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#11 Professor H

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Posted 14 February 2009 - 12:39 AM

nope. For n=3, it is 1/2, but your equation gives 1/3.


Now wait a second...does "a followed by a+1" include the possibility of having the last card as "a" then the first card as "a+1"? If you include that as a possibility, then you are right. Otherwise, my answer still stands...
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#12 EventHorizon

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Posted 14 February 2009 - 12:42 AM

Now wait a second...does "a followed by a+1" include the possibility of having the last card as "a" then the first card as "a+1"? If you include that as a possibility, then you are right. Otherwise, my answer still stands...

I think my labeling the cards a and a+1 made the problem a little confusing. There is no wrap around, and a is not a specific card, just one that has the card with the number one greater right after it.

Look at my example for n=3 in my previous post.
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#13 EventHorizon

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Posted 14 February 2009 - 01:31 AM

1) how many days is in their year?
2)10^(10^100)/2?
3)does n stand for a variable?

For 2, I said "nope...." when I should have said, you can get a better estimate...and to use spoilers ;)
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#14 Prof. Templeton

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Posted 14 February 2009 - 05:35 AM

Spoiler for No.1

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#15 Prime

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Posted 15 February 2009 - 12:06 AM

Just a thought...
Spoiler for B-days

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Past prime, actually.


#16 Magic_luver101

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Posted 15 February 2009 - 12:28 AM

Spoiler for answer

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I'm dyslexic.

Live. Laugh. Love.

 


#17 EventHorizon

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Posted 17 February 2009 - 09:11 PM

Spoiler for No.1

yup, there's the answer to the first question. It's fitting that you got it first.

Any ideas on the next two?
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#18 Prime

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Posted 20 February 2009 - 09:30 PM

...

And now, a somewhat related original puzzle...

(3) You are given a stack of cards numbered 1 to n, and their order is randomized. What is the probability that two consecutively numbered cards (lets say, a and a+1) are found consecutively (a immediately before a+1) somewhere in the stack?

I can calculate this probability as a series, but I haven't found a polynomial solution yet.
Spoiler for sequential probability

But I suppose, EH is looking for a polynomial solution. Is that so?
Or is the problem to calculate probability for two specific cards? But that would be too simple, not EH-like problem.
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Past prime, actually.


#19 EventHorizon

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Posted 21 February 2009 - 11:35 AM

I can calculate this probability as a series, but I haven't found a polynomial solution yet.

Spoiler for sequential probability

But I suppose, EH is looking for a polynomial solution. Is that so?
Or is the problem to calculate probability for two specific cards? But that would be too simple, not EH-like problem.

Very nice. I'll consider this one solved. Though here's an equation that will blow your mind ;)
Spoiler for simple equation for problem 3, and inspiration for the problem
So that just leaves question 2...
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#20 Prime

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Posted 23 February 2009 - 11:44 PM

Very nice. I'll consider this one solved. Though here's an equation that will blow your mind ;)

Spoiler for simple equation for problem 3, and inspiration for the problem
So that just leaves question 2...

That is interesting! The ubiquitous e. I suppose, whenever you get a series with a mixture of factorials and exponents, you can always look how to convert the series into f(ex).

A correction to my post with the birthday (question 2) equation.
Spoiler for B-day equation

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Past prime, actually.





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