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Author of Beta's Rule: "Never add to code that doesn't work."


Sep
24
awarded  Autobiographer
May
17
awarded  Informed
Mar
3
comment Why is not the answer to all probability questions 1/2.
@KaiSikorski: I respectfully disagree. That distribution ({1/n, 1/n, 1/n,...}) is useful in the zero-information case only if there can be other distributions when there is information. Just try applying Bayes's Theorem while holding to P(A|B)=P(A), and see where it gets you.
Mar
2
comment Why is not the answer to all probability questions 1/2.
You can define probability that way ("either is or isn't" => 1/2), but that kind of probability is useless. In the conventional probability theory, assuming that all possibilities are equally probable is simply incorrect, and the the theory proves useful.
Feb
22
comment Calculate arc length knowing its subtended chord and circumference diameter
There's a mistake in your algebra, going from the second to the third line.
Feb
21
answered Calculate arc length knowing its subtended chord and circumference diameter
Jan
21
comment How do I convince my students that the choice of variable of integration is irrelevant?
@neminem: If A is the act of putting on a shirt, and k is a number in [0,1], then k**A** is putting on a shirt with probability k, and AA is putting on a shirt and then putting on another shirt over it. The rest follows easily.
Jan
21
comment How do I convince my students that the choice of variable of integration is irrelevant?
Two of my best math teachers used "sailboat" (small drawing of a sailboat) and "potato" (sometimes a squiggle, usually the letter 'z'). I can still hear the phrases "integral of f of sailboat dee sailboat" and "e to the potato is the sum of potato to the k over k factorial" (the latter in a strong Swiss accent). The second teacher made it very clear that potato is not necessarily a number, as when he showed us how to take e to the power of the act of putting on a shirt.
Jan
8
comment There exists a power of 2 such that the last five digits are all 3's or 6's. Find the last 5 digits of this number
@Mr.Mindor: You're right, I was a little careless with the more/less language. As for adding to the answer, I felt it was getting too long so I tried for brevity.
Jan
8
comment There exists a power of 2 such that the last five digits are all 3's or 6's. Find the last 5 digits of this number
@Mr.Mindor: Let's call the 10^k digit of the jth row [j,k]. Given [j,k], there are two choices for [j+1,k], one odd and one even. The choice will determine whether [j,k+1] is odd or even. The 10's digit of n ([1,1]) is 3, so the 10's digit of n/2 ([2,1]) must be 1 or 6. If the 100's digit of n ([1,2]) is even then [2,1] is less than [1,1], namely 1; if [1,2] is odd then [2,1] is more than [1,1], namely 6. We already know it must be even, so it's 6, and [1,2] is odd (which is to say 3).
Jan
8
answered There exists a power of 2 such that the last five digits are all 3's or 6's. Find the last 5 digits of this number
Oct
1
revised Finding volume of a solid by integrating
deleted 22 characters in body
Oct
1
answered Finding volume of a solid by integrating
Apr
30
comment Expected Value of a Randomly decreasing function
If only there were a simple form for the probability that N=0 after k iterations...
Mar
22
comment Probability hit “bullseye”
@ldgorman: Yes. In that case the archer's aim is so bad that the arrow could land as easily anywhere on the plane, so the chance of hitting within an inch of the bullseye -- or within a gigaparsec of the bullseye -- is basically zero.
Mar
22
revised Probability hit “bullseye”
added 647 characters in body
Mar
22
awarded  Commentator
Mar
22
comment Probability hit “bullseye”
@LikeMike: which step is giving you trouble?
Mar
21
comment Probability hit “bullseye”
@Harold: Thanks!
Mar
21
answered Probability hit “bullseye”