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 Aug18 comment Simulate repeated rolls of a 7-sided die with a 6-sided die Nice and simple. Definitely the easiest way to do it with two physical dice. Aug18 comment Simulate repeated rolls of a 7-sided die with a 6-sided die @neverlastn - When dealing with physical dice, simplicity is to be preferred over efficiency. When dealing with dice simulators, efficiency is preferred over simplicity. Aug12 comment Hearthstone Arena Probability @Dgrin91 - That's true, but this is the realm of the Spherical Chicken. May2 comment I have the pattern: 1 + 2 + 3 + 4 + 5 + 6, but I need the formula for it @Cruncher - It's in the first sentence - he's writing software to do this comparison, and was just asking about estimating the time. May2 comment I have the pattern: 1 + 2 + 3 + 4 + 5 + 6, but I need the formula for it In computer science terms, this is a $O(n^2)$ operation - Doubling the number of users will take four times as long to do your comparison. See this blog for more on "Big O notation". May1 comment Factorial of 1,e+80 @derpy - Yes, you're right. Shows how much math terminology I remember. Apr30 awarded Commentator Apr30 comment Factorial of 1,e+80 @derpy - Given that multiplication is commutative, the problem could be broken down into arbitrarily small parallel tasks (Multiply the numbers n thru n+100000), which then can be multiplied to get the final result. Of course, then you'd need a supervisor program to figure out each range, farm it out, and get back the result, which itself may take a long time to run. Feb28 awarded Scholar Feb28 accepted Calculating odds of a dependent series Feb28 comment Calculating odds of a dependent series I get it now. Thanks! Put all that into an answer and I'll accept it. Feb28 comment Calculating odds of a dependent series I see, I think. All the other threes have P(1/4)*P(1/2), but MM flows to a P(1/4)*P(0) and a P(1/4)*P(1) instead. Right? Feb28 comment Calculating odds of a dependent series @user130512 - Why don't they? Does MMH count double because MMM isn't valid? That would seem to be the reasoning from #2, but I don't understand why. Feb28 awarded Student Feb28 asked Calculating odds of a dependent series Jan13 comment How to prove that a $3\times 3$ Magic Square must have $5$ in its middle cell? It's also worth pointing out that this question only applies to normal magic squares (i.e. 1...n). The general magic square might not have a 5 in it at all, let alone in the center. Sep17 awarded Good Answer Sep12 comment Interesting math-facts that are visually attractive @ssch - Thanks for some really nice graphics! Sep12 awarded Nice Answer Sep11 awarded Teacher