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Re: Spooky math problem

by lidden (Deacon)
on Sep 05, 2004 at 17:01 UTC ( #388632=note: print w/ replies, xml ) Need Help??


in reply to Spooky math problem

Suppose I have two envelopes. All you know is that they contain different numbers. I randomly hand you one of them. You open it, look at it, then hand it back. You now have sufficient information to, with guaranteed better than even odds, correctly tell me whether I gave you the envelope with the larger number. How?
No, that question actually is:
If I give you a number from a set of two diffrent numbers tell me if you got the small or large number?

My odds of doing that is exactly 50%.

If you sit down and do the algebra, you will find that your probability of being right turns out to be exactly 50% plus 1/2 the probability that you pick a number between my two.
Hmm. If I choose to guess myNumber + epsilon(where epsilon is small enough) I have 50% chance of being between your two numbers and hence have 75% chance of being correct, which is silly:-)

Ok four years late but I got here from this node.


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Re^2: Spooky math problem
by tilly (Archbishop) on Sep 09, 2004 at 01:29 UTC
    Suppose I have two envelopes. All you know is that they contain different numbers. I randomly hand you one of them. You open it, look at it, then hand it back. You now have sufficient information to, with guaranteed better than even odds, correctly tell me whether I gave you the envelope with the larger number. How?
    No, that question actually is:
    If I give you a number from a set of two diffrent numbers tell me if you got the small or large number?

    My odds of doing that is exactly 50%.
    You'd think that your odds of doing that are 50%. It turns out that they don't have to be. This violates common sense, which is what makes the problem interesting.

    Mathematicians are very interested in understanding situations where their intuitions differ from what happens. By examining those "pathological cases" you sharpen your intuition for how things work.

    Note that in these weird boundary cases it is important to be very precise in your thinking. Any sloppiness will cause you to misanalyze the problem to fit your preconceptions, not reality.

    If you sit down and do the algebra, you will find that your probability of being right turns out to be exactly 50% plus 1/2 the probability that you pick a number between my two.
    Hmm. If I choose to guess myNumber + epsilon(where epsilon is small enough) I have 50% chance of being between your two numbers and hence have 75% chance of being correct, which is silly:-)
    And that is what sloppiness looks like.

    The algebra only works if your method of choosing the other number is independent of the number that you get. As soon as you introduce a dependency you have to analyze that dependency, and it changes the answer.

    Since you don't seem inclined to try the algebra, allow me to demonstrate what it looks like. Suppose that the numbers that I have are x and y with x < y. Suppose that p(z) is the function that tells you for any number how likely you are to think that you got the larger one if you're handed that number. Then:

    P(You're right) = P(You're handed x)*P(You don't think that x is larger) + P(You're handed y)*P(You think that y is larger) = 0.5 * (1 - p(x)) + 0.5 * p(y) = 0.5 + 0.5*(p(y) - p(x))
    So far we haven't introduced any details about the method. With the method that I described, though, p is a monotonically increasing function, in fact p(y) - p(x) is the probability that you pick a number between x and y, so you're better than even odds by half the probability that your number is between my two.

    With the choosing method that you came up with, you always conclude that you're handed the smaller number. Therefore p(x) and p(y) are both 0.5 and your odds of being right remain at 50%. Slight difference!

    If your eyes are glazing over at the algebra, then please examine the following (bad) ASCII art version of the picture that I described in my root node.

    (Right if handed the larger) | <----------------------------| (even odds) (guaranteed) | (even odds) <-------------x------------------y---------------> | |-------------------------------> | (Right if handed the smaller)
    As you can see, no matter what number you independently come up with, you never have worse than even odds of being right, and you have some chance of guaranteeing that you're right. That chance gives you better than even odds overall.
      In order for the envelope receiver to receive the benefit of additional odds above 50%, I think the receiver must come up with a random number before viewing the number in the envelope. Then given the number in the envelope they can determine whether to say high or low.

      Using tilly's ascii chart above; my guess is z, assume my guess in between x and y. When I open the envelope to reveal x, which is smaller than z, I will then say that I was handed the smaller number and I will be right. If I open the envelope to reveal y, which is larger than z, I will say it is the larger number and I will be right. If my guess z is smaller than x, I will be wrong when ever I am handed x, and right whenever I am handed y. Similarly if z is larger than y, I will be right whenever I am handed x and wrong when handed y.

      Now, if I look at the number in the envelope first, as a human, I am incapable of picking a truly random number not biased by that number. (Hell, humans are incapable of picking a truly random number period). So, I now have a 50-50 shot at deciding to pick higher or lower than that number. Which means 50% chance of deciding whether to say the received envelope number is higher or lower than the number in the other envelope. It no longer matters whether my 'guess' falls between the actual numbers anymore, because I can not receive the other envelope. Therefore, the odds are 50%. Only by choosing a number before the envelope is handed to you can you increase your odds however slightly.

      -Scott

        It is possible to do it by choosing a number before you see the one in the envelope.

        But it is also possible to do it by choosing a number through a mechanical procedure (eg a series of coin flips), in which case it doesn't matter whether you see the number in the envelope first.

        However you're exactly right that the method of choosing the random number cannot depend on the number that you see.

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