1,221 reputation
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bio website morrellplumbing.com.au
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visits member for 1 year, 8 months
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Full time business manager part time programmer for in-house applications only - don't expect any polish!


2d
comment Probability of $j$ inversions in an array
From Wikipedia definition of Inversion: If $i < j$ and $A(i) > A(j)$, then the pair $(i, j)$ is called an inversion of $A$. If $A(i) = A(j)$ then this is not an inversion by this definition.
2d
answered Is it fair for all customers?
Sep
10
answered How to calculate t-value, given degrees of freedom and $\alpha$.
Sep
4
revised More computationally optimal way to solve probability of N or more empty buckets given B buckets and A balls
deleted 16 characters in body
Sep
4
answered More computationally optimal way to solve probability of N or more empty buckets given B buckets and A balls
Sep
4
answered Find the probability that A is the winner of the series given that A won the first game.
Sep
4
comment More computationally optimal way to solve probability of N or more empty buckets given B buckets and A balls
Should the question therefore be ported to "English Language & Usage" :-)
Sep
4
comment More computationally optimal way to solve probability of N or more empty buckets given B buckets and A balls
Glad we agree on our simulations. I agree that the answer is wrong and your question is nagging the hell out of me! Still thinking ...
Sep
3
revised More computationally optimal way to solve probability of N or more empty buckets given B buckets and A balls
added 1513 characters in body
Sep
3
comment More computationally optimal way to solve probability of N or more empty buckets given B buckets and A balls
OK added to answer
Sep
1
comment More computationally optimal way to solve probability of N or more empty buckets given B buckets and A balls
Slight error in my code - the actual p is about 0.045
Sep
1
comment More computationally optimal way to solve probability of N or more empty buckets given B buckets and A balls
I agree that the solution presented is not correct - I will think about it some more - I believe that the first equation (which I took from somewhere) is the problem. There is also something wrong with your algorithm - The expectation of 5+ empty buckets is about 0.08 based on 1,000,000 trials.
Aug
29
comment Prove $\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+d^2}+\frac{d}{1+a^2} \ge 2$ if $a+b+c+d=4$
So I did - I read positive integers
Aug
29
comment Prove $\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+d^2}+\frac{d}{1+a^2} \ge 2$ if $a+b+c+d=4$
Am I missing something: "a,b,c,d are positive real numbers and a+b+c+d=4" then a=b=c=d=1 - this is the only way the sum can be true with positive real numbers. It is then trivial to prove the inequality.
Aug
29
answered More computationally optimal way to solve probability of N or more empty buckets given B buckets and A balls
Aug
28
answered Probability in a knock-out tournament
Aug
28
comment Probability in a knock-out tournament
Intuitively, this cannot be right - A only has to win one game but C & D must win 2 - given that A and C were equally likely before A won a semi, A must now be more likely than B.
Aug
22
revised Variation of “The secretary problem”
added 131 characters in body
Aug
22
answered Variation of “The secretary problem”
Jul
23
answered Variance of a dice roll