I

The first gentleman will have a chance of not getting his umbrella of (n-1)/n. He will have the same chance of not getting his hat. So his chance of not getting both his umbrella and hat is ((n-1)^2)/n^2

 

Now I suppose it depends on whether the umbrellas and hats are distributed one by one, or if the gentlemen are randomly assigned umbrellas at the same time(is this even possible?).

 

In the second scenario the chance of no one getting his umbrella and hat back is ((n-1)^(2*n))/n^(2*n).

 

In the first scenario the solution is horridly complicated and I am unable to express my thoughts using the adequate mathematical signs and symbols, and I am even less able to type this solution into a text box in the internet.

Is diamond correct though?  Or has he found the probability that you get neither an umbrella or a hat? (the question asked for the probability you get both).

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What is the probability that no man gets back his hat and his umbrella?

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If the question asked the probability of everyone getting their hat and umbrella, the answer would be much more simple: (1/n!)^2.

 

 

OK, I guess my answer still isn't correct.

It askes the probability of everyone not getting either one of hat and umbrella.

Edited April 21, 200619 yr by Diamond

Probability of 1 gentleman getting his hat back: 1/n

Probability of 1 gentleman getting his brolly back: 1/n

Probability of two independent events happenning

P(hat ∩ brolly) = P(hat) P(brolly) = 1/n

∴ probability of 1 gentlemen retrieving his own hat = 1/n!

∴ probability of n gentlemen retrieving their hats = n/n! = 1/(n-1)!

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Probability of n gentlemen getting their hat is still 1/n!, as there is only one permutation out of n, where every element in its original position.

 

 

 

Anyway, everyone here is not correct (according to the answer Julian gave me ), I guess we're reading the question wrong in this part: "no man gets back his hat and his umbrella?". I think it doesn't mean "every gentleman doesn't get either of hat or umbrella", but rather "every gentleman gets at least one item wrong".

Well that's not what he asked.

 

Also:

Probability of 1 gentlemen retrieving his own hat = 1/n!

∴ probability of n gentlemen retrieving their hats

= a series of n elements, all being 1/n! and as it is any of them, it is an addition

= n/n!

= 1/(n-1)!

Let n = 3, so there are 3 gentlemen:

 

1, 2, 3

 

Then all possible (equally likely) permutations of hats are:

 

1, 2, 3 (1,2,3 get their hats)

1, 3, 2 (1 gets his hat)

2, 1, 3 (3 gets his hat)

2, 3, 1 (noone gets his hat)

3, 1, 2 (noone gets his hat)

3, 2, 1 (2 gets his hat)

 

Probability of 1 gentleman getting his hat:

4/6 (from example above)

4/6 ≠ 1/n! = 1/3! = 1/6

 

Probability n gentleman getting their hats, i.e. everyone gets their hat (if I understand you correctly):

1/6 (from example above)

1/6 ≠ 1/(n-1)! = 1/(3-1)! = 1/2! = 1/2

But,

1/6 = 1/n! (the only permutation where all elements are in their positions)

 

No?

I

Let n = 3, so there are 3 gentlemen:

 

1, 2, 3

 

Then all possible (equally likely) permutations of hats are:

 

1, 2, 3  (1,2,3 get their hats)

1, 3, 2  (1 gets his hat)

2, 1, 3  (3 gets his hat)

2, 3, 1  (noone gets his hat)

3, 1, 2  (noone gets his hat)

3, 2, 1  (2 gets his hat)

 

Probability of 1 gentleman getting his hat:

4/6 (from example above)

4/6 ≠ 1/n! = 1/3! = 1/6

 

Probability n gentleman getting their hats, i.e. everyone gets their hat (if I understand you correctly):

1/6 (from example above)

1/6 ≠ 1/(n-1)! = 1/(3-1)! = 1/2! = 1/2

But,

1/6 = 1/n! (the only permutation where all elements are in their positions)

 

No?

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Yellow bit is wrong.

 

Probability of all three gentlemen getting their hats back, in your example, is 1/6 (the top line out of the six). n! = 3 x 2 x 1 = 6, so 1/n! = 1/6.

 

But, the probability that one of the three gets his hat is

6 successful hat returns out of a total of 18 = 1/3, so the probability of none of the three gentlemen gets there hat back is 1-1/3 = 2/3.

 

I shall update this later ...