Paul
Posts:
257
Registered:
7/12/10
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Re: A good probability puzzle but what is the right wording?
Posted:
Feb 24, 2013 12:19 PM
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On Sunday, February 24, 2013 5:36:49 AM UTC, quasi wrote:
> pepstein5 wrote:
>
> >David C. Ullrich wrote:
>
> >> quasi wrote:
>
> >> > Paul wrote:
>
> >> >>
>
> >> >> Alice secretly picks two different real numbers by an
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> >> >> unknown process and puts them in two (abstract) envelopes.
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> >> >> Bob chooses one of the two envelopes randomly (with a fair
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> >> >> coin toss), and shows you the number in that envelope.
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> >> >> You must now guess whether the number in the other, closed
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> >> >> envelope is larger or smaller than the one you�ve seen.
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> >> >> Is there a strategy which gives you a better than 50%
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> >> >> chance of guessing correctly, no matter what procedure
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> >> >> Alice used to pick her 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
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> >> >probability 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 chooses 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 + (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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> >>
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> >> Huh. I thought the answer was obviously no. But this seems
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> >> right. Huh.
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> >
>
> >Rethinking this problem, I think your initial "obviously no"
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> >response may have been correct. There is an essential point here:
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> >the concept of "selection by an unknown process" is not readily
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> >translated into mathematics.
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>
>
> Right.
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>
>
> So any strategy which can defeat an unknown process better
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> not depend in any way on the nature of the process.
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>
>
> If you look carefully at the strategy I outlined for staying or
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> switching, the probability that Bob wins depends on the pair
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> (x,y) chosen by Ann, but that probability exceeds 1/2 regardless
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> of the choice of (x,y).
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>
>
> Said differently there is _no_ choice of (x,y) for which Bob
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> is not favored to win against _that_ selection.
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>
>
> Since Bob is favored to win regardless of Ann's chosen pair,
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> the method by which Ann chooses the pair (x,y) can't affect
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> the conclusion that Bob's probability to win exceeds 1/2.
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>
>
> >Whenever you pose maths-sounding-problems which have essentially
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> >non-mathematical components you easily achieve contradictions.
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>
>
> You're over-generalizing.
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>
>
> >For example, you can readily obtain paradoxical conclusions from
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> >the non-mathematical premise that A "selects a non-negative
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> >integer in such a way that each integer has an equal probability
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> >of being chosen."
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>
>
> That's a straw man argument.
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>
>
> Of course there's no way of selecting a nonnegative integer in
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> such a way that each integer has an equal probability of being
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> chosen.
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>
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> When the problem indicates that Ann can use any method, of course
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> that prohibits methods which are mathematically impossible.
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>
>
> But if that vagueness bothers you, then allow Ann to select a
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> pair (x,y) in R^2 using _any_ probability distribution on the
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> set {(x,y) in R^2 | x != y}.
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>
>
> >Rather than you saying that you were initially wrong, I think
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> >(tentatively think) that you were not wrong and that the argument
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> >above just illustrates that one can get paradoxical conclusions
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> >when one introduces mathematically incorrect or incoherent
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> >concepts. I believe that using the non-mathematical concept of
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> >"any process whatsoever" is essential to obtaining the paradoxical
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> >result in this thread. To make the question meaningful, there
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> >would need to be a well-defined set of random processes which
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> >selects the number and a well-defined means of choosing among
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> >this well-defined set of processes.
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>
>
> I gave you one such well defined set of processes:
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>
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> Ann can choose _any_ probability distribution on the set
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> {(x,y) in R^2 | x != y}. The set of such distributions is
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> certainly a well defined set.
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>
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> For any such distribution chosen by Ann, let Ann's pair (x,y) be
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> chosen at random subject to that distribution.
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>
>
> But regardless of her chosen distribution, Bob still wins with
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> probability greater than 1/2 against _any_ point (x,y) chosen
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> based on that distribution.
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>
>
> >But then, of course, we wouldn't get a result which seems
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> >surprising or counter-intuitive.
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>
>
> To my view it _was_ somewhat counter-intuitive that Bob has a
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> _single_ strategy which wins with probability greater than 1/2
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> against _all_ possible selection methods by Ann for the point
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> (x,y) (or if you prefer, _all_ possible probability
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> distributions on the set {(x,y) in R^2 | x != y}).
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>
>
> >At this point in time, my answer is that "no such method is
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> >possible to ensure a > 50% probability". The argument in this
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> >thread doesn't work because non-mathematics has been injected
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> >into it,
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>
>
> But you're wrong.
>
>
>
> Try to understand the strategy I outlined for Bob. I showed that
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> no pair (x,y) prevents Bob from being favored. Thus, the method
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> by which Ann chooses the point (x,y) won't affect that
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> conclusion,
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>
>
> >just as an argument which involves "the set of all sets" can not
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> >be trusted.
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>
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> It's not the same thing at all.
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>
>
> To dramatize the fact that the phrase "by any selection method"
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> is not automatically self-contradictory, consider the following
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> trivial example ...
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>
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> Two players, A,B.
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>
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> A chooses a negative real number a (by any method, it doesn't
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> matter how) and writes it on a card, placing it face down on
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> a table.
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>
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> B then announces a nonpositive real number b.
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> A's card is then turned up revealing the value of a.
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> B wins if b > a, otherwise A wins.
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> It's obvious that, regardless of A's selection method, B wins by
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> choosing b = 0.
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>
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> So the problem statement doesn't need to constrain the set of
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> possible selection methods for A since, assuming B plays
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> optimally, A's selection method doesn't matter.
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>
>
> quasi
I was indeed wrong and quasi's argument is correct. I hope that this acknowledgement of error satisfies the criteria of 1) Common courtesy, 2) Informing the readers that I am retracting the opinion that the problem is flawed, 3) Establishing credibility.
(The above were given as reasons for acknowledging error in a previous thread.)
Paul Epstein
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