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Re: Beating the Odds?
Posted:
Feb 1, 2013 3:15 PM
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On Feb 1, 1:32 pm, John Dawkins <artfldo...@aol.com> wrote:
> In article <Pine.NEB.4.64.1301300026190.3...@panix3.panix.com>,
> 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".
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