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#1 2016-03-13 16:59:53

912324434
Member
Registered: 2016-03-13
Posts: 2

about statistics

Problem 5
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, they will swap cards with the person closest to them in the room.
When everyone has a card with someone else’s 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 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.
(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. Present your results in a table.

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#2 2016-04-18 05:05:53

thickhead
Member
Registered: 2016-04-16
Posts: 1,086

Re: about statistics

We need to think of loops.5 people can form a loop of 5 among whom cards are exchanged.In this case no body will be left out when the names are called.Let the 5 people be A B C D &E.In order to find successful permutations start with A. He can have cards of any other 4.It makes 4 different ways.Suppose he gets card of C. for logical clarity now go to C.He should not have his card is already taken into account since A has selected it. In addition A's card is also taboo as it would prematurely close the loop and call for penalty. so he has only 3 choices. Suppose he chooses/gets card of E. e will have to avoid card of A,C and E. he has 2 choices. Say he gets B. Now only B is left over and he has to get the remaining card of A.This closely follows the rules given in the problem. A calls the name of C,C that of E,E that of B and finally B that of A. Even if somebody in the middle say C is picked up to start, it won't matter.So the number of ways of getting the cards comes out to be 4*3*2*1=4!=24 (factorial of one less than the number of people.) In order to find the number of ways which leads to penalty we have to think of smaller loops.
5 people can form 2 loops one of 3 and the other of 2. the number of ways these loops are formed is 5C3x2C2 =10;In each of these we have 2! ways for loop of 3 and 1! ways for the loop of 2.So total number of permutations with loops of 3 and 2 is 10x2!x1!=20.If the announcement of names comes from loop of 3 ,those in loop of 2 are left out and vice versa.There can not be loop of 4 because the 5th person will be having his own card.
So probability of not getting punishment is 24/(24+20)=24/44=6/11.
Extending this to any number of people n the no. of ways to avoid penalty is (n-1)!
however to find the number of people expecting penalty would be difficult as it it is not possible to scan through all loops of smaller size. A different algorithm is to be found out.

Last edited by thickhead (2016-04-18 19:03:49)


{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

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#3 2016-04-18 22:55:27

thickhead
Member
Registered: 2016-04-16
Posts: 1,086

Re: about statistics

To corroborate the valid number of cards (which I showed as 44) I made a chart for invalid draw of cards.
A  B  C  D  E  people at party
A  B  C  D  E (all 5 getting invalid cards)
3 people getting invalid cards (total 10 draws )
A  B  C  E  D
A  B  E  D  C
A  B  D  C  E
A  E  C  D  B
A  D  C  B  E
A  C  B  D  E
E  B  C  D  A
D  B  C  A  E
C  B  A  D  E
B  A  C  D  E
2 persons getting invalid cards (total  20 draws)
A  B  D  E  C
A  B  E  C  D
A  D  C  E  B
A  E  C  B  D
A  C  E  D  B
A  E  B  D  C
A  C  D  B  E
A  D  B  C  E
D  B  C  E  A
E  B  C  A  D
C  B  E  D  A
E  B  A  D  C
C  B  D  A  E
D  B  A  C  E
B  E  C  D  A
E  A  C  D  B
B  D  C  A  E
D  A  C  B  E
B  C  A  D  E
C  A  B  D  E
1 person getting invalid card ( chart is incomplete; it shows draw for only A to get invalid card;
Number here is 9 but altogether it is 45)
A  C  B  E  D
A  C  D  E  B
A  C  E  B  D
A  D  B  E  C
A  D  E  B  C
A  D  E  C  B
A  E  B  C  D
A  E  D  B C
A  E  D C  B

In all 76 card draws are invalid. However total no. of draws is 5!=120. So valid no. of draws =120-76=44 which was shown in the previous comment.

Last edited by thickhead (2016-04-19 00:00:54)


{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

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#4 2016-04-28 05:17:01

thickhead
Member
Registered: 2016-04-16
Posts: 1,086

Re: about statistics

This problem had been discussed in detail in earlier thread.
http://www.mathisfunforum.com/viewtopic.php?id=19100
However I have to add something.For a party of 5 No. of ways in which nobody gets punishment is 24. in the remaining 20 ways also some people escape punishment. I found out such no. ways for a person is 8.So the probability of not getting punishment =(24+8)/44=8/11. so the expected no of persons getting punishment is 4x(1-8/11)=12/11. strangely it coincides with the value given by bobbym. please note that i have multiplied probability by 4 to get expected no. getting punishment because it excludes host.


{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

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