Something about this game seems strangely familiar

Here's a small puzzle with an ingenious solution:

Consider the set of numbers {-4, -3, -2, -1, 0, 1, 2, 3, 4}. Two players alternately choose one number at a time from the set (without replacement).

The first player who obtains any three out of his or her selected numbers (this may happen after (s)he has chosen 3, 4 or even 5 numbers) that sum to zero wins the game.

Now, the question is that, does either player have a forced win? That is to say, can either player always choose in a way such that a win is guaranteed?

note: this problem is not original. I found the game on this blog .
http://recreational-math.blogspot.in

#CombinatorialGames #WinningStrategies #Logic #GameTheory #Logicalreasoning

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

Your game is analogous to Tic Tac Toe. So no, there is no forced win.

+3	-4	+1
-2	0	+2
-1	+4	- 3

Siddhartha Srivastava - 6 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$