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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: 386
Registered: 7/12/10
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
>
> >> >> 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 chooses 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 + (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.
>
> >
>
> >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.
>
>
>
> Right.
>
>
>
> So any strategy which can defeat an unknown process better
>
> not depend in any way on the nature of the process.
>
>
>
> If you look carefully at the strategy I outlined for staying or
>
> switching, the probability that Bob wins depends on the pair
>
> (x,y) chosen by Ann, but that probability exceeds 1/2 regardless
>
> of the choice of (x,y).
>
>
>
> Said differently there is _no_ choice of (x,y) for which Bob
>
> is not favored to win against _that_ selection.
>
>
>
> Since Bob is favored to win regardless of Ann's chosen pair,
>
> the method by which Ann chooses the pair (x,y) can't affect
>
> the conclusion that Bob's probability to win exceeds 1/2.
>
>
>

> >Whenever you pose maths-sounding-problems which have essentially
>
> >non-mathematical components you easily achieve contradictions.
>
>
>
> You're over-generalizing.
>
>
>

> >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."
>
>
>
> That's a straw man argument.
>
>
>
> Of course there's no way of selecting a nonnegative integer in
>
> such a way that each integer has an equal probability of being
>
> chosen.
>
>
>
> When the problem indicates that Ann can use any method, of course
>
> that prohibits methods which are mathematically impossible.
>
>
>
> But if that vagueness bothers you, then allow Ann to select a
>
> pair (x,y) in R^2 using _any_ probability distribution on the
>
> set {(x,y) in R^2 | x != y}.
>
>
>

> >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.
>
>
>
> I gave you one such well defined set of processes:
>
>
>
> Ann can choose _any_ probability distribution on the set
>
> {(x,y) in R^2 | x != y}. The set of such distributions is
>
> certainly a well defined set.
>
>
>
> For any such distribution chosen by Ann, let Ann's pair (x,y) be
>
> chosen at random subject to that distribution.
>
>
>
> But regardless of her chosen distribution, Bob still wins with
>
> probability greater than 1/2 against _any_ point (x,y) chosen
>
> based on that distribution.
>
>
>

> >But then, of course, we wouldn't get a result which seems
>
> >surprising or counter-intuitive.
>
>
>
> To my view it _was_ somewhat counter-intuitive that Bob has a
>
> _single_ strategy which wins with probability greater than 1/2
>
> against _all_ possible selection methods by Ann for the point
>
> (x,y) (or if you prefer, _all_ possible probability
>
> distributions on the set {(x,y) in R^2 | x != y}).
>
>
>

> >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,
>
>
>
> But you're wrong.
>
>
>
> Try to understand the strategy I outlined for Bob. I showed that
>
> no pair (x,y) prevents Bob from being favored. Thus, the method
>
> by which Ann chooses the point (x,y) won't affect that
>
> conclusion,
>
>
>

> >just as an argument which involves "the set of all sets" can not
>
> >be trusted.
>
>
>
> It's not the same thing at all.
>
>
>
> To dramatize the fact that the phrase "by any selection method"
>
> is not automatically self-contradictory, consider the following
>
> trivial example ...
>
>
>
> Two players, A,B.
>
>
>
> A chooses a negative real number a (by any method, it doesn't
>
> matter how) and writes it on a card, placing it face down on
>
> a table.
>
>
>
> B then announces a nonpositive real number b.
>
>
>
> A's card is then turned up revealing the value of a.
>
>
>
> B wins if b > a, otherwise A wins.
>
>
>
> It's obvious that, regardless of A's selection method, B wins by
>
> choosing b = 0.
>
>
>
> So the problem statement doesn't need to constrain the set of
>
> possible selection methods for A since, assuming B plays
>
> optimally, A's selection method doesn't matter.
>
>
>
> 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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