Bar game: place the last coaster
In honor of St. Patrick’s Day, here’s a classic puzzle that works well as a bar game.
The only thing you need is enough beer coasters to cover a table. You can ask a server for them or you can bring some from home (beer coasters are cheap
The rules
Here is how the game works:
–Someone goes first and places a coaster anywhere on the table
–The other person goes by placing a coaster anywhere else that’s open on the table
–The game continues with each player moving in turn to place a coaster on the table
–The winner of the game is the person who puts down the last coaster, i.e., there is no more open space on the table
To make it interesting, you can play with a rule that the loser has to buy the next round.
It’s a simple game, so what’s the best way to play? Is it better to go first or second? Is there a winning strategy?
I don’t think it matters if the table is round or rectangular, nor does it matter if the coaster is round or a square. My answer for the two-person game is in the comments.
Variations
The two person game is fun to analyze but perhaps that makes it less fun to repeat. Here are a few ways to spice up the game:
–Start the table with a random configuration of coasters and then flip a coin to see who goes first.
–Play this game with more than two players–the strategy quickly becomes more complicated!
–Try this game at a dinner party using small plates on a dining table.
Try this game and its variations out and let me know what you think!
(Just be careful if your opponent also reads this blog and you’re doing the two-person game)
Do you know of a cool bar game? Let me know and I’ll feature it so you can share your game.
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4 Responses to “Bar game: place the last coaster”
There is a winning strategy for the first player in a 2-person contest. Here is what to do:
Your first move is to place a coaster in the center of the table. Now, wherever your opponent places the coaster, you place yours symmetrically on the other side of the table. If they place a coaster in the southwest corner, you place yours in the analogous spot in the northeast corner. (If you imagine the center of the table as the origin, this is mathematically a reflection about the origin).
This strategy means you can always match your opponent’s move. The game ends when your opponent runs out of open sports, which equivalently means you have placed the last coaster.
By Presh Talwalkar on Mar 17, 2010
One thing that’s unclear is whether or not you can move coasters that have already been placed.
I like the solution and how it works:
Assuming optimal placement (each coaster touches another coaster) we would end up with a rectangular array of coasters. Since you placed an “anchor” at the center of the table, the array of coasters will have an odd number of rows and columns. Since an odd number times an odd number is also an odd number, there would be an odd number of rectangles. Since you went first, you will go last due to that odd number.
By Scott on Mar 17, 2010
Presh:
Your solution is an ‘involution’. I actually worked on a combinatorial game similar to this recently. It’s awesome to see this version of it! What a great way to introduce laypeople to game theory, and especially to show them how combinatorialists arrive at solutions.
If you’re interested in another form of the game check out F-Saturator, a game a few colleagues of mine worked on in the form of graphs.
http://scholar.google.com/scholar?cluster=5600507325285475104&hl=en&as_sdt=4000
By Craig on Mar 17, 2010
This assumes there’s always a matching spot. Holes in the table, or an irregular outline, would make that difficult. From the comments, I gather that once you no longer mirror, the strategy isn’t assured.
By bill on Mar 17, 2010