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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,882

What table?

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,605

1;

1, 1, 1;

1, 2, 4, 6, 6;

1, 3, 9, 24, 54, 90, 90;

1, 4, 16, 60, 204, 600, 1440, 2520, 2520;

1, 5, 25, 120, 540, 2220, 8100, 25200, 63000, 113400, 113400;

1, 6, 36, 210, 1170, 6120, 29520, 128520, 491400, 1587600, 4082400,

It is in the examples in the link I posted.

The line

1, 5, 25, 120, 540, 2220, 8100, 25200, 63000, 113400, 113400;

is the one of interest.

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Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,882

Hmmm, how do match that up with the pdf? Please demonstrate.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,605

Hi bobbym

Do you see his table for P(X=n) on page 18?

Let T_n be the nth number in the 6th line of the triangle

1;

1, 1, 1;

1, 2, 4, 6, 6;

1, 3, 9, 24, 54, 90, 90;

1, 4, 16, 60, 204, 600, 1440, 2520, 2520;

1, 5, 25, 120, 540, 2220, 8100, 25200, 63000, 113400, 113400;

1, 6, 36, 210, 1170, 6120, 29520, 128520, 491400, 1587600, 4082400,

.

.

.

Then P(X=n)=T_(n-3)*Binomial[n-1,2]/6^(n-1)

*Last edited by anonimnystefy (2013-02-16 05:10:07)*

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,882

Then what is i?

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,605

Sorry, wrong variable. Check again.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,882

Okay, you were supposed to check that but maybe you are not at your computer. I can check it. Supposing you are right is there an easier way to generate those numbers than he did?

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,605

I checked the for values of n from 3 to 9. I did not check the last 4.

I think it might be. I do not understand yet how the triangle was generated. I will have a look later.

Edit: OEIS gives a formula for those: T(k,n) = n![x^n](1+x+x^2/2)^k.

*Last edited by anonimnystefy (2013-02-16 04:43:26)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,882

Okay, I will need to work on it.

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,605

For this particular problem the formula is T(n) = n![x^n](1+x+x^2/2)^5, which is kinda obvious, considering the fact that it is the number of ways we can put n marbles into 5 jars, with both jars being different and order of putting in mattering.

I think we actually solved it!

*Last edited by anonimnystefy (2013-02-16 04:50:56)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,882

Hold on! You solved it, but so far it is not working.

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,605

Why not?

With the following code I get the correct answer!

`sum(i!/(2*6^(i-1))*coeff(ratexpand((1+x+x^2/2)^5),x^(i-3)),i,3,13),numer;`

*Last edited by anonimnystefy (2013-02-16 05:42:29)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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I was talking about the T[n] idea.

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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It is working for me. What's the problem?

I think it could be generalized for 4, 5 and more dice in a row.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,882

Calm down please.

You are getting this to work?

*Last edited by bobbym (2013-02-16 05:06:58)*

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,605

I am calm. Sorry if I sounded frustrated.

T(n-3), not T(n), my bad. Edited the original post, too.

*Last edited by anonimnystefy (2013-02-16 05:10:24)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,882

I am kidding you. Just being funny.

I have it working now. Okay, you say you can improve on that formula?

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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I think it is plausible. Give me 5 minutes.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,882

I am working on it also.

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,605

Hi bobbym

I am getting 11.21375706013234, or

in closed form.Edit: Here's my code:

`sum(i*(i-4)!/(6^(i-1))*binomial(i-1,3)*coeff(ratexpand((1+x+x^2/2+x^3/6)^5),x^(i-4)),i,4,19);`

*Last edited by anonimnystefy (2013-02-16 05:40:07)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,882

You are jumping ahead, what is that the answer to?

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,605

Expected number of throws till we get a number 4 times.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Please hold on with that. We are still working on the 3. I agree that it will be possible to generalize but first I would like to simplify how you are doing the three. If we can do that we can really present it to him as a superior solution.

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,605

Ok. So what do we do now?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Is there a way to generate the sequence T[5,n] that does not require us to expand a trinomial or a sum with a ceiling function in it?

*Last edited by bobbym (2013-02-16 05:47:45)*

**In mathematics, you don't understand things. You just get used to them.**

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