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20 replies to this topic

### #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

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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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