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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,860

here's a problem given to me by a professor of mine:

If two players are playing a game in which they draw one card each from two separate decks (shuffled) then what's the probability that they will draw at least one same card?

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: 82,724

Same type, ace, ten? Or exactly ace of hearts etc.

**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: 14,860

hi bobbym

exactly the same card.they have two standard decks without the jokers.

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: 82,724

Okay, thanks for that informatiuon. I am working on it.

**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: 14,860

how's it coming on bobbym?

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: 82,724

It is a standard problem, your professor is not very imaginative. Unfortunately I threw my solution away so please hold.

**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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**TMorgan****Member**- Registered: 2011-04-13
- Posts: 25

Do you mean on the first draw or do they work their way all through the decks?

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,860

hi bobbym

he didn't make it up.i think this problem was on a high school competition.

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

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,860

hi TMorgan

i mean through the decks.so if at any moment they draw the same card then it is a wanted event.

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: 82,724

Hi;

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: 14,860

how did you get that?

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: 82,724

Is it right?

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: 14,860

i think it is,but i am not sure.how did you get it?

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: 82,724

It is a standard derangement problem. Solved long ago. The probability of no fixed points in a permutation of {1,2,3,4,...n} approaches 1/ e.

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: 14,860

well how do you get the first probability,the 1-1/e one?

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: 82,724

That is not important because that is only an approximate answer. That is why I asked if you knew.

The exact answer I believe is:

The dn operator stands for the number of derangements.

The question was posed by Montmort in 1781 and answered by him and Bernoulli.

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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