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@iusemybrain@sh.itjust.works

Post #3386562

2026-06-15 20:53 UTC

hold on there, it actually does have strategic advantage from a statistical perspective. the basic notion is that for a probability of event A to occur it is proportional to the area of the event; so if larger area, larger probability, smaller area, smaller probability. if we take that idea and apply the same basis to battleship you could say the probability as the sum of each probability of each point which is 0℅ if we span the area to infinity. practically speaking, this is not true as you can’t span to an infinite scale, but you could say that the probability of hitting a point is 1℅ since battleship is a 10 x 10 grid so the probability is just 1/(10 * 10) = 0.01. Then the probability gets more complicated since you are being asked what is the probability of the second, third, fourth, etc… point being hit given that initial probability. the probability grows dependent on the first point being hit. I’m sure there is a way to find algorithmically an optimal method to finding in what location are the best positions in battleship, but generally speaking, no, there are worse conditions that have a higher probability of being guessed

Replies (1)

  • @binarytobis@lemmy.world 2026-06-16 02:05

    There’s a lower chance of getting hit on turn 1, but it takes more turns to sink a five space ship and all of the others than just one five space ship. The goal is to last more turns with at least one boat, not avoid getting hit for the longest. I don’t see the advantage. Unless you add an extra rule requiring you to shoot the same space multiple times.

    Open ##3386561