As stated in the problem, the game of fifteen is equivalent to random tic-tac-toe except in the winning conditions. The transformation is equivalent to numbering the tic-tac-toe board as a magic square. There are eight combinations of three numbers from the set one through nine which add to 15, precisely matching the lines in the magic square and the tic-tac-toe game. While the games actually end as soon as somebody gets a winning triple, if carried out to fill the board, there would be 9! = 362880 possible games. Games that end before the eighth move are more likely end states than others, since there is more than one of the 9! full-board games corresponding to this position. However, the final answer will be in the form N/9!, where N is an integer. I will analyze the results by looking at the full-board tic-tac-toe games, and determining the number of Fifteen games which lead to each, including (for those final states which have winning triples for both players) how many games lead to wins for each player. For any given final board, such as: BAA AAB BBA There are 5! ways of arranging A's moves, and 4! independent ways of arranging B's moves. Thus, there are 5!*4! = 120*24 = 2880 games of Fifteen which correspond to this filled tic-tac-toe board. In addition, there are up to 8 rotations and reflections of each board, which multiply this number further (to 23040 games for the board above). There are 3 full tic-tac-toe boards (distinct to rotations and reflections) which contain no winning line for either player. These are: BAA AAB (8 rot/ref) BBA ABA BBA (4 rot/ref) AAB BAB ABA (4 rot/ref) ABA These give 16*2880 = 46080 games of Fifteen in which neither player gets a winning triple, and as a result B wins. There are several full tic-tac-toe boards where only one player gets a winning line. These are wins for A: AAA BBA (4) BBA AAA BAB (4) BAB BAB AAA (1) BAB AAA BAB (8) BBA AAB BAA (4) BBA AAB BAB (4) BAA AAB AAB (4) BBA AAB BAB (8) ABA ABA BAB (1) ABA AAA ABB (8) BAB AAA ABB (8) BBA ABB AAA (8) BAB (Total 62) And these for B: BAA ABA (4) BAB BAA BBA (8) AAB (Total 12) This gives 62*2880 = 178560 games of Fifteen that win for A, and 12*2880 = 34560 games for B. These filled tic-tac-toe boards contain winning lines for both players. There can only be one line for each player, since two crossing lines for A, in any configuration, prevent B from getting a line, and B doesn't get enough plays to make two lines. AAA BBB (8) AAB AAA BBB (4) ABA AAA BAA (8) BBB AAA ABA (4) BBB BAA AAA (8) BBB ABA AAA (4) BBB (Total 36) This gives 36*2880 = 103680 games that may lead to a win for either A or B. Each set of 2880 has the same number of A wins and B wins, since only the order that the 6 numbers in the two winning lines occur matters. When the first three moves for A are the winning line (12/120), A wins. When the fifth move for A is part of the winning line, B wins (72/120). When the last of the winning line is A's 4th move (24/120), A wins unless B's winning line occurs in his first three moves (6/24). (So A wins this way 18/120 times and B wins this way 6/120 times.) Thus, A's overall chances of winning these cases is 30/120, or 1/4. So, A wins 25920 of these cases and B wins 77760 of them. This gives us a total of 178560 + 25920 = 204480 cases where A wins, and 46080 + 34560 + 77760 = 158400 of the 362880 total cases. Bet on A. Joseph DeVincentis ---------------------------------------------------------- Hi, Ed. Concerning the problem proposed by Sudipta Das, it seems that, among the 255168 possible matches: - 131184 times the match stops because A getts three balls totalling 15; - 77904 times the match stops because B getss three balls totalling 15; - 46080 times no one gets a winning triple. Thus player A wins 51.41% Ciao, Claudio Baiocchi.