harey 8 Posted March 2, 2018 Report Share 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 Link to post Share on other sites

1 Solution ThunderCloud 5 Posted March 2, 2018 Solution Report Share 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 Link to post Share on other sites

0 Donald Cartmill 3 Posted March 4, 2018 Report Share 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 Link to post Share on other sites

0 ThunderCloud 5 Posted March 4, 2018 Report Share 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 Link to post Share on other sites

## Question

## harey 8

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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