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Re: Beating the Odds?
Posted:
Feb 2, 2013 3:10 AM
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On Fri, 1 Feb 2013, Butch Malahide wrote:
> On Feb 1, 1:32 pm, John Dawkins <artfldo...@aol.com> wrote:
> > William Elliot <ma...@panix.com> wrote:
> >
> > > There is a fair coin with a different integer on each side that you
> > > can't see and you have no clue how these integers were selected. The
> > > coin is flipped and you get to see what comes up. You must guess if
> > > that was the larger of the two numbers or not. Can you do so with
> > > probability > 1/2?
> >
> > Yes, provided you have a random variable at your disposal, say a
> > standard normal random variable X. If the number showing on the coin
> > is less than X, then guess that the number on the other side of the
> > coin is the larger of the two. If the number showing is greater than
> > or equal to X, then guess that the number showing is the larger. Your
> > chance of being correct is
> >
> > (1/2)[1-Phi(x)] + (1/2)Phi(y) = 1/2 + (1/2)[Phi(y) -Phi(x)] > 1/2.
> >
> > Here Phi is the standard normal distribution function, and x < y are
> > the two numbers on the coin.
>
> Yes, this is (an instantiation of) the strategy I proposed upthread.
> With your strategy, for fixed x and y, the probability p(x,y) of
> guessing right is > 1/2, but the infimum over all x and y is = 1/2. So
> it's a semantic puzzle: what does it mean to say that "you can guess
> right with probability > 1/2"? A plausible interpretation would be
> that "there is a probability p > 1/2 such that you can guess right
> with probability p", so that the answer to the OP's vaguely specified
> question seems to be "no".
>
It didn't ask for a fixed probability p, nor a method that would choose
correctly with probability p. It hypothesized an single even and for that
single event, with the given strategy, a correct guess can be made with
probability > 1/2. How could that probability be calculated, given two
integers n,m in Z with n < m and the distribution of a random variable
over R?
Naive question. Can the distribution be uniform?
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