than 50% of the time. Can you think of a strategy?
    You are on a game show. The host has chosen two different whole numbers (integers) and has hidden them behind doors A and B. He allows you to open one of the doors, thus revealing one of the numbers. Then, he asks you: is the number behind the other door greater or smaller than the number you have revealed? Your task is to answer this question correctly with probability strictly greater than one half.
    Alice secretly picks two different real numbers by an unknown process and puts them in two (abstract) envelopes. Bob chooses one of the two envelopes randomly (with a fair coin toss), and shows you the number in that envelope. You must now guess whether the number in the other, closed envelope is larger or smaller than the one you’ve seen. Is there a strategy which gives you a better than 50% chance of guessing correctly, no matter what procedure Alice used to pick her numbers?
    I give here my own view on this, hidden to not influence anyone...


    We don't have any relevant information as to the relation between the second number and the first number. 50% is the best anyone rational should expect.


    EB

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    It's mathematically impossible to beat 50/50 here. There may be a non-mathematical (soft) solution, though.
    Most people cannot remember more than 7 digits, and one may suppose that a larger length number would not be chosen to write down by the average person.
    therefore, an artificial bound of the selection pool of 9999999 may be guessed with some defensibility.
    For the revealed number, if it is larger than 5555555, then one may have greater odds of success by guessing "lower", and the converse.

    But, strictly speaking, since the selection pool is theoretically infinite, there is no middle point for which to base your guess.

    Can we know the size of the piece of paper the number is written on and the size of the numerals written? Bounds for the size of the number can then be interpreted.
    Can we observe the writing of the number? The time taken to print the numerals (and other visual keys) can indicate relative length of the two numbers.

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    Contributor repoman's Avatar
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    It would be interesting to have world class and terrible poker players as either or both of the participants of this study. Then have raising antes or folds.

    Seeing how bluffs and reverse bluffs work would be fun.

    On the topic of method to generate the number, what about the decent chance of having each number generated by the same mathematical relation.

    For example if the first number was 64.81 i would guess lower because 36.49 or 25.64 might be the lower. Unless that would be a feint.

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    Put two balls in a box with different numbers on them. Reach in and pick one. What is the probability it is less than or greater than the one left in the box.

    I'd say in a series of random trials the probability is 50/50 that you always pick the greater or the lessor.

    If I always bet that the pick will be the greater I will be right 50% of the time.

    It is an easy experiment to do at the kitchen table.

    In the envelope puzzle it depends on how the player makes the choice. If he picks one condition all the time it is 50/50. If he randomly picks one or the other then I'd have to review joint probabilities. I believe the probabilities multiply so he would be right 25% of the time. If the player picks greater a percent of the time and lesser a different percentage it can be calculated, can;t do it off the top of my head.

    The experiment.

    Put two balls in a box with different numbers.

    First bet it is always the greater. Pick a ball from the two in a box 20 times and record how many times it is > or <.

    The flip a coin before each pick. Heads bet > tails bet <. Record how many times you would win based on the coin toss.

    Call the exercise an intro to probability and statistics for the inexperienced philosopher investigating how math work and how math develops. Games of chance were the germesis of probability and statistics.

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    We don't have any relevant information as to the relation between the second number and the first number. 50% is the best anyone rational should expect.
    Astounding insight. Does this mean if I randomly pick between two possibilities it will always be 50/50? Are you sure about that?.That would have serious implications.

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    Looks like I made a mistake.

    I took two coins and put an x on one and a o on the other. Then I put them in a box.

    On a third coin I put an x on one side and an o on the other, and put it in a second box.

    I shook the boxes and dumped the x/o coin on the table and drew one of the two coins in the other box. If the two coins matched it was a win, mismatch and a loss.

    My first run of 20 trials showed 35% win. I repeated with a better job of randomizing and it was close to 50%.

    It seems counter intuitive to me but the experiment says otherwise. I saw it as two independent uncorrelated random variables thinking the two probabilities should multiply.

