🚀 go-pugleaf

In article <23738@alice.att.com> ark@alice.UUCP () writes:
>In article <1992Sep27.161451.20136@cbnewsl.cb.att.com> ed@cbnewsl.cb.att.com (edward.m.hummel) writes:

>> Consider the game 1. Nf3 Nf6 2. Ng1 Ng8 3.Nf3 Nf6 4.Ng1 Ng8 5.Nf3 Nf6
>> and so on, forever; with neither player choosing to claim a draw.  I 
>> am asking for a proof that when collapsing the game tree to determine the
>> value of the initial position, this sequence must have the value of a draw.

>We can solve the problem in at least two ways.

>1. Define it away by saying that any position from which neither player can
>   force a win is a draw.

>2. Ask why, in the situation you mention, neither player claims the draw after
>   the third repetition.  Evidently the answer must be that they both think
>   that playing indefinitely is preferable to ending the game with a draw.
>   But in that case, the game is not zero-sum, which contradicts one of the
>   conditions necessary for the theorem to apply in the first place.

I think this is precisely Mr. Hummel's point.  Win, loss, and draw
are all defined, but it is possible that a nonterminating game has
value greater than a draw for both players.  Clearly that's a nonzero-
sum game.  But identifying and exploiting positive-sum games is a
major activity of a good life.  And it seems the rules of chess
assign no value to unfinished games.

Two points of my own:

1. If an unfinished game has a value for both players greater than
   that of a draw, a draw is not a normal outcome.  In fact, constructing
   a position where both players have to settle for a draw is nontrivial.
   Many games are drawn because neither player has any winning chances.
   Clearly real players do *not* value a continuing game more than a
   draw, although that isn't written into the rules.

2. Obviously this has no effect on chess being solvable.

-- Don Varvel (varvel@cs.utexas.edu)