Thread: Noughts and crosses question

Hiya folks,

finding it very hard to get my head around a question, i'm wonderinf if anyone can help me, at least to get started..

two computers play idiots noughts and crosses with x going first, o second. by idiot noughts and crosses i mean that neither is trying to win as such, they're placing their x or o in a random square of the 3x3 grid. one computer wins if it gets three xs or os in a straight line, be in horizontal, vertical or diagonal. what is the probabilty of a draw?

i know in total there are 16 ways to draw (8 for x and 8 for o) but i can't work out how many games there can be. what makes it harder is all nine squares haven't got to be filled- ie the game can last a mere five moves if x gets three in a row on his third move.

Any help would be hugely appreciated

Donsie

Originally Posted by Donsie

Hiya folks,

finding it very hard to get my head around a question, i'm wonderinf if anyone can help me, at least to get started..

two computers play idiots noughts and crosses with x going first, o second. by idiot noughts and crosses i mean that neither is trying to win as such, they're placing their x or o in a random square of the 3x3 grid. one computer wins if it gets three xs or os in a straight line, be in horizontal, vertical or diagonal. what is the probabilty of a draw?

i know in total there are 16 ways to draw (8 for x and 8 for o) but i can't work out how many games there can be. what makes it harder is all nine squares haven't got to be filled- ie the game can last a mere five moves if x gets three in a row on his third move.

Actually, you don't need to know how many games there are. Indeed, when the computers play, they don't choose their game uniformly from the set of all the possible games. They choose each move, one after another. The probability of each move is the inverse number of empty squares. Note that the games that finish sooner have greater probability.

A given draw situation (5 x's, 4 o's, no line) happens if computer "x" chooses one square among the 5 x's (probability 5/9), then computer "o" choose one square among the 4 o's (probability 4/8), etc., until computer "o" choose its last square (probability 1/2), and necessarily computer "x" fills the last square (probability 1/1). This event has probability...

Just multiply by the number of draw configurations, and you're done.