Theoretical Random Number Game

[Jazza]

The is between 2 players. Player A picks a number, 0 or 1, Player B tries to guess this number, winner wins say $1 from the loser. They repeat this many times.

Can player A choose his numbers in a non completely random way, such that player B cannot increase his EV above 0?

If you were player B and suspected he was choosing numbers in a not fully random way, what kind of algorithm would you use to try and guess the right number (using his past history of chosen numbers)?

One of the problems I see already is I don't know how to precisly define "a non completly random way"

[Brainwalter]

I don't think A can use any static strategy to prevent B from gaining an EV edge. Any nonrandom selection method will eventually be exploitable once B figures it out.

[Jazza]

but if A can figure out how B will try to exploit A, then A can exploit this, and so on...

[Brainwalter]

Yeah, I guess it doesn't matter that "A selects a number and B tries to guess it." Their selections are independent and for all practical purposes simultaneous.

If A selects his numbers randomly B cannot increase his EV above 0. So I think what you're really asking is whether A's optimal strategy is to select randomly (0 EV) or whether there is a strategy where B has forced -EV. This is different from your original question, and each player has the same (limited or long-run nonexistant) opportunity to outwit the other.

[Jazza]

[ QUOTE ]
This is different from your original question

[/ QUOTE ]

it is?

in any case it is true that if at least one player randomizes his strategy, 50/50, then the EV has to be 0.

but yeah i'm interested in when they try to outwit each other, for instance, if your opponent did the same thing 1000 times in a row, you would probably consider a strategy that is not randomly picking 0 and 1

[bighomage]

Neither player really has an advantage either way even if the numbers are picked non-randomly. Another way to rephrase the game is that if both players guess the same number, B wins; If the numbers are different, A wins. B may be able to catch a pattern of what numbers A is guessing, but A also can find a pattern of what B guesses. Neither player is at a disadvantage.

[Jazza]

[ QUOTE ]
Neither player really has an advantage either way even if the numbers are picked non-randomly. Another way to rephrase the game is that if both players guess the same number, B wins; If the numbers are different, A wins. B may be able to catch a pattern of what numbers A is guessing, but A also can find a pattern of what B guesses. Neither player is at a disadvantage.

[/ QUOTE ]

yes i'm aware of the symmetry, but i'm interested in the case where one suspects the other of playing non-optimally

[kitaristi0]

I could, and probably am wrong, but wasn't this exact example in some poker book? It had to do with discussing game theory. Maybe it was some other book I've read.