2. This Strategy does always work, because it exploits that you not only know (ex ante) there are two different numbers, but also that these two numbers are distinct. You just draw a RV Z~N(0,1). If Z is between the two numbers, you always decide right. Since on IR you have a positive probability to be in that interval, you decide with p>0.5 right.

You can use any measure which is absolutely continuous w.r.t lebesgue measure, because the 1-dim lebesgue measure give mass >0 to any non-empty interval on IR.

If the two numbers are very big (or small) and very near each other the probability to hit this interval will be small, but it stays always strictly positive.

If you use only positive integers, you may have to require that the distance of the two numbers is at least 2. That way you can construct an interval about the integer in the middle and round every random variable Z in this interval to this integer.

You may want to increase the probability by using a pdf other than the gaussian, because its tails decay so rapidly. The cauchy distribution might be better.

]]>Interesting suggestions. I like the idea of using a discrete power-law type distribution, where the higher the number is, the higher it is likely to be.

@Napo:

1. I considered the Cauchy and a couple other continuous distors, but working with any of these would make the code more complicated. For starters, I would have to round the answer off to an integer (A game of “guess the real number” might not be much fun, eh?). Next, since allowing negative numbers seems to violate the spirit of game and make it less fun, I would have to take the absolute value of the number. Unless the distro was centered at zero (and the center is supposed to move over time), this makes finding conditional probabilities harder since part of the tail gets added back in to the positive. Or I could truncate at zero, which might cause other issues like attraction to 0 over time, which is the problem with using the exponential, another possibility I considered. So far as the Cauchy itself goes…. that just seemed, well, kind of mean, don’t you think?

2. Thanks for the link. Interesting game, hadn’t heard of it. Seems like the strategy would only work if the person writing down numbers on the slips of paper did all their writing in advance, then shuffled the papers. Otherwise they could tweak the strategy of what they write down with each round.

]]>pick the largest of two numbers

http://www.stanford.edu/~cover/papers/paper73.pdf ]]>

And then scale it up to get a nice big median.

Or, since I am lazy, you could get close enough to that by generating from a Pareto and truncating.

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