What do you think?

gAr · 2011-09-02 19:34:50

Hi bobbym,

Okay.

I have not looked at the next one yet because I am still stuck on 3a.

Did you mean 4 a)? a and b look tough, may not be possible with a markov chain.

bobbym · 2011-09-02 19:37:37

Hi gAr;

This question.

3. If a die is rolled 100 times (say), what is the probability that all six sides have appeared at least one?

I would like to verify your markov chain with a simulation but can not.

gAr · 2011-09-02 19:42:37

Hi bobbym,

I verified for 10 rolls.
For 1000000 trials, it showed 0.271413.

From the matrix, it's 0.271812128486511.

I'll be back after a little break.

Last edited by gAr (2011-09-02 19:43:16)

bobbym · 2011-09-02 19:43:53

Hi gAr;

Yes that is what I was just about to do. Our only proof is that it is very close for less rolls.

Okay, it checks out for 20.

On to 4)!

gAr · 2011-09-02 21:16:45

Hi bobbym,

Okay, one more way to check for 100 rolls is by direct computation:

Wait, something's strange here. From the matrix, I get:

Last edited by gAr (2011-09-02 21:27:36)

bobbym · 2011-09-02 21:36:23

Different answers for 20 throws also.

For the 100 throws:

The matrix:

The formula:

gAr · 2011-09-02 21:56:22

Hi bobbym,

Yes.
Maybe the formula is wrong and not that simple.

bobbym · 2011-09-02 22:00:17

Hi;

I think the formula is wrong ( matrix is right ). For 20 throws it is off too much
from the simulation to be correct.

gAr · 2011-09-02 22:03:00

Yes, and worse, for 10 throws the formula gives a negative value!

bobbym · 2011-09-02 22:07:39

Hi gAr;

That is true. The matrix is keeping in step with the simulation.

gAr · 2011-09-02 22:11:21

Hi bobbym,

Yes.
To find a direct formula looks confusing.

bobbym · 2011-09-02 22:52:40

Hi gAr;

Try this:

For k = 6,7,8,9,...

gAr · 2011-09-02 22:59:57

Hi bobbym,

Looks good!
Did you derive it from the markov chain?

bobbym · 2011-09-02 23:02:07

Hi gAr;

Yes, it is from the row 1 column 7 of the markov chain. It looks like a
principle of inclusion - exclusion. Which is why the other formula did
not work. It overcounted!

gAr · 2011-09-02 23:14:06

Hi bobbym,

Okay, thanks for the formula.

bobbym · 2011-09-02 23:18:17

Hi gAr;

Your welcome!

bobbym · 2011-09-03 00:23:53

Hi gAr;

For the same problem.

gAr · 2011-09-03 01:11:12

Hi bobbym,

Thanks!
Do we find it by setting up simultaneous equations?

bobbym · 2011-09-03 03:47:49

Hi gAr;

That would be the way I would find the GF. Then you can use the
Wilf method from genratingfunctionology ( I think ) to get the
recurrence. Or maybe use simulutaneous equations on that too.

gAr · 2011-09-03 04:48:45

Hi bobbym,

Okay.
But I still couldn't get it...

bobbym · 2011-09-03 04:54:07

Hi gAr;

Which part?

gAr · 2011-09-03 05:05:57

Hmmm, I think I got the recurrence by solving simultaneous equations for row 1 column 7.

So all I need to do is to solve the linear recurrence relation for the closed form. Is that how you did it too?
Anyway, GF is remaining now...

bobbym · 2011-09-03 05:10:54

Hi gAr;

No, I just used mathematica commands.

To get the GF, you would fit for some specific form. Looking at the recurrence
would give you the hints you need.

gAr · 2011-09-03 05:21:06

Hi,

Okay.
You mean mathematica has built-in commands to get the recurrence from the matrix or you wrote a module for that?

For the GF part, I forgot that we can get the GF easily from a recurrence.
So everything is solved now!
Very useful these markov chains!

bobbym · 2011-09-03 05:25:15

Hi gAr;

I wrote some routines to do some of it. Mathematica does have
built in functions to get GF's, recurrences and
general terms of sequences.

Math Is Fun Forum

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