TheBlaarg
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 May20 awarded Nice Question Jun1 comment Odds of winning at minesweeper with perfect play I don't see why there should be a simple connection between the odds of solving a particular configuration and the odds of a particular square being a mine. How can I compute the odds of a particular configuration of mines being solvable given only the odds of each square being a mine? This may be exactly what you are asking. In any case, it seems to me a very difficult problem for all but the smallest cases. Jun1 comment Odds of winning at minesweeper with perfect play I don't see any reason why that should be significantly easier to compute than an actual algorithm. It seems to me that the problems arising when attempting to create an algorithm should also arise in any attempt to compute the odds of winning with perfect play in some form. I'd be happy to be proven wrong, though. Jun1 answered Odds of winning at minesweeper with perfect play Jun1 awarded Teacher Jun1 answered Publishing elementary proofs of theorems May8 awarded Supporter May8 awarded Scholar May8 comment Integrating $\frac{x^k }{1+\cosh(x)}$ A very nice solution. I wonder why Mathematica couldn't do it. May8 accepted Integrating $\frac{x^k }{1+\cosh(x)}$ May8 awarded Student May8 asked Integrating $\frac{x^k }{1+\cosh(x)}$