[bonanova]

She knew picking Friday the 13^th for her first day on the job was an invitation to disaster.

But with more valor than discretion, Ms. Jones showed up.

Her first task was stuffing 1000 envelopes with personalized letters.

She got the letters alphabetized correctly, but alas, the envelopes were fed into the stuffing machine in a completely random order.

"Well, maybe most of them went to the right person," she thought, aloud.

But her colleague Ms. Smith did a little calculation on the back of the 1001^st envelope[*] and thought otherwise.

How many did the sharp Ms. Smith predict would get to the correct recipient?

Do the math if you want, or just guess a percentage.

[*] The 1001st envelope was blank, and was not part of the mailing. Figure of speech.

[itachi-san]

Well since I can guess:

None of them would get filled correctly (most likely) because of the extremely low probability.

[Nikyma]

13%

[Guest]

.03 percent.

[bonanova]

No one has it yet.

What if the number of envelopes were smaller - say 4?

[unreality]

This is just like my Zarball Competition puzzle, right? with the properties of e

[Guest]

I think it is a very low number

For 4 it should be (1/4)(1/3)(1/2)(1)

Similarly for 1000

(1/1000)(1/999)....(1/2)(1)

[Guest]

No one has it yet.

What if the number of envelopes were smaller - say 4?

They would'nt all fit in - jaws refrence

Were gonna need a bigger boat bigger envelopes!

Edited April 6, 2008 by Lost in space

[Guest]

No one has it yet.

What if the number of envelopes were smaller - say 4?

I think ....

(1/4)*(1/3)*(1/2) = 1/24.

As the envelopes can be ordered in four different ways:

(1/24)*4 = 1/6 = 0.167 ... (and that is for four)

For 1000 envelopes, it is almost zero.

[itachi-san]

I think ....

(1/4)*(1/3)*(1/2) = 1/24.

As the envelopes can be ordered in four different ways:

(1/24)*4 = 1/6 = 0.167 ... (and that is for four)

For 1000 envelopes, it is almost zero.

Shouldn't the ordering of the envelopes decrease the probability?

Let's start with 4 like bonanova suggested. I agree the first envelope should be 1/4, but then for the second envelope it's not 1/3 because the second name may have gone into the first envelope and is no longer available to make the second envelope correct.

[Guest]

I need to think about this a bit, but everyone so far is missing the point that the question isn't "What is the chance ALL of them went to the right person?" but rather "What is the expected value of number of envelopes that went to the right person?" Expected value, meaning that if she repeated this process over and over, what is the average number of envelopes that would go to the right person.

[bonanova]

This is just like my Zarball Competition puzzle, right? with the properties of e

Kind of, but the "e" part has to do with incorrect matchings.

And today's question asks for something not asked for in the Zarball series.

I went back to check those numbers, and today's question is answered there if you combine the numbers correctly. Vague hint - the numbers are in the 2nd spoiler here.

For another "e" puzzle, check this one.

[bonanova]

I need to think about this a bit, but everyone so far is missing the point that the question isn't "What is the chance ALL of them went to the right person?" but rather "What is the expected value of number of envelopes that went to the right person?" Expected value, meaning that if she repeated this process over and over, what is the average number of envelopes that would go to the right person.

rhapsodize has stated the puzzle precisely.

Perhaps some of the lower estimates posted so far were answering the chances all of them were correct.

That probability indeed is small.

Today's answer, if you want a vague clue, is surprising.

[Guest]

rhapsodize has stated the puzzle precisely.

Perhaps some of the lower estimates posted so far were answering the chances all of them were correct.

That probability indeed is small.

Today's answer, if you want a vague clue, is surprising.

I believe the answer is 1, and I believe that is the answer no matter what the number of envelopes in this problem, whether its 1000 or 4 or whatever.

[bonanova]

I believe the answer is 1, and I believe that is the answer no matter what the number of envelopes in this problem, whether its 1000 or 4 or whatever.

rhapsodize has it ...

Interesting result!!

[Guest]

rhapsodize has it ...

Interesting result!!

Would love to see the mathematical proof of this. I know it has to do with n! = 1 + some crazy formula involving combinations, and must be able to be proved by induction, but I spent 10 minutes on trying to prove it before giving up.

[bonanova]

It's probably as simple as this:

Each of the N letters has a 1/N chance of going into the correct envelope.

The successes add because these are independent events.

That is, each letter has its own target envelope; none of the probabilities are conditional on the fate of the other letters.

Here's how it progresses for 1-4.

One:

1 -> 1! = 1 case; 1 correct. E = 1.

Two:

1 2

2 1 -> 2! = 2 cases; 2 correct. E = 1.

1 has a 1/2 chance of landing in column 1.

2 has a 1/2 chance of landing in column 2.

Three:

1 2 3

1 3 2

2 1 3

2 3 1

3 1 2

3 2 1 -> 3! = 6 cases; 6 correct. E = 1.

1 has a 1/3 chance of landing in column 1.

2 has a 1/3 chance of landing in column 2.

3 has a 1/3 chance of landing in column 3.

Four:

1 2 3 4

1 2 4 3

1 3 2 4

1 3 4 2

1 4 2 3

1 4 3 2

2 1 3 4

2 1 4 3

2 3 1 4

2 3 4 1

2 4 1 3

2 4 3 1

3 1 2 4

3 1 4 2

3 2 1 4

3 2 4 1

3 4 1 2

3 4 2 1

4 1 2 3

4 1 3 2

4 2 1 3

4 2 3 1

4 3 1 2

4 3 2 1 - 4! = 24 cases; 24 correct. E = 1.

1 has a 1/4 chance of landing in column 1.

2 has a 1/4 chance of landing in column 2.

3 has a 1/4 chance of landing in column 3.

4 has a 1/4 chance of landing in column 4.

Another way to look at it:

For any N,

Each number lands in the the correct column [N-1]! times.

Multiply that by N numbers and you have N! correct placements.

Since there are N! cases, E is always 1.