> Ballmer states that the answer is "No" for two reasons: firstly, because he can pick numbers that'll be the most difficult for you...
The article goes on to show that there are numbers where a binary search always has the guesser paying $1
> Ballmer states that the answer is "No" for two reasons: firstly, because he can pick numbers that'll be the most difficult for you...
The article goes on to show that there are numbers where a binary search always has the guesser paying $1
But he also knows that you know that he could pick numbers that will be the most difficult... So could then pick one of the numbers that actually are guessable within 5 guesses to trick you.
But then you also know that he knows that you know that he could pick difficult numbers too.
I'm not entirely sure if this invalidates Ballmer's advantage, but I would be interested to know what the 'perfect' strategy would be for this game considering the meta-game.
If we accept cheating is allowed, we can also potentially accept other 'cheating' scenarios where the player repeatedly punches Ballmer in the face until he discloses the number, thus winning. Or where the player just refuses to pay at the end.
IMO the problem is only interesting if we assume any form of cheating isn't allowed.
Here, you can play the game with me. Higher. Lower. Higher. Higher. Lower. Correct. Six guesses, you owe me $1.
if I started off by guessing 50 twice you're cooked. or 50 and 52.
> secondly because the expected value of the game (assuming Ballmer chooses randomly) is negative: you end up paying Ballmer