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mathgrant
A very tilted cell member
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 Posted: Mon Nov 07, 2011 4:53 pm Post subject: 1 |
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Inspired by the Visitor Games thread.
City Square Off is a game published by Gamewright. In the game, each player gets a 9x9 board with a hexomino-shaped city tile in the middle:
| Code: |
1 2 3 4
......... ......... ......... .........
......... ......... ......... .........
......... ......... ......... .........
....C.... ....CC... ....C.... ....C....
...CC.... ...CC.... ...CCC... ...CC....
...CCC... ...CC.... ...CC.... ...CC....
......... ......... ......... ...C.....
......... ......... ......... .........
......... ......... ......... ......... |
Each player has an identical set of 21 polyomino-shaped tiles (1 monomino, 1 domino, 2 distinct triominoes, 5 distinct tetrominoes, and 12 distinct pentominoes). Each turn, one of these 21 tiles is selected at random from a deck of cards, and each player must place that tile (rotated or reflected as desired) in a position that shares an edge with at least one other tile. If the tile cannot be placed on a particular player's board, the player is out of the game. Whoever can play the most tiles wins. Ties are broken by whoever has the largest contiguous group of empty spaces (so if one player has empty spaces distributed 6-1-1, and the other has them distributed 5-3, the 6-1-1 wins).
Question: is it possible to set up a legal board position where player A has over a 50% chance of beating player B, player B has over a 50% chance of beating player C, and player C has over a 50% chance of beating player A? Here's an example that isn't legal because the combined number of squares on the remaining pieces isn't the number of empty spaced plus 14:
| Code: |
AAA BBBB C
A A B CCCC
A A B C C
B
Pieces:
3
1 2 2 3
1111 222 333 |
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Zag
Tired of his old title
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| Posted: Mon Nov 07, 2011 10:30 pm Post subject: 2 |
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| mathgrant wrote: |
| Question: is it possible to set up a legal board position where player A has over a 50% chance of beating player B, player B has over a 50% chance of beating player C, and player C has over a 50% chance of beating player A? |
Sorry, I haven't actually read the rules, but wouldn't this usually be the case? In most 3-player games of this sort, the player who follows the first one to drop out has a big advantage. So, assuming they play A-B-C-A..., and assuming that they are equally skilled such that which player is the first to drop out is even, then
if A drops out first, B is more likely to beat C
if B drops out first, C is more likely to beat A
if C drops out first, A is more likely to beat B
Let me define
a = A's chance to win (given conditions)
b = B's chance to win (given conditions)
c = C's chance to win (given conditions)
(X) means X drops out first
We can extend the first line trivially
if (A), then b > c. if ! (A) b = c [because ! (A) implies either (B) or (C), which are equal probability]
if (B), then c > a. if ! (B) c = a
if (C), then a > b. if ! (C) a = b
Therefore, A is overall more likely to beat B, B is overall more likely to beat C, and C is overall more likely to beat A |
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mathgrant
A very tilted cell member
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| Posted: Tue Nov 08, 2011 12:56 am Post subject: 3 |
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| Zag wrote: |
| Sorry, I haven't actually read the rules. . . |
If you had, then you'd know that this is essentially a multiplayer solitaire where the players never directly interact with one another. Also, players play simultaneously; there is no turn order. Your argument about the "turn order" of the players who haven't been knocked out is irrelevant.
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Zag
Tired of his old title
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| Posted: Tue Nov 08, 2011 2:25 am Post subject: 4 |
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| LOL. Never mind, then. |
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