• restingOface@quokk.auOP
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      6 hours ago

      The grid is 10x10, so 100 spaces. Placed legally, your five ships take up 17 spaces. That means if your opponent picks a space at random, there is a 17% chance they will get a hit.

      It you cheat like this, your five ships only take up 5 spaces. That means if your opponent picks a space at random, there is a 5% chance they will get a hit.

      So, purely based on chance, this increases the odds of your opponent not beating you. You just have to make sure to also be on the offensive and sink all their battleships quickly, because if they happen to hit your battlestack just once, that means they are going to sink it all within the next 4-7 turns.

      • DrSteveBrule@mander.xyz
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        2 hours ago

        When the smallest ship (can’t remember what its called) is destroyed, you would then tell your opponent that they sunk it. It would be unlikely that they would continue to strike the next space in that row/column.

      • binarytobis@lemmy.world
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        2 hours ago

        If you had a ship on every possible space, they would have a 100% chance of landing a hit, so by your logic this would be worse. But it would take 100 turns to sink you, which is the best possible outcome.

        Having only one 5 space ship is objectively worse than having one five space ship and all the others. Unless of course you also created a new rule that they have to hit the same spot multiple times, once for each ship.

        At that point you’re just playing Calvin Ball, though, and you might as well put the ships under your chair and claim “You never said floor!”

        • restingOface@quokk.auOP
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          14 minutes ago

          If you had a ship on every possible space, they would have a 100% chance of landing a hit, so by your logic this would be worse.

          Yeah, because the challenge in Battleship is primarily the locating of the battleships. Having a 100% chance of being hit would indeed be bad.

      • iusemybrain@sh.itjust.works
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        2 hours ago

        I’ll add to this, it’s not just about finding the initial probability since the probability for event B given A, or event C given A and B, these would increase the liklihood of being found. I couldn’t tell you exactly what the probability is but say for example event B given A, it would be 1/n for n being the number of legal spaces around A. So if the ship was in the middle of the board it would have the smallest probability of being found but if you’re along a wall, there’s only 3 possible legal spaces so the probability increases to 33.3℅, and if A was sandwitched between two walls, the probability of B is 50℅.

        so if there’s a moral to this story, assert dominance and put all your pieces in the center.

      • EndlessNightmare@reddthat.com
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        5 hours ago

        Does 1 hit cause damage to all the ships in the stack that occupy the hit space? Or does only the top-most ship take the damage such that the top ships shield the bottom ones?

    • iusemybrain@sh.itjust.works
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      7 hours ago

      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

      • binarytobis@lemmy.world
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        2 hours ago

        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.

    • Allero@lemmy.today
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      7 hours ago

      Most iterations of the game assume you cannot place ships right next to each other. So, when you down a ship in a normal configuration, you immediately know the surroundings are free of ships.

      In this placement, all but these 4 cells will not give you such an advantage. Therefore, the tactic is advantageous as it keeps the opponent with less intel.