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harey

Sum up 15

Question

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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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.

 

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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

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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.

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