Logic/maths puzzles

[MoogleViper]

Here's one of my faves in case nobody has heard it...

 

You're on a quiz show. There are 3 doors. Behind 1 of the doors is a car. Behind the other two there is nothing. You randomly pick one of the doors. The host then opens one of the other empty doors. You are now given the option to stick with your choice or swap to the only other remaining door. Which is the best thing to do and why?

 

Quite a popular one so please don't spoil if you already know it!

 

I'm assuming that it's a trick question so everything I'm about to post is pointless, but I'm gonna do it anyway.

 

It's a 50/50 chance so it doesn't matter either way. But I'd feel worse if I changed to the wrong one than stuck with the wrong one. So I'd always stick.

 

This answer better be a good one and not something stupid like, "kill the host and open both doors."

 

[SPAMBOT4000]

Yeah, I promise it's not something stupid.

[MoogleViper]

Is it maths based?

[SPAMBOT4000]

It's not a trick question and is based on maths/logic.

[Supergrunch]

It's a famous probability problem, and no it's not a trick question (I'll delete such questions from this thread, they don't belong here). And your answer is wrong Moogle.

[MoogleViper]

It's a famous probability problem, and no it's not a trick question (I'll delete such questions from this thread, they don't belong here). And your answer is wrong Moogle.

 

I've just googled it and I disagree with the solution. I've only looked at the wiki article so correct me if I'm wrong but the solution they propose ignores the fact that it is a given that the opened door contains a goat.

[Supergrunch]

I've just googled it and I disagree with the solution. I've only looked at the wiki article so correct me if I'm wrong but the solution they propose ignores the fact that it is a given that the opened door contains a goat.

You and several maths professors at first, but the answer wiki gives as correct is provably right, although it's hell to explain why - I'm sure wiki gives better explanations than I ever could. You can also compuationally model the game show, and if you do enough trials, the success rate converges on the counterintuitive answer.

[jayseven]

Just to re-iterate grunch's post, mostly because I came in here to make the post only to find he'd already said it and I don't like to leave without making a scene;

 

This isn't a thread for trick answers. The other riddle thread is.

 

See? Reiterating, just with more words.

[MoogleViper]

You and several maths professors at first, but the answer wiki gives as correct is provably right, although it's hell to explain why - I'm sure wiki gives better explanations than I ever could. You can also compuationally model the game show, and if you do enough trials, the success rate converges on the counterintuitive answer.

 

Is it to do with the fact that the host doesn't open a door at random?

[SPAMBOT4000]

Is it to do with the fact that the host doesn't open a door at random?

 

Oui oui. But why.

Edited January 26, 2010 by SPAMBOT4000

[Charlie]

I'm pretty sure that same puzzle was in 'The Curious Incident Of the Dog In The Night Time' and they came up with the answer of sticking with your original choice.

 

there is a 1/3 possibility you chose the correct door to begin with. which means there is a 2/3 possibility you chose the wrong door. if you switch after being shown and empty door you have a 2/3 possibility of switching to the correct door.

 

 

Blatantly stolen from Yahoo Answers...

 

It's not that simple, surely?

 

[SPAMBOT4000]

Charlie, your first answer is wrong but the googled answer is right and it really is as simple as that.

[DuD]

I got the same as Supergrunch

 

I don't see as swapping would increase or decrease the probability of finding the car in Spambots question. Surely you just go from having a 1/3 chance before a door is ruled out to a 1/2 chance after a door is ruled out..?

[weeyellowbloke]

Moogle and Danny are presumably guessing the solution is n!, which is provably not the case - just look at line 3, where 3! gives 6 ways but there are only 4. In fact, I'm pretty sure the solution is just

2^(n-1), making the answer for analysis 128. The fact that the paths seem to cross over at some nodes just confuses things, and if we view each path on its own, it becomes clear that adding another letter doubles the total number of paths, giving the above formula.

 

Edit: This hastily photoshopped diagram shows you the difference between the 2 ways of looking at things, with the way on the right making the answer clear:

 

 

Yes, I am aware the file name is "analtree.jpg"

 

Supergrunch gets the right answer, 128. At each level you are given two choices, either left or right down to the next level. So basically the number of routes double at each level and seeing as you only have one starting point the question can just be simplified as 1x2x2x2x2x2x2x2=128. Alternatively, you could replace the letters with numbers, each one of which is the sum of the two above starting with 1, in which case you'd get.

 

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1

1 8 28 56 70 56 28 8 1

 

The sum of each row of number is a power of 2; 1, 2, 4, 8, 16, 32, 64 and the final row...128.

 

I wondered about that too. The answer is either 1 or 2 I think. Either way, trick question!

 

Man, there is an entire book on this question where I work. Think it's called the Monty Hall problem. Can't remember the answer at the moment as I'm too tired and slightly tipsy, but I think it's something still debated at maths parties (if such things exist).

 

 

Here's one of my faves in case nobody has heard it...

 

You're on a quiz show. There are 3 doors. Behind 1 of the doors is a car. Behind the other two there is nothing. You randomly pick one of the doors. The host then opens one of the other empty doors. You are now given the option to stick with your choice or swap to the only other remaining door. Which is the best thing to do and why?

