Paul
Posts:
492
Registered:
7/12/10


Re: A good probability puzzle but what is the right wording?
Posted:
Feb 23, 2013 5:33 PM


On Monday, February 4, 2013 2:33:12 PM UTC, David C. Ullrich wrote: > On Mon, 04 Feb 2013 08:15:59 0500, quasi <quasi@null.set> wrote: > > > > >quasi wrote: > > >>Paul wrote: > > >>> > > >>>The following puzzle is copied and pasted from the internet. > > >>> > > >>> 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 choose 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 + f(x)  f(y) > > > > > >I meant: > > > > > >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. > > > > >>Of course, it's not the case that probability of guessing > > >>correctly is more than c for any fixed c > 1/2, but the > > >>problem didn't require that. > > > > _If_ we assume in addition that Alice used some fixed > > probability distribution on {(x,y) : x < y} to choose > > x and y then it does give such a c, namely the > > expected value of 1/2 + 1/2(f(x)  f(y)). The expected > > value of a strictly positive random variable is > > strictly positive. > > > >
David,
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. Whenever you pose mathssoundingproblems which have essentially nonmathematical components you easily achieve contradictions. 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." 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. But then, of course, we wouldn't get a result which seems surprising or counterintuitive.
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, just as an argument which involves "the set of all sets" can not be trusted.
Paul Epstein

