
Re: Beating the Odds?
Posted:
Feb 2, 2013 3:10 AM


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)[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". > 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?

