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Topic: A good probability puzzle but what is the right wording?
Replies: 10   Last Post: Feb 24, 2013 12:19 PM

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Paul

Posts: 400
Registered: 7/12/10
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:
>
>
>

> >quasi wrote:
>
> >>Paul wrote:
>
> >>>
>
> >>>The following puzzle is copied and pasted from the internet.
>
> >>>
>
> >>> 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?
>
> >>
>
> >>Yes.
>
> >>
>
> >>Let R denote the set of real numbers and let (0,1) denote
>
> >>the open interval from 0 to 1.
>
> >>
>
> >>Let f : R -> (0,1) be a strictly decreasing function.
>
> >>
>
> >>Use the following strategy:
>
> >>
>
> >>If the initially exposed value is t, "switch" with probability
>
> >>f(t) and "stay" with probability 1 - f(t).
>
> >>
>
> >>Suppose Alice chooses the pair x,y with x < y (by whatever
>
> >>process, it doesn't matter). After Alice choose that pair,
>
> >>then, by following the strategy I specified above, the
>
> >>probability of guessing the highest card is exactly
>
> >>
>
> >> (1/2)*f(x) + (1/2)*(1 - f(y))
>
> >>
>
> >>which simplifies to
>
> >>
>
> >> 1/2 + f(x) - f(y)
>
> >
>
> >I meant:
>
> >
>
> >which simplifies to
>
> >
>
> > 1/2 + (1/2)*(f(x) - f(y))
>
> >
>
> >>and that exceeds 1/2 since f is strictly decreasing.
>
>
>
> Huh. I thought the answer was obviously no. But
>
> this seems right. Huh.
>
>
>

> >>Of course, it's not the case that probability of guessing
>
> >>correctly is more than c for any fixed c > 1/2, but the
>
> >>problem didn't require that.
>
>
>
> _If_ we assume in addition that Alice used some fixed
>
> probability distribution on {(x,y) : x < y} to choose
>
> x and y then it does give such a c, namely the
>
> expected value of 1/2 + 1/2(f(x) - f(y)). The expected
>
> value of a strictly positive random variable is
>
> strictly positive.
>

> >
>

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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