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Subject: "Some Help Please:)"     Previous Topic | Next Topic
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Ariyan
Member since Feb-22-04
Mar-07-04, 09:32 AM (EST)
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"Some Help Please:)"
 
   Hello,
Please excuse my bad English. My name is Ariyan and I live in Holland. I'm a great fan of your website and I visit it every day, just like your forum.
I have been having some problems with a problem that I saw in a magazine. I don't know if it has ever been put on this forum but I don't think so. Here goes.

You have a big cake, you're going to receive 100 guests. The first guest gets 1% of the entire cake. The second guest gets 2% of what is left. The third guest gets 3% of what is left.

My question:
1) Which guest gets the biggest piece?
2) Is there a formula to calculate how big the piece is for the 'n-th' person?

I have been working on question 2 mostly but I'm just getting a very very long row. It is something like:

n) n/100(1-1/100-2/100(1-1/100)-3/100(1-1/100-2/100(1-1/100)-...-(n-1)/100(1-1/100-2/100(1-1/100)...(n-2)/100(1-1/100-2/100(1-1/100)-...(n-3)/100(1-1/100-2/100(1-1/100)-...and so on...

I came up with something, but I don't if it is correct. It goes as following:
Name every pre-calculation 'a' So the formule would get something like

n/100(1-1/100-2/100(a2)-3/100(a3)-...-(n-1)/100(a(n-1))-...and so on... and a2>a3>a4>a(n-1)

Well all of this is really work that hasn't got much to with the answer but still I like it

I appreciate all the help I get.
Thank you


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  Subject     Author     Message Date     ID  
Some Help Please:) Ariyan Mar-07-04 TOP
  RE: Some Help Please:) Ariyan Mar-07-04 1
  RE: Some Help Please:) alexb Mar-07-04 2
     RE: Some Help Please:) jvwert Mar-07-04 3
         RE: Some Help Please:) JJ Mar-08-04 4
             RE: Some Help Please:) KL_GB Mar-08-04 5
                 RE: Some Help Please:) Ariyan Mar-08-04 6

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Ariyan
Member since Feb-22-04
Mar-07-04, 05:27 PM (EST)
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1. "RE: Some Help Please:)"
In response to message #0
 
   I see I'm missing some words and comma's here and there. Maybe I should have previewed it...

but I don't KNOW if it is correct


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alexb
Charter Member
1227 posts
Mar-07-04, 05:33 PM (EST)
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2. "RE: Some Help Please:)"
In response to message #0
 
   >n)
>n/100(1-1/100-2/100(1-1/100)-3/100(1-1/100-2/100(1-1/100)-...-(n-1)/100(1-1/100-2/100(1-1/100)...(n-2)/100(1-1/100-2/100(1-1/100)-...(n-3)/100(1-1/100-2/100(1-1/100)-...and
>so on...
>
>I came up with something, but I don't if it is correct.

From your explanation, I do believe that this awful expression is correct, although I am incapable of perusing it.

The sequence is not monotonic, although at the beginning it appears to be.


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jvwert
Member since Jan-7-02
Mar-07-04, 09:44 PM (EST)
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3. "RE: Some Help Please:)"
In response to message #2
 
   My rough computation (via Excel) indicates the tenth person gets the largest piece - about 6.28 percent of the whole cake, and each person from then on gets a diminishing piece that gets very, very small as the the number gets above about twenty. For example, the fiftieth person gets 1.53*10^-5 percent of what is left, which is coincidentally 1.53*10^-5 percent of the original cake.

jvwert


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JJ
guest
Mar-08-04, 09:46 AM (EST)
 
4. "RE: Some Help Please:)"
In response to message #3
 
   The formula is :
The n-th gets : (99! / (100-n)!) (n / (100^n))
:

The first guest gets (99! / 99!) (1 / (100^1)) = 0.01

The second : (99! / 98!) (2 / (100^2)) = 0.0198

The third : (99! / 97!) (3 / (100^3)) = 0.02106

etc.

The 9th. : (99! / 91!) (9 / (100^9)) = 0.0621253...

The 10th. : (99! / 90!) (10 / (100^10)) = 0.0628156... which is the maximum.

The 11th. : (99! / 89!) (11 / (100^11)) = 0.06218749...

etc.

The 100th. : (99!) (100 / (100^100)) = 9.33262 / (10^43)


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KL_GB
Member since Feb-13-04
Mar-08-04, 03:07 PM (EST)
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5. "RE: Some Help Please:)"
In response to message #4
 
   If C(i) is the amount of cake left after the i-th persons allotment,

then

C(i) = M <100!/((99-i)!)><1/100^(i+1)>

where M is the size of the cake at the start (i.e. C(0) = M)

The i-th person gets to eat C(i-1) * (i/100) which is

M <100!/((100-i)!)>

KL

KL


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Ariyan
Member since Feb-22-04
Mar-08-04, 03:39 PM (EST)
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6. "RE: Some Help Please:)"
In response to message #5
 
   Yes, thank you all.


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