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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,980

The expected value is 1/probability is it not?

The matrix when computed should be 13.7

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,883

http://www.wolframalpha.com/input/?t=cr … 5%2F6))%5D

How is the expected value 1/probability?

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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: 82,980

Hi;

You did not enter the problem correctly

Matrix Inverse( IdentityMatrix[6] - {{0,1,0,0,0,0 }, {0,1/6,5/6,0,0,0} ,{ 0,0,1/3,2/3,0,0} ,{ 0,0,0,1/2,1/2,0} ,{ 0,0,0,0,2/3,1/3} , {0,0,0,0,0,5/6 }}).{{1},{1},{1},{1},{1},{1}} gives the right result.

If the probability is 1 / 6 of something happening then we expect it to happen once in 6 times.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,883

My result is correct. I just didn't sum the first row.

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: 82,980

When I entered yours it says the trace is 14.7. The trace is the sum of the diagonal elements. It may or may not be correct.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,883

The trace and the sum of the first row are the same for that matrix.

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: 82,980

Yes, but are you sure that will always be the case?

This way spits the answer out in a form that is easy to understand.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,883

No, I sum the first row. WA is the one who thinks you need the trace.

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,980

Hi;

When I run your link, I get this result. That is clearly not the right answer.

For one thing he did not interpret the identitymatrisx command correctly. For another the inverse was not calculated...

I would do it the way I showed in post#128.

Taking a little break.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,883

That is not the same result I am getting when I press the link.

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,980

Supposing others get my result? Isn't my way more reliable, since we are both getting the same answer. Also, my syntax is much closer to the package's syntax.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,883

You just had multiplication by the vector (1,1,1,1,1,1) and I did the multiplication by hand. It is the same method.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,980

Hi;

I am talking about the syntax of the commands.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,883

New problems?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,980

** A deck of cards are dealt out in a circle. What is the expected number of pairs of adjacent cards which are both black?**

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,883

A standard French deck?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,980

Hi;

It is a regular deck of playing cards.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,883

Okay.

What do you get as the answer for decks of 4 and 6 cards (half of which are black)?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,980

E(4 card deck) = 2 / 3

E(6 card deck) = 6 / 5

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,883

And E(8)?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,980

E(8 card deck) = 12 / 7

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,883

I haven't been able to find a solution...

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,980

Hi;

Okay, I can give the answer if you want it.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,883

I'd rather get a method than an answer, but either is okay...

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,980

Hi;

I'd rather get a method than an answer,

Hmmmm. You should already have the answer or darn close to one. Have you forgotten how we work?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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