Can I choose the larger of 2 arbitrary numbers with P>0.5 after only seeing 1? 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?
 A: 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.
A: 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 $\mathbb P(p(A,B)>0)=1$. 
