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Ojo Rojo

Even this much (ε) interest?

MathJames

Yes, ε is greater than zero, so you are all going to pay now, MWHAHAHAHA! I will thrill you to sleep with mathematical tales of high dullness... as soon as i return from the "Face of Victoria's Secret" Model search which i was forced to judge.

tom

This talk of infinite sets and whatnot causes me to recall something I learned in high school Algebra II.

Having programmed computers since age 7, I was pretty sure that any positive real number divided by 0 was infinity, which my Apple ][ obviously couldn't handle.

But per my teacher, not so. He insisted that it was undefined.

To which I said "Ah, bullshit. I'm already learning some calculus, and I can tell you that as 1/.01 is big. 1/.001 is bigger. 1/.000000000001 is even bigger. Continue, and as it approaches zero, you get larger and larger results, with no upward bound. Thats infinity, bro."

To which Mr. Loewer responded: "Nice, but remember for something to be =, the inverse must be maintaied. In other words, for 1/0 = infinity to be true, then 0*infinity = 1 must also be true.

"Which it ain't. Anything zero times is zero." Duh.

UNDEFINED == PWNED

(all quotes above from a malt liquor-addled memory of high school and may not reflect actual contents of discussion)

MathJames

Ah, but with Mr. Conway's generous new set, omega is not only be defined in a new way, but can be used in addition/negation, multiplication/division, powers/roots. It so happens that 1/ω = ε, and can proved. But not only can you divide by infinity you could, say, take the infinite root of a number!!!

MathJames

Surreal Numbers == RoXoRz VVi7h 1+5 c0x0rZ

tom

Hey! Are you coming to my birthday party? Did you get the evite?

Pembrose

It's a bit OT but, I just read this interesting probability study....

www.montyhallproblem.com

P.

tom

Ahhh, the Lets Make A Deal problem where the probability shifts after one solution is revealed.

I've seen this argued at bars over beers, on blogs and newsgroups, in classrooms, and even in the pages of Parade magazine (in the Marilyn Vos Savant column, who allegedly has the world's highest IQ).

To sum it up:

If you've got three doors, and one of them has a prize behind it, you win 1 out 3. If you pick 1, but then I reveal 3, and give you the option to change your choice, you always should as your new choice will be right 1/2 times instead of 1/3.

I'm not an genius or a hyperintelligent shade of blue or anything, so I'll give you my layman's perspective on this one. In the immortal words of Geddy Lee, however, "If you choose not to decide you still have made a choice."
If I have an option to choose again, and choose the original door, I've got a 1 in 2 chance. Therefore, not changing your pick when given an option gives you the same 1/2 that changing your choice would have given you.

P.S. Erik "Schwarzburg" wrote a c++ program during his EE degree at UT to test this problem, and oddly enough it indicated that I'm wrong. Go figure.

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