
Re: Beating the Odds?
Posted:
Feb 1, 2013 3:15 PM


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)[1Phi(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".

