Battleship and Configuration probabilities

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SUMMARY

The discussion focuses on calculating the probabilities of different ship configurations in the game Battleship, which utilizes a 10x10 grid. Participants outline the ship placements: one ship of length 5, one of length 4, two of length 3, and one of length 2, with configurations being either horizontal or vertical. The calculations suggest approximately 10^10 possible combinations when considering the placement of each ship and their intersections. A random simulation is proposed as a potential method for simplifying the analysis of these configurations.

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  • Understanding of combinatorial mathematics
  • Familiarity with probability theory
  • Knowledge of the game mechanics of Battleship
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Mathematicians, game designers, educators studying probabilistic reasoning, and anyone interested in combinatorial analysis in gaming contexts.

cecih
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Hi All,

I'm doing a study that looks at how children think probabilistically while playing battleship. I'm trying to figure out the probabilities of the different configurations of the ships that can exist. So for those that don't know battleship--there is a 10 x 10 grid and you have to hide 5 ships ( one of length 5, 1 of length 4, 2 of length 3, and 1 of length 2). The ships can traditionally only be placed horizontally or vertically. This leads to the following possible configurations: all 5 ships being placed horizontally, 4 ships being placed horizontally and 1 vertically, 3 ships being placed horizontally and 2 vertically, 2 ships being placed horizontally and 3 vertically, 1 ship being placed horizontally and 4 vertically, and all ships being placed vertically. Its been a while since I've done this type of math--so I'm at a bit of a loss as to where to start--and the variation in ship size as well as space constraints are throwing me off as well--any suggestions on how to approach this problem?

Many thanks in advance,

Cecilia
 
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This is a tricky one - 120 ways to place the 1x5 ship (60 horiz, 60 vert), 140 ways to place the 1x4 etc ... multiplied together and subtracting the few that intersect, this would be about 10^10 combinations altogether. I'm not sure if there's a way to simplify this to a manageable number of separate cases, so maybe a random simulation would be the way to go?
 

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