What do you think?

bobbym · 2013-02-16 04:18:10

What table?

anonimnystefy · 2013-02-16 04:24:59

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.

bobbym · 2013-02-16 04:30:41

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

anonimnystefy · 2013-02-16 04:34:38

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)

bobbym · 2013-02-16 04:36:32

Then what is i?

anonimnystefy · 2013-02-16 04:38:28

Sorry, wrong variable. Check again.

bobbym · 2013-02-16 04:39:56

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?

anonimnystefy · 2013-02-16 04:42:02

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)

bobbym · 2013-02-16 04:44:39

Okay, I will need to work on it.

anonimnystefy · 2013-02-16 04:50:38

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)

bobbym · 2013-02-16 04:52:28

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

anonimnystefy · 2013-02-16 04:55:53

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)

bobbym · 2013-02-16 04:57:25

I was talking about the T[n] idea.

anonimnystefy · 2013-02-16 05:02:29

It is working for me. What's the problem?

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

bobbym · 2013-02-16 05:06:41

Calm down please.

You are getting this to work?

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

anonimnystefy · 2013-02-16 05:09:17

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)

bobbym · 2013-02-16 05:13:59

I am kidding you. Just being funny.

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

anonimnystefy · 2013-02-16 05:34:24

I think it is plausible. Give me 5 minutes.

bobbym · 2013-02-16 05:35:08

I am working on it also.

anonimnystefy · 2013-02-16 05:37:22

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)

bobbym · 2013-02-16 05:40:12

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

anonimnystefy · 2013-02-16 05:41:32

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

bobbym · 2013-02-16 05:43:38

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.

anonimnystefy · 2013-02-16 05:45:32

Ok. So what do we do now?

bobbym · 2013-02-16 05:47:26

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)

Math Is Fun Forum

#2151 2013-02-16 04:18:10

Re: What do you think?

#2152 2013-02-16 04:24:59

Re: What do you think?

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Re: What do you think?

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