Yeah, explain that for us...Yep you got em now, bonanova! ;D so nobody got the trick with

e?

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Guest Message by DevFuse

Started by unreality, Feb 21 2008 02:00 AM

16 replies to this topic

### #11

Posted 23 February 2008 - 11:07 AM

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #12

Posted 23 February 2008 - 01:28 PM

5! = 1 * 2 * 3 * 4 * 5 = 120

120/e = 44 (there's an infinite string of decimals following it but that doesn't matter)

44/120 are the number of combinations where no hat is given back to its previous owner.

it works with other numbers, such as 10! or 17!, etc

just lop off the decimal part and keep only the integer

1/e = 0.something

so there are 0 combinations to give back the hats without giving one to its previous owner if there is only 1 person, obviously

2/e = .7something, which would be 0... even though with 2 hats there would be 1 combination where neither would get his hat back. So it's not perfect, though it's a pretty good ratio between e and what we're looking for

though since we wanted AT LEAST ONE then we had to subtract that from 1

(n!-integer(n!/e))/n!

for the 5 people:

(120-44)/120 = 76/120 = .6333333 chance that at least one person gets their hat back

120/e = 44 (there's an infinite string of decimals following it but that doesn't matter)

44/120 are the number of combinations where no hat is given back to its previous owner.

it works with other numbers, such as 10! or 17!, etc

just lop off the decimal part and keep only the integer

1/e = 0.something

so there are 0 combinations to give back the hats without giving one to its previous owner if there is only 1 person, obviously

2/e = .7something, which would be 0... even though with 2 hats there would be 1 combination where neither would get his hat back. So it's not perfect, though it's a pretty good ratio between e and what we're looking for

though since we wanted AT LEAST ONE then we had to subtract that from 1

(n!-integer(n!/e))/n!

for the 5 people:

(120-44)/120 = 76/120 = .6333333 chance that at least one person gets their hat back

### #13

Posted 26 February 2008 - 11:03 PM

So yeah. I hope that was clear

### #14

Posted 27 February 2008 - 09:56 AM

Yeah, I guess. Wondering why, tho.

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #15

Posted 27 February 2008 - 10:35 PM

Same ;D

looking at it:

integer(n!/e)/n! = permutations where no hat is given back to its previous owner. Doesn't work with 1 or 2 hat-owners, works with 3 or more though.

consider it NOT using the integer function (which is there, to simplify it cuz of probability)

that would make:

(n!/e)/n!

which is:

n!/(e*n!)

which is:

1/e

however it's not 1/e every single time with 5 hats or with 10 hats, it varies slightly. Because of the taking-the-integer-only part.

So anyway, consider it in pure probability's sake, 1/e chance that nobody gets their hat back.

After some searching the internet, I found a section of an article on wikipedia that explains the property. Check this out:

Bernoulli trials

(read the section on Bernoulli trials)

The integer-taking part approximates e's limit to the limit with the slots-machine example

looking at it:

integer(n!/e)/n! = permutations where no hat is given back to its previous owner. Doesn't work with 1 or 2 hat-owners, works with 3 or more though.

consider it NOT using the integer function (which is there, to simplify it cuz of probability)

that would make:

(n!/e)/n!

which is:

n!/(e*n!)

which is:

1/e

however it's not 1/e every single time with 5 hats or with 10 hats, it varies slightly. Because of the taking-the-integer-only part.

So anyway, consider it in pure probability's sake, 1/e chance that nobody gets their hat back.

After some searching the internet, I found a section of an article on wikipedia that explains the property. Check this out:

Bernoulli trials

(read the section on Bernoulli trials)

The integer-taking part approximates e's limit to the limit with the slots-machine example

### #17

Posted 28 February 2008 - 11:24 PM

Yeah. The hat check problem. I didnt know it was on wikipedia, I saw it elsewhere. It's a good'un of course e is cool and all, especially with e^(theta•i) = cos(theta) + i•sin(theta) equalling the special case e^(pi*i) = -1

even that's not as cool as the golden ratio and its much neglected other root of the quadratic equation that forms it (x^2-x-1=0), which is its negative inverse and also 1-phi, and shares its amazing properties, most of which I figured out on my own, others I had to look up. But the golden ratio is amazing... you think a few cool properties are amazing- but then you soon realize they are just specific cases of its much bigger and more awesome properties, mostly connected to the Fibonacci numbers... yeah lol I love the golden ratio. And the Fibonacci numbers of course, which go hand in hand with phi

*exhales long breath* hehe i'm a bit overenthusiasitic about that i have such nerdy pasttimes that nobody knows about lol

but yeah.

even that's not as cool as the golden ratio and its much neglected other root of the quadratic equation that forms it (x^2-x-1=0), which is its negative inverse and also 1-phi, and shares its amazing properties, most of which I figured out on my own, others I had to look up. But the golden ratio is amazing... you think a few cool properties are amazing- but then you soon realize they are just specific cases of its much bigger and more awesome properties, mostly connected to the Fibonacci numbers... yeah lol I love the golden ratio. And the Fibonacci numbers of course, which go hand in hand with phi

*exhales long breath* hehe i'm a bit overenthusiasitic about that i have such nerdy pasttimes that nobody knows about lol

but yeah.

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