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Bad news from Belgium

Posted By: Rich Munitz
Date: Saturday, 9 October 2010, at 3:12 a.m.

In Response To: Bad news from Belgium (Steve Mellen)

"If the second player chooses randomly, why is it impossible for him to win?"

The skillful player will choose 1.0.

What is the probability that a randomly chosen real number in [0,1] will be 1.0? Zero.

A finitely progressive game is one in which the rules are such that the game must eventually end. Othello is an obvious example, as all games must end in 60 moves or less. Chess is also finitely progressive as it has clear rules preventing repeating positions and the number of possible positions are finite. Backgammon is not finitely progressive.

In any finitely progressive game, there is a non-zero probability that random play will choose perfect play. In fact, that remains true for any arbitrarily long but finite series of such games.

The only question then is whether perfect play is sufficient to win. For example, a finitely progressive game may produce a tie with perfect play by both sides, or may result in a win for whoever moves first or second. Also, games can be rigged in favor of one player via handicaps. I would wager that a chess game in which random-player spots skilled player a queen, there is a zero probability that random-player can avoid losing the game no matter who moves first. But certainly we can say that given a fixed initial game setup, if the player to move first is chosen randomly, the probability of random-player losing will be non-zero in any finitely progressive game.

So what about games that are not finitely progressive? Go lacks the strong repeat position rule of chess (it prohibits immediate repeats, but not delayed ones) and is therefore not finitely progressive. A configuration called triple ko can arise in which it is legally possible to repeat a sequence of moves indefinitely. Go therefore provides an interesting dilemma for our random player. If a triple ko threat arises in an otherwise undecided position, perfect play by both sides may require that the kos be cycled through indefinitely, because otherwise any player that concedes the ko fight and breaks the cycle will lose. What is the probability that the random player will infinitely repeat a sequence of moves? Zero. Therefore in such a situation, random play is not capable of perfect play. So imagine that we create a new game that uses the rules of Go, but the starting position contains such a triple ko condition. In that game, random-player has a 100% probability of losing to skilled-player.

Lets construct a simpler game based on this principle. Rock-paper-scissors. Except with slightly different rules. Players alternate turns so that you get to choose your move knowing what your opponent just put down. You must win by two. Player to go first chosen randomly. If my opponent chooses "rock", I choose paper (1 for me). Now my opponent knows that I chose "paper" and so he chooses "scissors" for his next move (1 for him). The next correct reply to "scissors" is "rock". Perfect play is easy in this game. Probability of random-player winning this game? Zero.

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