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visits member for 2 years, 7 months
seen Dec 11 '14 at 4:35

Aug
18
comment Simulate a 7-sided die with a 6-sided die
Nice and simple. Definitely the easiest way to do it with two physical dice.
Aug
18
comment Simulate 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.
Aug
12
comment Hearthstone Arena Probability
@Dgrin91 - That's true, but this is the realm of the Spherical Chicken.
May
2
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.
May
2
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".
May
1
comment Factorial of 1,e+80
@derpy - Yes, you're right. Shows how much math terminology I remember.
Apr
30
awarded  Commentator
Apr
30
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.
Feb
28
awarded  Scholar
Feb
28
accepted Calculating odds of a dependent series
Feb
28
comment Calculating odds of a dependent series
I get it now. Thanks! Put all that into an answer and I'll accept it.
Feb
28
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?
Feb
28
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.
Feb
28
awarded  Student
Feb
28
asked Calculating odds of a dependent series
Jan
13
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.
Sep
17
awarded  Good Answer
Sep
12
comment Interesting math-facts that are visually attractive
@ssch - Thanks for some really nice graphics!
Sep
12
awarded  Nice Answer
Sep
11
awarded  Teacher