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who pays for the drinks?



A simple game to determine who picks up the tab involves drawing six vertical lines on a sheet of paper.

We can call the lines A B C D E and F. P 11.11 / 2/3.4/5=.53+ 40

Holding the paper out of sight of Player 2, the first player joins pairs of upper ends of the lines.

He then folds the paper in half, to expose only the lower ends and passes the paper to Player 2.

Player 2 joins pairs of lower ends, hoping to make the six lines form a single closed loop.

Here is a configuration that wins for the second player.

Top ends Fold Bottom ends

A +----------------------------|--------------------------------+

B | +-----------------------|----------------------------+ |

C | +-|-----------------------|----------------------------+ |

D | | +-----------------------|--------------------------+ |

E | +-+-----------------------|--------------------------|-----+

F +----------------------------|--------------------------+

  1. Is this a fair game? If not, which player has the advantage?
  2. As the number of lines (always an even number) increases, the chances of a closed loop decrease.
    How many lines must be drawn for Player 2's probability of winning to be less than 0.2?
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I don't understand your example. It seems to me that

C is joined to E is joined to A is joined to F is joined to D is joined to B,

but B and C aren't joined. So, it seems that player 2 didn't win because

a loop wasn't made! Can you clear up my misunderstanding?

On top, the joined pairs are AF BD and CE

The bottom pairs are AE DF and BC.

There is not room for a vertical stroke between the adjacent lines B and C, it's implied.

Sorry, a limitation of the typed characters.

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