The Wrong Hats Problem

Agnishom · 2012-05-19 23:30:00

It has been many days since I last posted. Feeling to post again

Anyway,
There had been eight men and eight hats. They were so much drunk that they could not recognise their own hats. So they picked them up in a random manner. Now what is the chance that no man will choose his own hat?
Of Course the first step is to find the factorial of 8 which is equal to the number of ways they can pick up the hats in. But it is such a large number that it would take years to write all of them down?

Can You give me the answer????

I know the answer but I dont know how to derive it. So a help would be nice

bobbym · 2012-05-19 23:41:21

Hi;

Agnishom · 2012-05-19 23:47:48

How do you get it?

anonimnystefy · 2012-05-19 23:49:32

If I am correct,than this is the derangement problem for 8 elements. I think you can find it on Wikipedia.

bobbym · 2012-05-19 23:51:26

Yes, this is a derangement problem. You can first use the approximation of

There are many formulas to generate the number of derangements.

http://en.wikipedia.org/wiki/Derangement

Agnishom · 2012-05-19 23:58:22

Thnks I will read the article but will you explain what is approximation of 1/e?

bobbym · 2012-05-19 23:59:19

As n approaches infinity the probability of a derangement approaches 1 / e.

http://en.wikipedia.org/wiki/Derangement

Agnishom · 2012-05-20 00:10:15

http://upload.wikimedia.org/wikipedia/en/math/3/f/6/3f619a61786028a59a524136391f52f0.png

How do you prove the above?
Please explain step by step
Can't Understand

bobbym · 2012-05-20 00:17:02

Hi;

I am sorry but I can not see that picture clearly.

anonimnystefy · 2012-05-20 00:20:24

Hi bobbym

It says !n=(n-1)*(!(n-1)+!(n-2))

Last edited by anonimnystefy (2012-05-20 00:20:43)

Agnishom · 2012-05-20 00:21:22

The formula is:
!n=(n-1) (!(n-1) + !(n-2))

Last edited by Agnishom (2012-05-20 00:21:58)

bobbym · 2012-05-20 00:25:28

Hi Agnishom;

That is the recurrence relation for the number of derangements. Did you read this:

http://en.wikipedia.org/wiki/Derangement

About half way down the page they will describe the process. It is difficult.

Agnishom · 2012-05-20 00:34:12

They wrote:

Suppose that there are n persons numbered 1, 2, ..., n. Let there be n hats also numbered 1, 2, ..., n. We have to find the number of ways in which no one gets the hat having same number as his/her number. Let us assume that first person takes the hat i. There are n − 1 ways for the first person to choose the number i. Now there are 2 options:
1. Person i does not take the hat 1. This case is equivalent to solving the problem with n − 1 persons n − 1 hats: each of the remaining n − 1 people has precisely 1 forbidden choice from among the remaining n − 1 hats (i's forbidden choice is hat 1).
2. Person i takes the hat of 1. Now the problem reduces to n − 2 persons and n − 2 hats.
From this, the following relation is derived:
!n = (n - 1) (!(n-1) + !(n-2)).\,

Now would you please explain point 1 and 2

bobbym · 2012-05-20 00:38:55

Hi Agnishom;

That is a standard book way of getting a recurrence form. I know it as East-West analysis from a famous book written by the late Herbert S. Wilf. Though I have read it many times I am unable to do it or even explain it. I am sorry about that, wish I could do more here.

Agnishom · 2012-05-20 00:46:04

Dear Bobbym,
Do you own a copy of Amusement in Mathematics by Henry. E. Dudeney?

bobbym · 2012-05-20 00:49:05

Hi;

Yes.

Agnishom · 2012-06-24 14:32:09

Actually I don't understand these two lines:
1. Person i does not take the hat 1. This case is equivalent to solving the problem with n − 1 persons n − 1 hats: each of the remaining n − 1 people has precisely 1 forbidden choice from among the remaining n − 1 hats (i's forbidden choice is hat 1).
2. Person i takes the hat of 1. Now the problem reduces to n − 2 persons and n − 2 hats.

and why hat 1? Why dont we extend the list with Hat 2 and Hat 3 and forever???

anonimnystefy · 2012-06-24 14:46:30

You could've said that we start from Person 2 for the casework, but that wouldn't change much except a few indexes. The point of that casework is to see what are the values weget for each case that exists.

Agnishom · 2012-06-27 16:43:01

Actually I still don't get it

biffboy · 2012-07-04 21:20:23

Agnishom wrote:

It has been many days since I last posted. Feeling to post again
Anyway,
There had been eight men and eight hats. They were so much drunk that they could not recognise their own hats. So they picked them up in a random manner. Now what is the chance that no man will choose his own hat?
Of Course the first step is to find the factorial of 8 which is equal to the number of ways they can pick up the hats in. But it is such a large number that it would take years to write all of them down?
Can You give me the answer????

I know the answer but I dont know how to derive it. So a help would be nice

I would have thought the answer was simply 7/8*6/7*5/6*4/5*.3/4*2/3*1/2*1/1=3/4

bobbym · 2012-07-04 22:36:48

Hi biffboy;

Welcome to the forum! Please check the link in post #12.

Agnishom · 2012-07-09 13:48:55

Hi biffboy
Why do you calculate that way?
Pls Explain

anonimnystefy · 2012-07-09 14:07:18

Hi Agnishom

biffboy's solution isn't correct.

bobbym · 2012-07-09 20:48:00

Hi Agnishom;

The calculating part is a derangement problem. biffboy's solution is not correct.

Math Is Fun Forum

#1 2012-05-19 23:30:00

The Wrong Hats Problem

#2 2012-05-19 23:41:21

Re: The Wrong Hats Problem

#3 2012-05-19 23:47:48

Re: The Wrong Hats Problem

#4 2012-05-19 23:49:32

Re: The Wrong Hats Problem

#5 2012-05-19 23:51:26

Re: The Wrong Hats Problem

#6 2012-05-19 23:58:22

Re: The Wrong Hats Problem

#7 2012-05-19 23:59:19

Re: The Wrong Hats Problem

#8 2012-05-20 00:10:15

Re: The Wrong Hats Problem

#9 2012-05-20 00:17:02

Re: The Wrong Hats Problem

#10 2012-05-20 00:20:24

Re: The Wrong Hats Problem

#11 2012-05-20 00:21:22

Re: The Wrong Hats Problem

#12 2012-05-20 00:25:28

Re: The Wrong Hats Problem

#13 2012-05-20 00:34:12

Re: The Wrong Hats Problem

#14 2012-05-20 00:38:55

Re: The Wrong Hats Problem

#15 2012-05-20 00:46:04

Re: The Wrong Hats Problem

#16 2012-05-20 00:49:05

Re: The Wrong Hats Problem

#17 2012-06-24 14:32:09

Re: The Wrong Hats Problem

#18 2012-06-24 14:46:30

Re: The Wrong Hats Problem

#19 2012-06-27 16:43:01

Re: The Wrong Hats Problem

#20 2012-07-04 21:20:23

Re: The Wrong Hats Problem

#21 2012-07-04 22:36:48

Re: The Wrong Hats Problem

#22 2012-07-09 13:48:55

Re: The Wrong Hats Problem

#23 2012-07-09 14:07:18

Re: The Wrong Hats Problem

#24 2012-07-09 20:48:00

Re: The Wrong Hats Problem

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