I'm not smart, but I felt something off about their smartness.Ah yes, another classic case of dumb people writing smart people.
The correct way to solve this is not to calculate the probability of each individual number, which is impossible, but simply to recognize that the outcome is functionally just a normal distribution and estimate the probabilities accordingly.
Ah yes, another classic case of dumb people writing smart people.
The correct way to solve this is not to calculate the probability of each individual number, which is impossible, but simply to recognize that the outcome is functionally just a normal distribution and estimate the probabilities accordingly.
In a standard bingo board, every square can be a part of either 2 or 3 bingos (other the center, which can be a part of 4, which is why it’s free). In this particular game it’s also possible for a square to be a part of 0 or 1 possible bingos because certain numbers are impossible to get. So basically you want to create a group of squares with 3 possible bingos with high probability, and put the low probability or impossible numbers in a way that doesn’t interfere.I think rather than putting the same number all over the board, it would've been safer to put a line of each of the most likely numbers.
btw, @afortuneforetold wich numbers would you have put and where in the board? Soeey if my english is kinda bad.
If you look at this image you can see how it approaches a normal distribution over time. And you can estimate it by understanding the proportionate heights of the normal distributionno way it's like binomial or something right?? i swear i learned that in stats
normal distr is continuous anyway, how would you use that for a discrete case like dice rolls?
also also, does it really matter if the smart people aren't actually smart? it might be funnier if it just str8 up makes up whatever bs to get the job done
Okay now I get it, thanks. I guess we needed the rule to not repeat numbers for that to be the best arrangement, haha.Basic
In a standard bingo board, every square can be a part of either 2 or 3 bingos (other the center, which can be a part of 4, which is why it’s free). In this particular game it’s also possible for a square to be a part of 0 or 1 possible bingos because certain numbers are impossible to get. So basically you want to create a group of squares with 3 possible bingos with high probability, and put the low probability or impossible numbers in a way that doesn’t interfere.
The author again fails at this because their most likely number (14) has one of its 3 bingo paths made impossible by a number you can’t roll (1). It’s good to put 1 and 4 (both bad numbers) together because you don’t lose anything from sabotaging a line a second time, but bad to put them in a place where they interfere with likely numbers, they should be on the other side of the board.
I’m too lazy to map this out the whole way but hopefully you get the picture. Basically you want to minimize the number of lines sabotaged by bad numbers while simultaneously making the good numbers go together and hit as many lines as possible (by putting them on diagonals not sabotaged by bad numbers)