Paul
Posts:
406
Registered:
7/12/10


Re: Beating the Odds?
Posted:
Feb 2, 2013 5:38 AM


On Friday, February 1, 2013 8:15:00 PM UTC, Butch Malahide wrote: > 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".
Your challenge to the OP to explain more formally what "there is a probability p > 1/2 ..." means is a good one. However, I disagree that your interpretation is plausible. I will demonstrate the implausibility of your interpretation by analogy. You will do an exam consisting of N multiple choice questions each question having two options A and B, exactly one of which is correct. Before writing the paper, the examsetter uses an unbiased coin. For each question 1 to N, the examsetter does a coin toss. If coin toss number x is heads, the exam is set so that A is the correct answer to question x. It's B if tails. N is an odd positive integer, but you have no idea what odd number N will be. Also, most unfortunately, you have absolutely no understanding of the subject being examined and have no understanding of any of the questions in the paper. However, somehow the answer to question 1 has been leaked by a reliable source and you know that this answer is A. Puzzle. Do you have a strategy enabling you to score > 50% with a probability of > 50%? Surely, the normal answer is "yes"  in fact any strategy does this so long as you answer 1A. However, by your logic, your answer would be that there is no such strategy because the above 50% bounds can't be improved without knowing N. To this ultrasimple analogy, if you accept my answer  answer 1A and then fill the rest of the paper however you want  then, it seems you have to accept the intended solution of the OP's problem.
Paul Epstein

