I did look at some of the infinite chess board questions and my question is kind of different. Say, two players are playing on a infinite chess board ( Say on an infinite Montglane Service :-)). There is a third player who actually makes the moves.
1) The third player does not know the game so far and has no way of knowing who
is in an advantageous position or not.
2) He is given a set of $k_{n,black}$ or $k_{n,white}$moves for the current player's $n^{th}$turn (Black or White) out of which he has to pick one move to play.
3) He can survey the game from his start to his current move and decide who is in an advantageous position at a given state of the Game.
The question is can he, bias it in such a way whether he makes the move so Black or White can win? Call the favored player $c$, and the other player $d$
My thought is, yes, if he can study the game so far and can foresee,say, the last 10 moves of the end Game before a check mate,
- Then he can either not make winning move for White or Black and extend the
game, but now the Game continues on and he would no longer know the new
Game's winner yet! - Or he can move the standard set of moves that lead up to his biased Winner.
assuming it is a forced checkmate ( which means the set of moves that he
thought of is in the given set).
Now,
For the last 10 moves of the game, say the third player did not foresee, then what would be the probability that he makes the last 10 moves correctly?
Probability of picking the correct move at $n$ is $P(n,c)=\frac{1}{k_{n,c}}$.
Probability of picking the in-correct move at $P(n, d)$$= 1 – \frac{1}{k_{n,d}}$.
Since, he has to do all the moves up-to the last move, the probability of winning becomes $P(1,c)*P(1,d)*P(2,c)*P(2,d)….P(10,c)$.
Is there a general strategy that maximizes his probability of winning for $c$?
Or, am I too way off mark…? Thanks!
Best Answer
I have been thinking about this and arriving at the following strategy.. The idea is to keep increasing the probability of Winning for every move for $c$, say and $f$ foreseeable moves... where $g = min(f, max(k_{n,black}, k_{n, white}))$ and play it in such a way such that he chooses a move such that $P(f) = P(1,c)*P(1,d)*...P(f,c)$ is maximum for the set $(k_{n,c}, k_{n,d})$ whether $c$ chooses black or white. This ensures that the move to choose for e.g for $d$ is such that the $k_{n+1,d}$ moves for the $n+1^{th}$ move always supports an increase of $P(f)$. Like, for e.g the third player chooses a move that $k_{n+1, c}$ keeps decreasing for e.g such that $\frac{1}{k_{n+1,c}}$ increases at the same time $k_{n+1, d}$ will keep increasing. So, it my seem that $c$ is being cornered to make fewer number of moves when in-fact $c$'s probability of winning increases...Appreciate your thoughts...