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Tic-Tac-Toe Toss Tactics? (Posted on 2004-10-04) Difficulty: 3 of 5
Two expert and jaded tic-tac-toe players, after drawing for the n-th time, decided to add some randomness to their favorite game.

First, they used a coin to decide who would start. Then, that player would pick his initial move randomly. Next, the other player would also pick his answer randomly. Finally, from then on the game went on as usual, with each player playing in the best possible way.

For each player, what are the odds of winning, losing, or drawing?

  Submitted by Old Original Oskar!    
Rating: 3.6667 (3 votes)
Solution: (Hide)
If the first player starts in a corner (4/9 odds) and the second player doesn't answer in the center (7/8 odds), the first one will win; otherwise, it will be a draw.

If the first player starts in the center (1/9 odds) and the second doesn't answer in a corner (4/8 odds) the first will win; otherwise, it will be a draw.

Finally, if the first player starts in one of the four other possible squares (4/9 odds), if the second plays next to his (3/8 odds) or opposite (1/8 odds) it will be a draw; otherwise, the first player will win.

Thus, the odds of the first player winning are 4/9x7/8 + 1/9x4/8 + 4/9x4/8 = 2/3, and the odds of a draw are 1/3.

If a player goes first (1/2 odds) he will win with 2/3 odds, and tie with 1/3 odds. If he goes second (1/2 odds) he will tie with 1/3 odds, and lose with 2/3 odds. So, the odds of winning are 1/2x2/3=1/3; tying, 1/2x1/3+1/2x1/3=1/3, and losing, logically also 1/3.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionI finally wrote my tic-tac-toe playing gamePenny2004-10-27 11:37:39
Some Thoughtsre(5): Solutione.g.2004-10-05 21:22:41
re(4): SolutionPenny2004-10-05 16:06:02
re(3): SolutionCharlie2004-10-05 13:46:37
re(3): SolutionCharlie2004-10-05 13:35:34
re(2): SolutionPenny2004-10-05 11:45:51
Solutionre: SolutionCharlie2004-10-05 09:52:21
SolutionSolutionPenny2004-10-05 09:06:49
Solutionre: Here's a tryFederico Kereki2004-10-04 12:41:13
SolutionHere's a tryRajal2004-10-04 12:33:00
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