View Full Version : Very tricky logic question

Claudius

02-26-2011, 08:35 PM

Suppose the manager of a company has 6 letters to send. He gives his secretary 6 letters and 6 envelopes and asks her to put each letter into the correct envelope.

The secretary, being sneaky, puts the letters randomly into the 6 envelopes (assuming 1 letter per envelope).

What is the probability that exactly 5 letters were put into the correct envelopes?

billnepill

02-26-2011, 08:51 PM

0,00139 % ?

DeShaun

02-26-2011, 09:00 PM

The sixth letter (and the only remaining envelope) would be correctly matched, if the other five already were. It would be impossible to match "exactly" five out of six correctly.

borg number one

02-26-2011, 09:00 PM

I would think the answer is 1/6 raised to the 6 power, since you have a 1/6 chance with each envelope and you have to get all six envelopes right if you are to get 5 right. How could you get 5 right and not 6? So, my answer would be .166 x .166 x .166 x .166 x .166 x .166= .0000214335.

Claudius

02-26-2011, 09:06 PM

DeShaun is correct...

Steady Eddy

02-26-2011, 09:07 PM

What is the probability that exactly 5 letters were put into the correct envelopes?The probability is 0%.

GetBetterer

02-26-2011, 09:33 PM

Here's another question: Who killed Hamlet's father, (Old) Hamlet? O_o

Polaris

02-26-2011, 10:24 PM

The sixth letter (and the only remaining envelope) would be correctly matched, if the other five already were. It would be impossible to match "exactly" five out of six correctly.

The probability is 0%.

Bingo! We have 2 winners.

Sentinel

02-26-2011, 11:37 PM

Here's another question: Who killed Hamlet's father, (Old) Hamlet? O_o

Isn't Hamlet the name of the dog that Hagar's son owns ?

EndLy

02-27-2011, 12:04 AM

I would think the answer is 1/6 raised to the 6 power, since you have a 1/6 chance with each envelope and you have to get all six envelopes right if you are to get 5 right. How could you get 5 right and not 6? So, my answer would be .166 x .166 x .166 x .166 x .166 x .166= .0000214335.

even though the answer has already been stated, I took a different approach by first taking 1/6 then assuming that is right, then there would be a 1/5 chance to get the next one correct so and the next 1/4 and so on so (1/6)(1/5)(1/4)(1/3)(1/2) = 0.1375% if I'm not mistaken, but obviously if you got the first 5 right, then the last would be correct so you can't get 5 correct. only 4 max... big DUH moment to myself

fruitytennis1

02-27-2011, 05:41 AM

Bob and Rob are going to school.

Bob runs half the time and walks half the time

Rob runs half the distance and walks half the distance

Assuming they both walk and run at the same speed who will reach school first?

borg number one

02-27-2011, 05:56 AM

even though the answer has already been stated, I took a different approach by first taking 1/6 then assuming that is right, then there would be a 1/5 chance to get the next one correct so and the next 1/4 and so on so (1/6)(1/5)(1/4)(1/3)(1/2) = 0.1375% if I'm not mistaken, but obviously if you got the first 5 right, then the last would be correct so you can't get 5 correct. only 4 max... big DUH moment to myself

You are right, we are on the same page. Yet, I ran that calculation again. It's not 1/6 raised to the 6th power as I thought (if you're trying to see the probability of getting all 6 right, not 5). It is as you stated, 1/6 x 1/5 x 1/4 x 1/3 x 1/2 (or you could also add the x 1 with the 6th envelope). Yet, when you do the math, I get that the probability of getting all 6 envelopes correct is .001389, so the probability of getting all 6 is much less than 1%. Very interesting. This is tricky, because it is impossible to get exactly five right. Great thread.

Steady Eddy

02-27-2011, 06:12 AM

Bob and Rob are going to school.

Bob runs half the time and walks half the time

Rob runs half the distance and walks half the distance

Assuming they both walk and run at the same speed who will reach school first?Bob would get there first.

Bob goes at his higher speed (running) half the time. Does Rob spend half his time running as well? No, he spends less than half his time running. Rob walks half the distance, and runs half the distance. Since walking is slower, he'll spend more time walking to cover the same ground as running. He is then walking over half the time. So Bob makes better time.

fruitytennis1

02-27-2011, 06:17 AM

Bob would get there first.

