Nine cards numbered 1 to 9 lie face up; players alternately take one, and the first to hold three cards...

competition/games/fifteen.game New

New entry, edited from material collected for the archive and not yet in a released version.

Problem

Nine cards numbered 1 to 9 lie face up; players alternately take one,
and the first to hold three cards adding up to exactly 15 wins.

A certain game is played as follows:

9 cards are placed face up on the table.

Players take turns removing a card from the table, and adding it to
their hand.

The first player that has three cards in their hand that add up to
exactly 15 wins the game.  (That player may have more than three cards
in their hand at the time.)

Assuming perfect play from both players, who will win?

(Puzzle from one of Martin Gardner's books.)

[ Editor: the first paragraph is the editor's summary; the rest is the
  original posting by Ian Woollard. ]

Show solution

In the following, I assume that the cards are numbered from 1 through 9:

1.  Notice that the digits from 1 through 9 can be arranged in a magic
    square where all the horizontal, vertical, and diagonal lines sum
    to 15.

2.  Notice also that, for at least one such square, *all* sets of
    three cards that sum to 15 are in a straight line.

3.  Notice that picking cards that sum to 15 is equivalent to playing
    the game of tic-tac-toe on the grid.  Players alternate picking
    cards.  Cards are never returned.  Getting three in a row wins.

4.  Recall that the player that plays first in tic-tac-toe can never
    lose, but may not win.

Therefore, playing first is best.

This solution comes from the mind of Michael Golden (thanks).

-- Waleed Meleis <waleed@eecs.umich.edu>

[ Editor: the posting says who should play first rather than who will
  win; by its point 4 the first player at least cannot lose. ]