iusemybrain

joined 1 week ago
[–] iusemybrain@sh.itjust.works 1 points 1 day ago* (last edited 1 day ago)

*jumps in dumpster

*looks inside

"oh shit there's 1 week old chicken that clearly smells like ammonia! you're a keeper."

[–] iusemybrain@sh.itjust.works 0 points 2 days ago

I guess you didn't read my last part, humans are deterministic by nature; they draw trends based on patterns. if you wanted to win with one ship just place that ship in a place that is completely counterintuitive to what they expect. Having one ship does have strategic implications, that it narrows the possibility of being hit. Combine that with counterintuitive placement of said ship and your opponent is as good a player as a fish.

so does probability play a role in battleship? yes, it removes the statistical likeliness of being hit (ignoring the premise of humans drawing generalizations by pattern recognition). It is drastically less likely to hit one ship than 5, all you need to do is place that one ship in a position that is counterintuitive to the trend to remove any biases of your opponent who has played several games already.

[–] iusemybrain@sh.itjust.works 0 points 2 days ago (2 children)

what are you going on about, that probability depends on the number of turns you get? last I checked it, no it does not depend on that.

the argument I'm trying to make is it's statistically less likely to hit that 5 point battleship (probability is 5%) compared to the 17 points in ships spread abitrarily (probability is 17℅). IT DOES NOT DEPEND ON THE NUMBER OF TURNS IT TAKES TO SINK ALL 5 BATTLESHIP'S. The probability does depend after the first event has occurred (where you've hit one of those pieces). Whether you get a hit on your first turn or the first hit in 30 turns, it does not matter. probability does not depend on the number of turns it takes.

technically speaking the probability calculated at the beginning is when the number of trials (theoretically) approach to infinity. The 1/6 probability to role a six sided die on a number (for example, lets say 6) is after doing trial and error of rolling a 6. Now whether you get the 6 on the first try or the 100th try these are approximations of the 1/6 th probability. As you do more and more trials it does eventually converge to the 1/6 th probability.

So yes, technically speaking someone could hit the standalone stacked battleship in one try, probability and statistics doesn't concern that. So if that concerns you, psychologically manipulate your openent by putting your battleship in the first row.

[–] iusemybrain@sh.itjust.works 1 points 3 days ago* (last edited 3 days 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.

[–] iusemybrain@sh.itjust.works -1 points 3 days ago* (last edited 3 days ago) (5 children)

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

[–] iusemybrain@sh.itjust.works 1 points 6 days ago* (last edited 6 days ago)

here's another one if you use chatGPT to solve the first one. Anyone who is fully optimized should be able to solve analytically shrodingers second order partial differential equation.

start with those problems, then we'll talk if you are fully optimized.

[–] iusemybrain@sh.itjust.works 1 points 6 days ago (2 children)

find all possible eigenvalues and eigenvectors for the following 3 x 3 square matrix [1/0 1/0 1/0; 1/0 1/0 1/0; 1/0 1/0 1/0]

[–] iusemybrain@sh.itjust.works 1 points 1 week ago

prostitution and sex trafficking. But besides that not much...

oh wait I forgot, he doesn't support the independence of women; just sex trafficking.

[–] iusemybrain@sh.itjust.works 1 points 1 week ago

you could approximate it using Taylor expansions, although this generally isn't a rapidly convergent series. You might take a fancy for some other numerical method that would get really precise digits really quickly...

[–] iusemybrain@sh.itjust.works 1 points 1 week ago

loves to see the world burn

[–] iusemybrain@sh.itjust.works 1 points 1 week ago

I think having both the indentation and the braces surrounding code is much better than just indentation or braces, since it removes any ambiguity of what code goes to what if, for, or while loop/condition. we already have to memorize precedence to define ambiguous expressions. Why have another set of rules that could otherwise be prevented with braces and indentation?

[–] iusemybrain@sh.itjust.works 2 points 1 week ago* (last edited 1 week ago)

a dolphin poisoned by pollution and carbon monoxide gas.

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