harey 7 Report post Posted March 2, 2018 There are 9 cards on the table, numbered 1 to 9, face up. You and I alternatively pick one card. The first one who can produce exactly 3 cards summing up exactly 15 wins. What is your strategy? Quote Share this post Link to post Share on other sites
1 ThunderCloud 5 Report post Posted March 2, 2018 In essence... Spoiler I'd play Tic-Tac-Toe on a magic square: 2 7 6 9 5 1 4 3 8 If I get to go first, I'll start with the "middle square" and choose 5. Next I'll choose a corner -- 2, 4, 6, or 8 -- depending on what you pick. If you chose one of the edges, I choose a corner adjacent to it; if you chose a corner, I choose a neighboring corner. Presumably, your next choice will be forced so as to prevent me from scoring 15 on a diagonal, after which I can guarantee a win for myself by choosing another corner number, thus leaving me with two ways to win on my next turn. Quote Share this post Link to post Share on other sites
0 Donald Cartmill 2 Report post Posted March 4, 2018 I'd play Tic-Tac-Toe on a magic square: 2 7 6 9 5 1 4 3 8 If I get to go first, I'll start with the "middle square" and choose 5. Next I'll choose a corner -- 2, 4, 6, or 8 -- depending on what you pick. If you chose one of the edges, I choose a corner adjacent to it; if you chose a corner, I choose a neighboring corner. Presumably, your next choice will be forced so as to prevent me from scoring 15 on a diagonal, after which I can guarantee a win for myself by choosing another corner number, thus leaving me with two ways to win on my next turn. Your solution is invalid. Your opponent chooses a 4; you choose a 2 or an 8; opponent chooses either 2 or 8; You are now forced ,you cannot take the remaining corner or your opponent score a 15 across the bottom ;if your 2nd was a 2; Therefore you must go a 3 to block ; he then must go 7 to block ....No winner Quote Share this post Link to post Share on other sites
0 ThunderCloud 5 Report post Posted March 4, 2018 8 hours ago, Donald Cartmill said: I'd play Tic-Tac-Toe on a magic square: 2 7 6 9 5 1 4 3 8 If I get to go first, I'll start with the "middle square" and choose 5. Next I'll choose a corner -- 2, 4, 6, or 8 -- depending on what you pick. If you chose one of the edges, I choose a corner adjacent to it; if you chose a corner, I choose a neighboring corner. Presumably, your next choice will be forced so as to prevent me from scoring 15 on a diagonal, after which I can guarantee a win for myself by choosing another corner number, thus leaving me with two ways to win on my next turn. Your solution is invalid. Your opponent chooses a 4; you choose a 2 or an 8; opponent chooses either 2 or 8; You are now forced ,you cannot take the remaining corner or your opponent score a 15 across the bottom ;if your 2nd was a 2; Therefore you must go a 3 to block ; he then must go 7 to block ....No winner Indeed, if both players play the ideal strategy for Tic-Tac-Toe, the game always results in a draw. It is nonetheless the ideal strategy. Quote Share this post Link to post Share on other sites
There are 9 cards on the table, numbered 1 to 9, face up. You and I alternatively pick one card.
The first one who can produce exactly 3 cards summing up exactly 15 wins.
What is your strategy?
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