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See: https://www.quantamagazine.org/20150722-solution-information-from-randomness

The article reasons as follows:

Given 2 arbitrary numbers A < B.
Choose a number G.

Case 1. G < A < B
You see a number larger than G. Select it. You'll be right 0.5 of the time.

Case 2. A < B < G
You see a number less than G. Chose the other. You'll be right 0.5 of the time.

Case 3. A < G < B. If you see a number < G choose the other. If > G, keep it. You're right 100% of the time.

Therefore the total probability of being right is > 0.5.

I think the problem is that the probability of, say, Case 3 occurring is (B-A)/Infinity or undefined. So the weighted sum of probabilities is undefined. So the article is wrong.

Is there a flaw in my logic or theirs?

z5h
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    Or is the flaw in the logic is that "There is no uniform distribution on the natural numbers." - http://math.stackexchange.com/a/14169/37159 – z5h Jul 23 '15 at 14:55
  • Nobody said those numbers are random, they are arbitrarily chosen. – fkraiem Jul 23 '15 at 15:36
  • I changed the wording. We still need P(case 3) > 0. It's not clear to me how we can guarantee that. – z5h Jul 23 '15 at 16:08
  • Basically, just because something can happen doesn't mean it has non-zero probability. I think the original problem is not very interesting, it is more useful to consider the "bonus question" where the numbers are chosen from ${0,1,\dots,10}$. – fkraiem Jul 23 '15 at 16:09
  • Yes! Exactly. So if we don't know case 3 has nonzero probability, then the articles claim is wrong. No? – z5h Jul 23 '15 at 16:12
  • Well, yes and no. The basic idea is sound, as demonstrated by the discrete case where you do get an advantage. It's just that going to the reals introduces complications (which I am not qualified to discuss in depth). It is possible that a rigorous argument will yield a "nice" generalisation of the discrete case. – fkraiem Jul 23 '15 at 16:14

2 Answers2

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Apparently both sides pick their numbers according to some (necessarily non-uniform) distribution on the set of feasible numbers ($\mathbb N$? $\mathbb N_0$? $\mathbb Z$? $\mathbb R$? I'll assume the latter). Your argument that case 3 does not occur a.s. corresponds to claiming that the two distributions do not "overlap". This of course depends on the distributions picked, but you can unilaterally ensure that an overlap exists, for example by picking $G\in\mathbb R$ according to a standard normal distribution.

Hagen von Eitzen
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  • We are told in the article "You have absolutely no idea how I generated these two numbers." My argument is that if there is a distribution over all of R, then the chance of me randomly selecting G where A<G<B is undefined. Not that it can't occur. – z5h Jul 23 '15 at 15:10
  • @z5h: But once you choose your distribution to pick $G$, the probability isn't undefined at all. Using Hagen's example of the standard normal distribution, for any given real $A,B$ with $A\lt B,$ there is a positive probability that your $G$ lies strictly between $A$ and $B$. You don't know what the probability is, but you do know that it's positive. – Cameron Buie Jul 24 '15 at 12:45
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Interesting problem!

Let $A,B\in\mathbb R$ be fixed with $A<B$.

I can just choose a distribution on $\mathbb R$ and pick out a $G\in\mathbb R$ with respect to this distribution.

For convenience I choose a continuous distribution that assigns positive probabilities to intervals. There are plenty of them of course.

Then $p:=\mathbb{P}(A<G<B)>0$ and the probability of selecting the hand with $B$ in it is: $$\frac12(1-p)+1p=\frac12+\frac12p>\frac12$$

So yes it works!

I don't see why it should not work if $A$ and $B$ are somehow generated on some mysterious way unknown to me.

Then $p$ can then be looked at as a random variable $p(A,B)$ with $\mathbb P(p(A,B)>0)=1$.

drhab
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