Paul
Posts:
763
Registered:
7/12/10


Re: A good probability puzzle but what is the right wording?
Posted:
Feb 24, 2013 12:19 PM


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 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
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

