
Re: A good probability puzzle but what is the right wording?
Posted:
Feb 4, 2013 9:33 AM


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 youve 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. > >quasi

