quasi
Posts:
10,903
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 mathssoundingproblems which have essentially >nonmathematical components you easily achieve contradictions.
You're overgeneralizing.
>For example, you can readily obtain paradoxical conclusions from >the nonmathematical premise that A "selects a nonnegative >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 nonmathematical 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 welldefined set of random processes which >selects the number and a welldefined means of choosing among >this welldefined 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 counterintuitive.
To my view it _was_ somewhat counterintuitive 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 nonmathematics 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 selfcontradictory, 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

