TheBlaarg
Reputation
Next privilege 250 Rep.
 May 20 awarded Nice Question Jun 1 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. Jun 1 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. Jun 1 answered Odds of winning at minesweeper with perfect play Jun 1 awarded Teacher Jun 1 answered Publishing elementary proofs of theorems May 8 awarded Supporter May 8 awarded Scholar May 8 comment Integrating $\frac{x^k }{1+\cosh(x)}$ A very nice solution. I wonder why Mathematica couldn't do it. May 8 accepted Integrating $\frac{x^k }{1+\cosh(x)}$ May 8 awarded Student May 8 asked Integrating $\frac{x^k }{1+\cosh(x)}$