identical Battleship boards odds
Hello, numbers gods.
I need someone to calculate the odds of two people (in this scenario they are drunk, though I don't know that that has any influence) setting up -exactly- identical Battleship game boards with no foreknowledge of their opponent's tactics. And you know, obviously without cheating.
This really happened to me back in 2006. Once I figured it out I won, but the real winning can happen today with your help! Please explain your answer. I'd figure all this out myself if I a) had the game to go from and b) more than an elementary grasp of mathematics.
Please and thank you.
-Adeleine
I need someone to calculate the odds of two people (in this scenario they are drunk, though I don't know that that has any influence) setting up -exactly- identical Battleship game boards with no foreknowledge of their opponent's tactics. And you know, obviously without cheating.
This really happened to me back in 2006. Once I figured it out I won, but the real winning can happen today with your help! Please explain your answer. I'd figure all this out myself if I a) had the game to go from and b) more than an elementary grasp of mathematics.
Please and thank you.
-Adeleine
Last edited on
Just figure out how many possible ways you can lay out the board. Then from there you can calculate the possibility of getting a particular layout, then the possibility of a particular layout twice in a row.
Divide 1 by the number of possible combinations plus any invalid combinations that can be created as the users are drunk. Ultimately there are too many variables to consider. If they were drunk enough they could have been influenced by an outside factor to make them the same for a laugh.
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