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**lucik1900****Member**- Registered: 2013-03-15
- Posts: 12

You are holding a party at home, and everyone is about to participate in the following game.

Each person will write their name on a card. All the cards will then be collected and randomly redistributed (one per person). If anyone gets back the card with their own name then all the cards will be collected and randomly distributed again, and this process will be repeated until no-one is holding the card with their own name.

When everyone has a card with someone elses name on it, you will call out the name on your card. The called person will then call out the name on their card, and so on, until finally your own name is called out.

If anyones name does not get called out at some stage during this game, they will have to drink a whole 1 litre bottle of vodka by midnight.

(a) Suppose that there are five people at your party (including yourself).

Find the probability that no one will have to drink 1 litre of vodka by midnight.

Then find the expected number of people

who will have to drink 1 litre of vodka by midnight.

(b) Derive general formulas for the probability and expectation in (a), ones which are correct for any number of people attending your party (i.e. 2, 3, 4, etc).

Then apply these two formulas to the cases where there are 2, 3, 4, 5, 10 and 100 people at your party, respectively.

Really need help~

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

Hi;

That is a lot of stuff up there. Looks like an entire course.

*Last edited by bobbym (2013-03-15 19:28:22)*

**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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**cmowla****Member**- Registered: 2012-06-14
- Posts: 57

This problem doesn't make sense.

lucik1900 wrote:

You are holding a party at home, and everyone is about to participate in the following game.

Each person will write their name on a card. All the cards will then be collected and randomly redistributed (one per person).

Think of the "randomly distributed" deck of cards be a random permutation scramble of *n* objects, where *n* is the number of people you have in your group. If no one gets a card with his or her own name on it, then this is a **derangement**.

The number of possible ways in which a deck of cards can be distributed to the party members at random is *n*!. The number of derangement distributions/scrambles is subfactorial of *n* or !*n*.

Now this is where the problem doesn't make sense to me.

lucik1900 wrote:

When everyone has a card with someone else's name on it, you will call out the name on your card.

Okay, so this is telling me that only when there is a derangement will any names will be called out.

lucik1900 wrote:

If anyones name does not get called out at some stage during this game, they will have to drink a whole 1 litre bottle of vodka by midnight.

The phrasing of this is odd because by the quoted text above this one, no one's name will be called when there isn't a derangement. So when the first round occurs where at least someone has his or her own card, then no body's name will be called and therefore everyone will have to drink before midnight (is this question a joke or an encrypted message saying that everyone is going to a drink vodka before midnight at a party?).

Since we weren't given a specific number of rounds the game has, and since the name card distributions are done at random, we can definitely assume that when enough rounds are executed, everyone will have to drink before midnight.

So no matter what, every round that "counts" towards determining the fate of any given players blood alcohol levels is one in which everyone will have to drink vodka.

So all I can say is the probability that no one person gets back his or her card in a single distribution round is subfactorial(n)/n!, which is a general formula which holds true for any *n*.

So if we were given that there were 15 rounds in the game, and we had say, 5 players, then the probability that 0 persons will not have gotten back his or own card in any of those 15 rounds would be

which pretty much guarantees that all people will have to drink before midnight.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

Hi;

*Last edited by bobbym (2013-03-15 20:38:15)*

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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**lucik1900****Member**- Registered: 2013-03-15
- Posts: 12

Thanks for your work. I still do that .

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

Hi;

*Last edited by bobbym (2013-03-16 19:43:19)*

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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**lucik1900****Member**- Registered: 2013-03-15
- Posts: 12

wow..~That is a reasonable formulae. I am checking it. Thanks for your work. Discuss it with you later.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

That is a reasonable formulae.

I am forced to disagree. If this problem came out of some real world application I would gladly bet my dough on the answer anywhere in town. But since it probably came from academia it is highly unlikely to be correct. They just do not have the imagination to create a problem that is as bizarre as one of my solutions.

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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**lucik1900****Member**- Registered: 2013-03-15
- Posts: 12

bobbym wrote:

Hi;

That is a lot of stuff up there. Looks like an entire course.

how do you get that? can you show me the working?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

I will show you what I did please hold on.

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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**lucik1900****Member**- Registered: 2013-03-15
- Posts: 12

bobbym wrote:

Hi;

show me the working PLZ

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

Hi;

For a) I wrote a program that counted them all up for k = 3, 4, 5, 6, 7, ... and played spot the pattern. The pattern was easy to discover and you will have to prove it by induction.

Anyways it produced a formula which now answers two questions.

(b) Derive general formulas for the probability in (a)

Plugging in k = 5 answers a.

*Last edited by bobbym (2013-03-19 00:01:13)*

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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**lucik1900****Member**- Registered: 2013-03-15
- Posts: 12

hi

Can you show me the working of part( a) without using programming.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

The programming allows me to play spot the pattern much more than without it. Basically without it the general form produced is very difficult if not impossible for all but the best of the best.

But to answer a) when stuck it is not a disgrace to list them all and do it by hand. Anything beats no answer!

Calling each person 1 to 5, the number of different ways to arrange them with no fixed points is called derangements. For 5 of anything there are 44 of them.

*Last edited by bobbym (2013-03-19 00:30:39)*

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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**NobodyButYou****Member**- Registered: 2013-03-23
- Posts: 8

Hi, would you be able to show me the working for b, for the general formula of expected number of drunk please? thank you very much .

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**NobodyButYou****Member**- Registered: 2013-03-23
- Posts: 8

bobbym wrote:

Hi;

If possible, can you please write this in a simplier format? as I don't quite understand these notations. Thank you very much.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

Hi;

I used the idea of spotting the pattern. This is in the style of the new experimental mathematics or should I say the old.

We could then work on a method to prove the formula, say by induction.

Very often in combinatorics formulas and answers are found in this manner.

For your second question, that is the simplest form I know. What do you need explained?

*Last edited by bobbym (2013-03-23 20:33:02)*

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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**NobodyButYou****Member**- Registered: 2013-03-23
- Posts: 8

I think there's a gamma distribution in there? Unfortunately I haven't learned that yet. Would be okay if you rewrite the formula without the gamma distribution? thank you very much.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

Hi;

here is something slightly less complicated inasmuch as there is no gamma function.

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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**NobodyButYou****Member**- Registered: 2013-03-23
- Posts: 8

Hi, sorry for bothering again, but what is j in this case?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

Hi;

It is no bother.

J is what is called a dummy variable. It is an index of summation.

I can try to get a closed form for the whole expression.

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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**NobodyButYou****Member**- Registered: 2013-03-23
- Posts: 8

Oh yes please. Thank you .

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

Then please hold on, it will take awhile.

Sorry, but the closed form is gigantic. Would a recurrence relation be okay?

*Last edited by bobbym (2013-03-23 21:35:07)*

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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**NobodyButYou****Member**- Registered: 2013-03-23
- Posts: 8

That is quite okay, thank you.

Although I do not understand the 2nd summation in the numerator. What does it compute? the probability that n people (who are gonna get drunk) don't have each other's name?

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**NobodyButYou****Member**- Registered: 2013-03-23
- Posts: 8

and why did we have to time that by (k-1)!?

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