Bob goes at his higher speed (running) half the time. Does Rob spend half his time running as well? No, he spends less than half his time running. Rob walks half the distance, and runs half the distance. Since walking is slower, he'll spend more time walking to cover the same ground as running. He is then walking over half the time. So Bob makes better time.

Yep. There is a more difficult version where you add in bogus variables and something else but I'm too lazy to make that up.

Andres

02-27-2011, 06:40 AM

If the company is a big company, the chances of the secretary being fired rises to 89.7%

If the company is not so big, the secretary has a rough 43% chance of keeping her job.

On the other hand, if the secretary is the manager's wife, there's a 77% chance all she'll get is a slap on the wrist and a smile. And of course, the manager won't get laid that night.

So that'd be it, give or take

BobFL

02-27-2011, 06:56 AM

0,00139 % ?

This is the correct answer. It doesn't get any simpler than that:

1/6! = 1/(6x5x4x3x2x1) = 0.001388888888888888888...

fruitytennis1

02-27-2011, 07:13 AM

This is the correct answer. It doesn't get any simpler than that:

1/6! = 1/(6x5x4x3x2x1) = 0.001388888888888888888...

No. If you get 5/6 envelops right you obviously got the 6th envelop right too. So its not possible to get exactly 5/6 right.

borg number one

02-27-2011, 07:15 AM

No. If you get 5/6 envelops right you obviously got the 6th envelop right too. So its not possible to get exactly 5/6 right.

Thats' the right calculation for getting all six envelopes right though. It's 0.0014 for getting six right (rounded, basically a tenth of 1%). Yet, to get five right is technically impossible, since if you have five right, you must necessarily have six right.

BobFL

02-27-2011, 07:23 AM

No. If you get 5/6 envelops right you obviously got the 6th envelop right too. So its not possible to get exactly 5/6 right.

Please...that's irrelevant for this math problem. That was not the question. The question was about probability and they said 5/5 just to add some additional trickiness :) The answer given in #2 is correct.

BobFL

02-27-2011, 07:36 AM

Thats' the right calculation for getting all six envelopes right though. It's 0.0014 for getting six right (rounded, basically a tenth of 1%). Yet, to get five right is technically impossible. It is impossible to get exactly five right though, since if you have five right, you must necessarily have six right.

What do you mean by "technically impossible"? Everything is possible (think Verdasco winning the slam) and it is all about probability. It can be super-low or very high but it is always a possibility. Some of you guys are focusing on irrelevant part of this question. Of course that if you match 5 you will match 6 but that is not important at all. I mean, it is important to get the right number becasue at least 5 out of 10 people will use 5! instead of 6! and that is the main and only trick about this question. It is 1 divided by 6 factorial. I am 100% sure (probabilty that I am not sure is 0 ;)).

borg number one

02-27-2011, 07:51 AM

What do you mean by "technically impossible"? Everything is possible (think Verdasco winning the slam) and it is all about probability. It can be super-low or very high but it is always a possibility. Some of you guys are focusing on irrelevant part of this question. Of course that if you match 5 you will match 6 but that is not important at all. I mean, it is important to get the right number becasue at least 5 out of 10 people will use 5! instead of 6! and that is the main and only trick about this question. It is 1 divided by 6 factorial. I am 100% sure (probabilty that I am not sure is 0 ;)).

What I mean is that if you want to look at it technically, you can't get exactly 5 right. You can get 4 right, or 6 right, but not 5 right. So, the right answer does not leave room for the assumption that yes, to have 5 right, you must necessarily have all six envelopes correct. So, to have exactly five right, and not that sixth one as well is impossible. The probability is 0.0% for 5, which answers the call of the central, question posed by the OP. I am 100% sure and so are you, because we're both saying the same thing lol.

BobFL

02-27-2011, 08:15 AM

What I mean is that if you want to look at it technically, you can't get exactly 5 right. You can get 4 right, or 6 right, but not 5 right. So, the right answer does not leave room for the assumption that yes, to have 5 right, you must necessarily have all six envelopes correct. So, to have exactly five right, and not that sixth one as well is impossible. The probability is 0.0% for 5, which answers the call of the central, question posed by the OP. I am 100% sure and so are you, because we're both saying the same thing lol.

