[mulur]

Hmm I'm not exactly good with maths (or puzzles for that matter) but here is a blind stab:

By break even, I think you mean that the game is played multiple times, as long as necessary. If you are not following a martingale like strategy (edit: you can't anyways, the bets are fixed) and respond randomly (or with full faith that this is your lucky day, doesn't matter really), you are expected to break even with 50% chance.

How can we improve and make better than 50%?

'A' chooses two distinct integers each time. They can choose among infinite number of integers. For the first guess, I can't do better than 50% chance but if I start to keep history of the numbers that were picked it feels to me like I can do better than chance if I assume A chooses their distinct integers uniformly random by expecting a uniform distribution. I don't have much ideas about how to judge uniformness in an unbounded set of integers though but maybe something like: if first hand is less than the average of all previous numbers, I'd say "other number is higher", otherwise I'd say "other number is lower". I feel like this would improve my chances to more than 50% in the very long term (at infinite plays perhaps?) but am not able to prove it.

If A is not choosing their numbers randomly with a true RNG then I'd play many games at 50% chance then run their choices through a neural network to extract patterns then I'd do better than random for sure.

[sirclueless]

You can do better than that. You can guarantee there is > 50% chance you win on the first guess.

[Hoagy]

Wow, that is very counterintuitive..

I suppose you could set any threshold x, and play so that if the first number is bigger than x you stick, otherwise you switch. This would be strictly better than random (break even) if x is within the range of numbers that A is picking from and the same as random if not.

It depends on when you measure your chances: before A has picked a generator this gives you >50% chance but if A has already picked a range to draw from that doesn't contain x then you're back to 50/50. Still, I think this would count as guaranteeing >50%?

[mundo]

> You can guarantee there is > 50% chance you win on the first guess.

You mean "greater than" and not "greater than or equal to"?

The latter is easy; lots of naive strategies accomplish that. But the former implies that you can declare a strategy and the puzzle offerer cannot then pick a distribution that defeats it. It's hard to imagine how you could get >50% for example if the puzzler's strategy is to pick N and N+1 for very large N.

[mulur]

Ok, my solution is needlessly complicated because unbounded numbers between -Infinity and Infinity has a mean of 0, so if we just pick 0 and say "if hand1 < 0 then next is bigger, otherwise smaller" strategy already gives 75% accuracy long term.

pyk's solution above works 66.6% even for the first try...

[mulur]

Oh that's interesting. Thank you for the clarification.