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.
What you calculated are not the chances of the opponent not beating you, you calculated the chance of getting the first hit. But it is the last ship being hit that will make you lose.
Imagine player A using a regular strategy, and B using this stacking strategy. Now imagine that B gets insanely lucky and sinks all of A’s ships except their largest one, without missing any at all (while A keeps missing all their shots). So now, even after all that insane luck from player B, who has a better chance of winning? At this point, they both have only 5 spaces which can be hit, and any hit will quickly end the game, so they both have 50% chance to win right?
This means that clearly if B doesn’t get insanely lucky in the first turns of the game, then A will have much better chances to win. A’s chance to win reduces to it’s absolute minimum at 50% when it has only one single 5-length ship left.
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!”
You know what, i’m in :)
On my field, i’ll have a ship on every single square, so no matter where you aim, you score a hit :)
And on your field, you just have 1 ship of 5 long, so i have only a 5% chance to hit :).
Do remember though, the game ends when all our ships are sunk.
So you will have to do 100 shots every round to win (covering every single square)
While i just have to find your ship (with some basic strategies, i’d estimate not even needing 20 shot to do that), and then i win :).
…
starting to get it? XD
you’re focusing so hard on how easy it is to hit a ship, that you forget the goal of the game: sink every ship.
If my entire playfield is ships, you’ll indeed hit with every single shots, but you’ll also need the maximum of 100 shots every game to ‘win’ XD.
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.
According to the rules (which also state you can’t stack ships. Lol) you’re supposed to state which ship was hit.
This means that you would have to say the names of all the ships that just got hit. Also, if you were trying to skirt that rule by not saying the names of all the ships underneath, and only saying the name of the top ship, the odds that both if your opponents first guesses that struck would hit the 2 spot ship are pretty low.
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?
By marking design only allowing one peg to mark places you’ve called out, it would strongly be implied that it would only make sense for the hit to happen on all the ships under the spot.
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.
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.
What you calculated are not the chances of the opponent not beating you, you calculated the chance of getting the first hit. But it is the last ship being hit that will make you lose.
Imagine player A using a regular strategy, and B using this stacking strategy. Now imagine that B gets insanely lucky and sinks all of A’s ships except their largest one, without missing any at all (while A keeps missing all their shots). So now, even after all that insane luck from player B, who has a better chance of winning? At this point, they both have only 5 spaces which can be hit, and any hit will quickly end the game, so they both have 50% chance to win right?
This means that clearly if B doesn’t get insanely lucky in the first turns of the game, then A will have much better chances to win. A’s chance to win reduces to it’s absolute minimum at 50% when it has only one single 5-length ship left.
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!”
Yeah, because the challenge in Battleship is primarily the locating of the battleships. Having a 100% chance of being hit would indeed be bad.
You know what, i’m in :) On my field, i’ll have a ship on every single square, so no matter where you aim, you score a hit :) And on your field, you just have 1 ship of 5 long, so i have only a 5% chance to hit :).
Do remember though, the game ends when all our ships are sunk.
So you will have to do 100 shots every round to win (covering every single square) While i just have to find your ship (with some basic strategies, i’d estimate not even needing 20 shot to do that), and then i win :).
… starting to get it? XD
you’re focusing so hard on how easy it is to hit a ship, that you forget the goal of the game: sink every ship.
If my entire playfield is ships, you’ll indeed hit with every single shots, but you’ll also need the maximum of 100 shots every game to ‘win’ XD.
Lol. I’m starting to wonder if you ever really played Battleship. You sound like you’d be terrible at the game.
How so? Am I wrong?
On a very fundamental and obvious level. It’s why you have all those down votes.
Can’t expect users on the shitpost community to understand statistics.
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.
According to the rules (which also state you can’t stack ships. Lol) you’re supposed to state which ship was hit.
This means that you would have to say the names of all the ships that just got hit. Also, if you were trying to skirt that rule by not saying the names of all the ships underneath, and only saying the name of the top ship, the odds that both if your opponents first guesses that struck would hit the 2 spot ship are pretty low.
TIL I’ve never played by the rules.
Haha. So just like Monopoly and Uno.
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?
Depends on how hard you want to cheat/ragebait your opponent
By marking design only allowing one peg to mark places you’ve called out, it would strongly be implied that it would only make sense for the hit to happen on all the ships under the spot.
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.