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 anyone’s 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

Thanks for your work. I still do that .

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

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?

bobbym wrote:

Hi;

show me the working PLZ

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

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 .

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 (2013-03-24 19:33:02)

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

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.

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 (2013-03-24 20:35:07)

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