Paul
Posts:
258
Registered:
7/12/10
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Re: A good probability puzzle but what is the right wording?
Posted:
Feb 23, 2013 5:33 PM
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On Monday, February 4, 2013 2:33:12 PM UTC, David C. Ullrich wrote:
> On Mon, 04 Feb 2013 08:15:59 -0500, quasi <quasi@null.set> wrote:
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>
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> >quasi wrote:
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> >>Paul wrote:
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> >>>
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> >>>The following puzzle is copied and pasted from the internet.
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> >>>
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> >>> Alice secretly picks two different real numbers by an unknown
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> >>> process and puts them in two (abstract) envelopes. Bob
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> >>> chooses one of the two envelopes randomly (with a fair coin
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> >>> toss), and shows you the number in that envelope. You must
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> >>> now guess whether the number in the other, closed envelope
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> >>> is larger or smaller than the one you�ve seen. Is there a
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> >>> strategy which gives you a better than 50% chance of guessing
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> >>> correctly, no matter what procedure Alice used to pick her
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> >>> numbers?
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> >>
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> >>Yes.
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> >>
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> >>Let R denote the set of real numbers and let (0,1) denote
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> >>the open interval from 0 to 1.
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> >>
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> >>Let f : R -> (0,1) be a strictly decreasing function.
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> >>
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> >>Use the following strategy:
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> >>
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> >>If the initially exposed value is t, "switch" with probability
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> >>f(t) and "stay" with probability 1 - f(t).
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> >>
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> >>Suppose Alice chooses the pair x,y with x < y (by whatever
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> >>process, it doesn't matter). After Alice choose that pair,
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> >>then, by following the strategy I specified above, the
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> >>probability of guessing the highest card is exactly
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> >>
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> >> (1/2)*f(x) + (1/2)*(1 - f(y))
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> >>
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> >>which simplifies to
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> >>
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> >> 1/2 + f(x) - f(y)
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> >
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> >I meant:
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> >
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> >which simplifies to
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> >
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> > 1/2 + (1/2)*(f(x) - f(y))
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> >
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> >>and that exceeds 1/2 since f is strictly decreasing.
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>
>
> Huh. I thought the answer was obviously no. But
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> this seems right. Huh.
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>
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> >>Of course, it's not the case that probability of guessing
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> >>correctly is more than c for any fixed c > 1/2, but the
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> >>problem didn't require that.
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>
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> _If_ we assume in addition that Alice used some fixed
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> probability distribution on {(x,y) : x < y} to choose
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> x and y then it does give such a c, namely the
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> expected value of 1/2 + 1/2(f(x) - f(y)). The expected
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> value of a strictly positive random variable is
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> strictly positive.
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> >
>
David,
Rethinking this problem, I think your initial "obviously no" response may have been correct. There is an essential point here: the concept of "selection by an unknown process" is not readily translated into mathematics. Whenever you pose maths-sounding-problems which have essentially non-mathematical components you easily achieve contradictions. For example, you can readily obtain paradoxical conclusions from the non-mathematical premise that A "selects a non-negative integer in such a way that each integer has an equal probability of being chosen."
Rather than you saying that you were initially wrong, I think (tentatively think) that you were not wrong and that the argument above just illustrates that one can get paradoxical conclusions when one introduces mathematically incorrect or incoherent concepts.
I believe that using the non-mathematical concept of "any process whatsoever" is essential to obtaining the paradoxical result in this thread. To make the question meaningful, there would need to be a well-defined set of random processes which selects the number and a well-defined means of choosing among this well-defined set of processes. But then, of course, we wouldn't get a result which seems surprising or counter-intuitive.
At this point in time, my answer is that "no such method is possible to ensure a > 50% probability". The argument in this thread doesn't work because non-mathematics has been injected into it, just as an argument which involves "the set of all sets" can not be trusted.
Paul Epstein
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