Rand and Rubio are playing a game in which they each take turns to pick a digit between 1 and 9, without replacement (i.e. all digits chosen are distinct). If one of them manages to get three digits which sum to 15, then he wins. If neither player achieves this, they both lose.

Rand goes first. Does he have a winning strategy? If so, what is it? If not, prove it.

Source: Peter Winkler's Mathematical Puzzles: A Connoisseur's Collection.

  • 2
    $\begingroup$ Yes, he said "you take 2, 6 and 7, and I'll take 1, 5, and 9. No one said first to 15 wins so we'll both win this way" $\endgroup$ Jan 18 '17 at 2:59
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    $\begingroup$ Of course Rand wins - he wouldn't be stupid enough to play against Rubio if he didn't already have a winning strategy worked out! $\endgroup$
    – YowE3K
    Jan 18 '17 at 3:40
  • $\begingroup$ Does the game end as soon as one of the players get 15? Apparently it doesn't, but still. $\endgroup$
    – Nautilus
    Jan 19 '17 at 19:25




This game is equivalent to playing noughts and crosses on a 3 x 3 magic square. With perfect play, noughts and crosses is a draw, so the first player does not have a winning strategy.


Exactly eight triplets of different digits between 1 and 9 sum to 15:

1 5 9
1 6 8
2 4 9
2 5 8
2 6 7
3 4 8
3 5 7
4 5 6


Exactly eight lines of three numbers - horizontal, vertical, or diagonal - appear on any 3 x 3 magic square. No two are the same.


If the magic square contains all of the digits between 1 and 9, each line contains a different one of the above triplets and all triplets appear. This proves equivalence.


The solution:


The reason:

They're really playing tic-tac-toe on this grid! (All possible combinations with a sum of 15 are contained in it.)

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    $\begingroup$ Are "not losing" and "winning" the same thing? There are plenty of games where neither wins; is it required that Rand have a strategy that can always beat me, or does never losing to me suffice? $\endgroup$
    – Rubio
    Jan 18 '17 at 1:12
  • $\begingroup$ @Rubio Winning means actually winning. Edited to clarify. $\endgroup$ Jan 18 '17 at 1:17
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    $\begingroup$ Hmmm... If this is the case, the only way to win is not to play.... $\endgroup$
    – user30599
    Jan 18 '17 at 8:04
  • $\begingroup$ I'm not sure if this answer came before @TheEarth's one, but this one is way simpler to understand. Nice! $\endgroup$
    – Xenocacia
    Jan 18 '17 at 9:34
  • $\begingroup$ @Xenocacia This answer was first, but both are good answers and deserve upvotes. $\endgroup$ Jan 18 '17 at 12:18

Simple answer.

The only way to win for sure would be to have a combination of numbers than can withstand 2 blocks from the opponent. Which would require a number which can branch off into at least 2 winning combinations which themselves can also branch off into 2 winning combinations.
Example : 1-2-6, 1-2-7, 1-3-8, 1-3-9
But it is impossible to have a combination of 2 numbers to branch of into 2 possibilities that equals 15.
No matter what, the second player can block you on his second move.
Therefore it is impossible to win for sure.


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