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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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 quasi Posts: 12,067 Registered: 7/15/05
Re: A good probability puzzle but what is the right wording?
Posted: Feb 24, 2013 12:36 AM

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 youï¿½ve 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.

>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

Date Subject Author
2/4/13 Paul
2/4/13 quasi
2/4/13 quasi
2/4/13 quasi
2/4/13 David C. Ullrich
2/23/13 Paul
2/24/13 quasi
2/24/13 Paul
2/4/13 RGVickson@shaw.ca
2/5/13 Paul
2/5/13 David Petry