Hahaha, that is 100% true! ;) You are smart guy and you saw that 5 matched = 6 matched hence you saw the main 'trick' about this math-problem :)

Claudius

02-27-2011, 08:18 AM

Alright, here's new one.

A man stares at a portrait and says "Brothers and sisters I have none, but this man's father is my father's son.". Who is he staring at?

Steady Eddy

02-27-2011, 08:21 AM

Alright, here's new one.

A man stares at a portrait and says "Brothers and sisters I have none, but this man's father is my father's son.". Who is he staring at?

It's his son.

Take the phrase "my father's son", that is the man himself because he has no siblings. Since "my father's son" = "me" replace those three words with the one word. "this man's father is me", then it's clear that the man must be his son.

GetBetterer

02-27-2011, 08:40 AM

Sentinel:

Isn't Hamlet the name of the dog that Hagar's son owns ?

I'm talking about Shakespeare's play.

Claudius

02-27-2011, 08:57 AM

Jeebus....

Okay, here's something that may take awhile.

Which of the following words does not belong?

act bone came deified eerie gore later maid no nuclear oodles pot use wary

Steady Eddy

02-27-2011, 11:05 AM

Jeebus....

Okay, here's something that may take awhile.

Which of the following words does not belong?

act bone came deified eerie gore later maid no nuclear oodles pot use waryTaking the first letter of each word we have: a b c d e g l m n n o p u w. That's so close to the alphabet that I think I'm on to something. There's two n s in a row, so is the 2nd n the one that doesn't belong? I'll guess it's "nuclear".

Claudius

02-27-2011, 11:55 AM

^No. Think about the structure of each word.

GetBetterer

02-27-2011, 12:14 PM

It's bone.

If you switch the first 2 letters of each word, you get another word except for bone.

Polaris

02-27-2011, 01:49 PM

Please...that's irrelevant for this math problem. That was not the question. The question was about probability and they said 5/5 just to add some additional trickiness :) The answer given in #2 is correct.

No, unfortunately, you are wrong here. The question clearly specified that you had to find the probability of getting exactly 5 correct. This means 5 correct and 1 wrong. That combo is impossible.

If the questioner had asked for the probability of getting at least 5 correct, then your answer would be fine.

borg number one

02-27-2011, 02:05 PM

It's bone.

If you switch the first 2 letters of each word, you get another word except for bone.

Good catch there. That must be it.

Jeebus....

Okay, here's something that may take awhile.

Which of the following words does not belong?

act bone came deified eerie gore later maid no nuclear oodles pot use wary

I'm a simpleton, so my basic logic tells me the two words that don't have a vowel as the second letter. Therefore: Act & Use:confused:

BobFL

02-27-2011, 02:15 PM

No, unfortunately, you are wrong here. The question clearly specified that you had to find the probability of getting exactly 5 correct. This means 5 correct and 1 wrong. That combo is impossible.

If the questioner had asked for the probability of getting at least 5 correct, then your answer would be fine.

What is the correct answer then?

Polaris

02-27-2011, 02:17 PM

What is the correct answer then?

Correct answer is what DeShaun and Eddy gave. The probability of getting exactly 5 correct is zero.

Claudius

02-27-2011, 02:37 PM

GetBetterer got it.

Now this:

ample crib dress gender grate here may mute pound peach tract vex

GetBetterer

02-27-2011, 02:56 PM

That one's easier than the first.

It's crib.

All the other words can get some sort of prefix and still make another letter (adhere, immute, impound, impeach, intract etcetera).

fruitytennis1

02-27-2011, 04:30 PM

That one's easier than the first.

It's crib.

All the other words can get some sort of prefix and still make another letter (adhere, immute, impound, impeach, intract etcetera).

Genius!

Here is a well known riddleish thingy

A farmer was going to town with a fox, a goose and a sack of corn. When he came to a stream, he had to cross in a tiny boat, and could only take across one thing at a time. However, if he left the fox alone with the goose, the fox would eat the goose, and if he left the goose alone with the corn, the goose would eat the corn. How does he get them all safely over the stream?

