The reason it's counterintuitive is because, in the game, you don't know that the two numbers are random. In fact, they are chosen by an adversary. You flip a 2: do you think this is most likely the lower number, or do you think the other player picked 1 and 2 just to mess with you?

That's the point of picking a random pivot: it reduces your opponent's ability to influence the situation to only placing their numbers as close to each other as possible.

I think this distracts from the actual problem. They have a more generalized case, (-inf, inf).

In your case, if we don't allow double selecting, 2 is the lowest number between 1 and 10. So you have a 100% chance of guessing right. You're using the bound to help you. Being unbounded you still have an infinite set of numbers smaller than the number you see and an infinite number higher. To be more analogous, that would be like flipping over 5 every single time (on [0,10]). You have {1,2,3,4} below and {6,7,8,9} above. EVERY TIME.

Of course, but I'm having trouble understanding why this works on an infinite number line. Say you flip over a 2,387,290,723,013,172,348,238,987: what do you think is the right guess?

Its the largest because its positive, 75% I’m right.

And if we are only looking at positive numbers in an infinite axis I’d need to borrow your generator for a test to approximate its midpoint (which obviously it doesn’t have), which I then weigh against the original number with the logic “if it’s larger than what I got, so is your other number. If it’s smaller so is the other number.” Now since the chance for my number being the middle of the three drawn is 1/3(only case I lose), that gives me 66% chance of winning that bet.