The new dice pdf is out!

anonimnystefy · 2013-02-17 00:33:02

Be sure to tell if you get an idea!

bobbym · 2013-02-17 05:54:10

I have something. It will take some time.

anonimnystefy · 2013-02-17 06:03:10

Can you post just the beginning steps?

bobbym · 2013-02-17 07:10:03

Calculation is running, it will take a while.

anonimnystefy · 2013-02-17 07:11:20

Post when it's done.

bobbym · 2013-02-17 07:27:57

Hi;

Procedure failed, I do not know why.

anonimnystefy · 2013-02-17 07:28:33

What procedure?

bobbym · 2013-02-17 07:29:43

An attempt to develope an asymptotic form that would have eliminated the coefficient part of your maxima code.

anonimnystefy · 2013-02-17 07:30:10

Can you elaborate?

bobbym · 2013-02-17 07:34:31

Instead of picking out the i - k power of x out of the expansion. For large k, I figured that expansion would approach (e^x)^5. The coefficient of x^(i-k) of that is 5^(i-k) / (i-k)!. For large k this would have replaced expanding a big polynomial raised to the 5th power. Unfortunately, it did not work.

anonimnystefy · 2013-02-17 07:35:32

Why not?

bobbym · 2013-02-17 07:41:03

I do not know. I tested it for k = 20, 30 , 40 , 50, 60, 100, 200. It did not asymptotically approach the exact answer.

anonimnystefy · 2013-02-17 07:43:18

I guess that there are too many terms that contribute to the x^(i-k) term.

bobbym · 2013-02-17 07:57:13

I am not sure, I will think about it.

anonimnystefy · 2013-02-17 12:06:09

Hi bobbym

I got a reply. He said he does not need the solution if it just involves the standard bunch of methods.

Personally, I really do not think that that problem is the best one for showing how Markov chains can be used, but, it is his work, and I will not insist on it anymore.

I would still like to continue working on an analityc form here.

bobbym · 2013-02-17 21:05:56

Personally, I really do not think that that problem is the best one for showing how Markov chains can be used, but, it is his work, and I will not insist on it anymore.

I am not following you. He does not always use Markov chains. Your method is a standard method, even though I do not know what he means by that.

anonimnystefy · 2013-02-17 21:08:41

Hi bobbym

What I meant to say is, if he wanted to show how Markov chains work, he should've done it on a smaller and bit simpler problem.

bobbym · 2013-02-17 21:16:01

Hi;

I am not suggesting that you persist in publishing. That is a decision that is up to you, but the purpose of that pdf is not to just demonstrate Markov chains. It is to solve dice problems.

anonimnystefy · 2013-02-17 21:24:42

Again you misunderstood me. He said that, if the answer is a classical one, then he doesn't need it. He obviously wants the second part of 10th problem to be with Markov chains.

bobbym · 2013-02-17 21:28:00

That is impossible, the next one would require a 4097 x 4097 matrix!

anonimnystefy · 2013-02-17 21:30:35

I meant for the 3 case.

gAr · 2013-02-17 21:34:22

Hi bobbym,

I think we can't approximate it using e^(5x), since we are using coefficients beyond the upper limit of the egf. Below is the probability plot for n=45.

Hi anonimnystefy,

Standard bunch of methods? Did he mean he has solution for that already, but did not publish in his pdf?

Last edited by gAr (2013-02-17 21:35:14)

bobbym · 2013-02-17 21:34:44

He already solved the 3 case. It took a 730 x 730 matrix. He will never get a general answer as you did using Markov chains.

Last edited by bobbym (2013-02-17 21:35:33)

anonimnystefy · 2013-02-17 21:41:08

bobbym wrote:

He already solved the 3 case. It took a 730 x 730 matrix. He will never get a general answer as you did using Markov chains.

Yes, but it seems he wanted to show how Markov chains work on particularly that problem.

Anyway, there is no point in discussing it any more. Have you found anything else that could help us with the GF?

bobbym · 2013-02-17 21:46:17

Hi;

I am not unhappy with your decision, as I said it is your choice. You should be advised though that someone from another forum will eventually see your solution, change it a bit and call it his own.

I could possibly shorten the computation a bit but I do not want to do that. I would prefer to eliminate the gf or to just leave it as it is.

Math Is Fun Forum

#26 2013-02-17 00:33:02

Re: The new dice pdf is out!

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