December 14th, 2008, 12:31 pm
Actually, both 1 and 2 are stable solutions in the sense that one can't win by choosing a different number if everyone else picks 1 or if everyone else picks 2. This result implies we need a tiebreaker. I can think of four tiebreakers:1. Simplicity: 1 is the first number and so many people might think it is "more natural" as a choice.2. Contextual information: If the company is named Two Guys and an Algorithm LLC, then 2 might be the right answer. Or people in a Cantonese-speaking region might prefer "2" because that number is a good number and sounds like the work "easy."3. Broken Symmetry: Consider the possible distribution of choices of 1s and 2s, the calculation of the average, and the role of the round-up rule (e.g., that 1.5 rounds to 2). Of the 91 possible values for the fraction of "1"s in the population of 90 choices, 50.55% of them favor choosing a "2". Round-up breaks the symmetry between the two choices. In fact, the closer one thinks the distribution will be 50-50, the greater the chance of winning with a "2". 4. Robustness to mistakes: If one believes that some of the 90 "geniuses" are are not that smart (e.g., those losers in sales who couldn't integrate x*dx if their lives depended on it), then we might anticipate a few mistaken answers that are much higher than 1. The answer "2" would be more robust to rare mistakes. Of course, if one expects a non-negligible fraction of mistakes, then answers higher than 2 may be warranted (see vxs's post for an example of that).Factors 3 and 4 suggest a choice of "2", factor 1 suggests a choice of 1, and factor 2 depends on context. EDIT: "2" isn't stable. It looks like I should have read the problem statement more closely (or drank my coffee before posting)!
Last edited by
Traden4Alpha on December 13th, 2008, 11:00 pm, edited 1 time in total.