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    Contributor Speakpigeon's Avatar
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    Quote Originally Posted by Gun Nut View Post
    There may be a non-mathematical (soft) solution, though.
    Most people cannot remember more than 7 digits, and one may suppose that a larger length number would not be chosen to write down by the average person.
    therefore, an artificial bound of the selection pool of 9999999 may be guessed with some defensibility.
    For the revealed number, if it is larger than 5555555, then one may have greater odds of success by guessing "lower", and the converse.

    But, strictly speaking, since the selection pool is theoretically infinite, there is no middle point for which to base your guess.

    Can we know the size of the piece of paper the number is written on and the size of the numerals written? Bounds for the size of the number can then be interpreted.
    Can we observe the writing of the number? The time taken to print the numerals (and other visual keys) can indicate relative length of the two numbers.
    We could always modify the premises and a realistic investigation would certainly try to do that but I don't think we could determine the probability that each number be picked out, even if we assumed 0 above some particular number deemed too big to think of or to write down etc.

    I agree with your idea about 7 digits but that would only work for most people, as you say.

    Quote Originally Posted by Gun Nut View Post
    It's mathematically impossible to beat 50/50 here.
    That's what I think. However, one interesting twist is that there is an infinity of numbers to choose from for both the first and the second number so that I don't see how we could make any rationally justified calculation of probability. The probability for any one number to be selected is zero or infinitely small if that's different. So, I would say that the best we can do is just trust our intuition, which says 50/50.

    Do you have a calculation to justify 50/50?
    EB

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    Contributor Speakpigeon's Avatar
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    Quote Originally Posted by repoman View Post
    It would be interesting to have world class and terrible poker players as either or both of the participants of this study. Then have raising antes or folds.

    Seeing how bluffs and reverse bluffs work would be fun.
    And some people seem to believe they have a strategy to beat the 50% mark. They would loose their shirt.

    Quote Originally Posted by repoman View Post
    On the topic of method to generate the number, what about the decent chance of having each number generated by the same mathematical relation.

    For example if the first number was 64.81 i would guess lower because 36.49 or 25.64 might be the lower. Unless that would be a feint.
    Any way we could suss the probability of the second number given the first would do. But I think this is what we don't have in this case.
    EB

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    Contributor repoman's Avatar
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    Quote Originally Posted by Speakpigeon View Post
    Quote Originally Posted by repoman View Post
    It would be interesting to have world class and terrible poker players as either or both of the participants of this study. Then have raising antes or folds.

    Seeing how bluffs and reverse bluffs work would be fun.
    And some people seem to believe they have a strategy to beat the 50% mark. They would loose their shirt.

    Quote Originally Posted by repoman View Post
    On the topic of method to generate the number, what about the decent chance of having each number generated by the same mathematical relation.

    For example if the first number was 64.81 i would guess lower because 36.49 or 25.64 might be the lower. Unless that would be a feint.
    Any way we could suss the probability of the second number given the first would do. But I think this is what we don't have in this case.
    EB
    I misread the scenario. But if A knew what both numbers were and B picked a random envelope and A also saw that number then this would be a good case for a research study. It has probaly already happened.

    I wonder how much of a winning spread A could get if skilled at "head games" versus a skilled or nervous opponent B.

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    Contributor Speakpigeon's Avatar
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    Quote Originally Posted by repoman View Post
    I misread the scenario. But if A knew what both numbers were and B picked a random envelope and A also saw that number then this would be a good case for a research study. It has probaly already happened.

    I wonder how much of a winning spread A could get if skilled at "head games" versus a skilled or nervous opponent B.
    You mean like in the old cartoons of Fritz the Cat, where A could read the figures on the eyes of B as if reading on a jackpot machine the dollars he expects to win?

    That would seem a bit far fetched but then, the question is whether there's a strategy to beat the 50% mark. Some dudes think they've demonstrated there is such a strategy to win 50% of the time plus a delta which is infinitesimally small.
    EB

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