12
$\begingroup$

Nine cards numbered 1 through 9 are kept facing up on a table. Alice and Bob are playing a game wherein they pick up cards one at a time, alternatively. The first person to have three cards with sum 15 wins the game.

Further, it is upto Bob to choose who picks the first card. Can Bob ensure a win?

Note that

(7, 8) is not a winning hand, because two cards have a sum 15
(1,2,3,9) is not a winning hand because four cards have sum 15
(2, 6, 3, 7) is a winning hand because three cards(2, 6, 7) have sum 15

$\endgroup$
  • 1
    $\begingroup$ I swear I've seen this somewhere on the site before. $\endgroup$ – Joe Z. Dec 18 '15 at 8:12
  • 3
    $\begingroup$ (or it could just have been my post on PPCG of the same puzzle.) $\endgroup$ – Joe Z. Dec 18 '15 at 8:13
18
$\begingroup$

This game is actually...

Tic-Tac-Toe (or Noughts and Crosses).
4 9 2
3 5 7
8 1 6
If the numbers are arranged in this 3x3 magic square, all rows, columns, and diagonals (and ONLY those) are sets of 3 that sum to 15.

Therefore,

Bob cannot ensure a win.

$\endgroup$
  • $\begingroup$ That was quick. $\endgroup$ – iamwhoiam Dec 18 '15 at 5:08
  • $\begingroup$ @Manal: I've been told that before on many of my answers. ;) $\endgroup$ – Deusovi Dec 18 '15 at 5:08
  • $\begingroup$ That's a "weird" logical jump... How did you determine what the answer "actually" is? Or is this one of those "I know the trick to that..." that you heard somewhere else? $\endgroup$ – WernerCD Dec 18 '15 at 13:49
  • $\begingroup$ @WernerCD: When I say "actually", I mean that it's isomorphic - put simply, the games have the same structure. The solution to the more common game in the spoiler block applies to the card game. You could prove the solution using the same tactic but in terms of the cards (without any reference to the simpler game), but it would be more complicated. $\endgroup$ – Deusovi Dec 18 '15 at 13:53
  • $\begingroup$ I'd like to see another solution for this game in order to be able to use it to show that there is no winning strategy for tic-tac-toe! $\endgroup$ – JiK Dec 18 '15 at 13:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.