Genius!

Here is a well known riddleish thingy

A farmer was going to town with a fox, a goose and a sack of corn. When he came to a stream, he had to cross in a tiny boat, and could only take across one thing at a time. However, if he left the fox alone with the goose, the fox would eat the goose, and if he left the goose alone with the corn, the goose would eat the corn. How does he get them all safely over the stream?

Won't answer cause I know the answer, but these guys will tear this one up.

RoddickAce

02-27-2011, 04:42 PM

Genius!

Here is a well known riddleish thingy

A farmer was going to town with a fox, a goose and a sack of corn. When he came to a stream, he had to cross in a tiny boat, and could only take across one thing at a time. However, if he left the fox alone with the goose, the fox would eat the goose, and if he left the goose alone with the corn, the goose would eat the corn. How does he get them all safely over the stream?

Tell the goose to fly/swim over while he brought the corn over, go back and tell the goose to fly/swim back, then bring the fox, and tell the goose to fly/swim to the other side again?

We'd need a pretty obedient goose though lol.

edit: WAIT got it, he can carry the goose over and get back to get the corn over, but on his way back, bring the goose back, and bring the fox over, then go back and get the goose over.

fruitytennis1

02-27-2011, 04:44 PM

Tell the goose to fly/swim over while he brought the corn over, go back and tell the goose to fly/swim back, then bring the fox, and tell the goose to fly/swim to the other side again?

We'd need a pretty obedient goose though lol.

Actually really close...

The goose cant fly/swim though.

Tell the goose to fly/swim over while he brought the corn over, go back and tell the goose to fly/swim back, then bring the fox, and tell the goose to fly/swim to the other side again?

We'd need a pretty obedient goose though lol.

edit: WAIT got it, he can carry the goose over and get back to get the corn over, but on his way back, bring the goose back, and bring the fox over, then go back and get the goose over.

Yeah, 4 trips.

Claudius

02-27-2011, 05:01 PM

How about this:

Take the goose over, comeback and take the corn over, take the goose back, bring the fox over, and then come back to bring the goose..

Claudius

02-27-2011, 05:05 PM

There are 20 people in an empty, square room. Each person has full sight of the entire room and everyone in it without turning his head or body, or moving in any way (other than the eyes). Where can you place an apple so that all but one person can see it?

There are 20 people in an empty, square room. Each person has full sight of the entire room and everyone in it without turning his head or body, or moving in any way (other than the eyes). Where can you place an apple so that all but one person can see it?

On top of one person's head?

Claudius

02-27-2011, 05:15 PM

Yes 10char

Yes 10char

Yeah, I'm okay with the visual ones, it's the code breaker one's that I'm too simple for.

BobFL

02-27-2011, 05:24 PM

Yeah, the same thing as putting the label on one person's forehead. Everyone can read it but the person with the label :)

fruitytennis1

02-27-2011, 05:40 PM

"This statement is false"

fireice

03-01-2011, 04:10 PM

Let's try one more. This probably won't take you that long, BUT...

A man was born before his father, killed his mother, married his sister, and was considered perfectly normal by everyone he knew. Why?

fruitytennis1

03-01-2011, 04:43 PM

Let's try one more. This probably won't take you that long, BUT...

A man was born before his father, killed his mother, married his sister, and was considered perfectly normal by everyone he knew. Why?

Everyone he knew didn't know this?

tribunal4555

03-01-2011, 04:57 PM

Let's try one more. This probably won't take you that long, BUT...

A man was born before his father, killed his mother, married his sister, and was considered perfectly normal by everyone he knew. Why?

He was adopted? Only thing I can think of for being born before his father lol

edit: WAIT I THINK I GET IT!

A man was born before his father- the father was there during his mom's labor? as in, born in front of his father?

killed his mother- the mother then would have had to die during labor

married his sister- he became a priest and "married" her, as in performed the ceremony?

that makes more sense than adoption, at least... although what if the sister was a stepsister because his father remarried? then it would be satisfied too wouldn't it? pretty sure about the first two though

fruitytennis1

03-01-2011, 05:02 PM

^Good search rocks don't it^

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