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John and Jane are playing a game. They both go to the bank and get 1000 euros in 2 euros. So each of them has 500 euros at stake. They then find a large rectangle table and decide who wins the following game gets to keep all of the coins currently at the table. The rules are as follows:

  • It is a turn-based game, so John and Jane take turns alternatively (1 coin per turn)
  • The coins are placed on the table and they cannot be put on top of each other.
  • Coins may not be moved once put on the table
  • Assume the table isn't large enough to hold all the coins
  • If there isn't any room to put coins on the table, the person who placed the last possible coin wins.

What is the winning game strategy?

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  • $\begingroup$ How many coins can they place per turn? $\endgroup$ – Stormenet Oct 6 '14 at 11:47
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    $\begingroup$ without knowing the size of the table it is impossible to know $\endgroup$ – Elgert Oct 6 '14 at 12:16
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    $\begingroup$ There also seems to be a vocabulary issue here: "Get 1000 euros in 2 euros, so each has 500 euros at state". I don't know the terminology, but this problem seems simplifiable to "500 coins of equal size and value". $\endgroup$ – Mooing Duck Oct 6 '14 at 19:07
  • $\begingroup$ What if the table is sufficiently large to place all coins? Does player 2, as the last player to place a coin, win the challenge? What if the table is non-symmetric? What if gravity fails and the coins don't stay on the table? We need answers! $\endgroup$ – FreeAsInBeer Oct 6 '14 at 19:22
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The winning strategy for the first player is to put their coin in the dead center of the table. Then whatever move their opponent makes, they exactly mirror it, around the center.

e.g. If the second player puts their first coin 1 inch to the left of the center coin, the first player mirrors this by putting their coin 1 inch to the right of the center coin.

It's relatively easy to see why this works. If a spot is free for player 2, then its mirroring spot must also be free for player 1 because after each turn player 1 takes, there will be no unmirrored spots left. This means that wherever player 2 goes, there's guaranteed to be a spot left for player 1. So the only person who can possibly reach a state where there's no spot to go is player 2, who inevitably loses (unless they didn't bring enough coins to cover their table!)

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  • $\begingroup$ gratz! that was fast. +1 $\endgroup$ – Rafe Oct 6 '14 at 12:16
  • $\begingroup$ Exactly! Great job! $\endgroup$ – Kevin Oct 6 '14 at 12:39
  • $\begingroup$ What about the second player (lets just assume that the first player is not aware of the winning strategy for player 1) $\endgroup$ – Kristoffer Sall-Storgaard Oct 6 '14 at 13:45
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    $\begingroup$ @KristofferSHansen: I am not aware of a proof of who wins if the first player does not play at the center. I suspect it depends on the dimensions of the table (measured in coin diameters). $\endgroup$ – Ross Millikan Oct 6 '14 at 14:36
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    $\begingroup$ @Stacey If I could win significant amount of money all I need is a measuring tape and I would measure it all out. $\endgroup$ – Kevin Oct 7 '14 at 15:07

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