 

Quite a popular one so please don't spoil if you already know it!

[Supergrunch]

Supergrunch gets the right answer, 128. At each level you are given two choices, either left or right down to the next level. So basically the number of routes double at each level and seeing as you only have one starting point the question can just be simplified as 1x2x2x2x2x2x2x2=128. Alternatively, you could replace the letters with numbers, each one of which is the sum of the two above starting with 1, in which case you'd get.

 

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1

1 8 28 56 70 56 28 8 1

 

The sum of each row of number is a power of 2; 1, 2, 4, 8, 16, 32, 64 and the final row...128.

Yep, the joys of Pascal's triangle, which has all sorts of crazy properties.

Man, there is an entire book on this question where I work. Think it's called the Monty Hall problem. Can't remember the answer at the moment as I'm too tired and slightly tipsy, but I think it's something still debated at maths parties (if such things exist).

Well, the actual solution is no longer debated as you can computationally demonstrate it, but some people (including mathematicians) aren't really happy with any of the proofs. It's just so counterintuitive.

[MoogleViper]

Supergrunch gets the right answer, 128. At each level you are given two choices, either left or right down to the next level. So basically the number of routes double at each level and seeing as you only have one starting point the question can just be simplified as 1x2x2x2x2x2x2x2=128. Alternatively, you could replace the letters with numbers, each one of which is the sum of the two above starting with 1, in which case you'd get.

 

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1

1 8 28 56 70 56 28 8 1

 

The sum of each row of number is a power of 2; 1, 2, 4, 8, 16, 32, 64 and the final row...128.

 

Well you didn't specify that it had to be a joined up path like that. I thought I could choose any letter on the row.

[Supergrunch]

Well you didn't specify that it had to be a joined up path like that. I thought I could choose any letter on the row.

Hmm, I interpreted it the other way. For your interpretation, n! would be right.

[MoogleViper]

Hmm, I interpreted it the other way. For your interpretation, n! would be right.

 

I'm just going to go ahead and take that as a victory.

[Supergrunch]

Alrighty, time for another well-known but awesome problem, although the setting is a bit contrived:

 

There is an island with 201 people on it, each of whom are "perfect logicians" - if there is a logical deduction that can be made, they make it instantly. 100 of them have blue eyes, 100 have brown, and 1 has green. They are not aware of these proportions, but are of course capable of seeing what colour eyes other people have. If a person works out what colour eyes they have, they must leave the island that night. On day 1, the green eyed person speaks to everyone, saying:

 

"I can see someone with blue eyes."

 

Aside from this, no communication is possible between anyone, and people can't check their eye colour in mirrors or anything similarly stupid. The question is:

 

Who leaves the island, and when?

Edited January 27, 2010 by Supergrunch

[SPAMBOT4000]

Man, there is an entire book on this question where I work. Think it's called the Monty Hall problem. Can't remember the answer at the moment as I'm too tired and slightly tipsy, but I think it's something still debated at maths parties (if such things exist).

Wow an entire book, really? Yeah it is called the Monty Hall Problem

 

Well, the actual solution is no longer debated as you can computationally demonstrate it, but some people (including mathematicians) aren't really happy with any of the proofs. It's just so counterintuitive.

I know originally a lot of people weren't happy with it but still? Really?

 

I don't see as swapping would increase or decrease the probability of finding the car in Spambots question. Surely you just go from having a 1/3 chance before a door is ruled out to a 1/2 chance after a door is ruled out..?

Nope. If you hadn't have made a choice in the beginning and entered into the game after the host had opened one of the doors THEN it would be a 1/2 chance.

Given true randomness you are 2/3 more likely to choose a door with nothing behind than the door with the car. The host then opens a door which has nothing behind. You are still 2/3 more likely to have chosen a door with nothing at this point, that fact doesn't change.

So if you have chosen an empty door (2/3 chance), the host must open the other empty door meaning the switch door must be the car (2/3)

 

So if you switch at this point you'll get a car, if you stick you'll get sod all. Granted if you originally choose the car and switch, you'll get nothing but the odds of originally choosing the car are 1/3.

 

It's easier demonstrated with more doors. Say you have 1000 doors, 999 with nothing behind and 1 with a car.

You choose one door to begin with. The odds of choosing the car are 1/1000. The host then opens up 998 empty doors leaving you with two left. One must be the car, one must be empty. Your original odds of choosing a car door have not changed (1/1000). There's a much higher chance (999/1000) that you picked an empty door, so if you switch you get the car.

 

 

 

Grunch, with your problem does one person speak up each day and if so is it a random person each day?

[Will]

Grunch, with your problem does one person speak up each day and if so is it a random person each day?

 

With the eye problem it's only just the first person that makes the statement and then nothing else.

[Supergrunch]

I know originally a lot of people weren't happy with it but still? Really?

Well, generally it's only students and so on, but occasionally more famous people (like Paul Erdős) have got irritated about the proofs.

With the eye problem it's only just the first person that makes the statement and then nothing else.

Yep. And it's directed at everyone.

Edited January 27, 2010 by Supergrunch

[SPAMBOT4000]

Surely with no communication either nobody leaves (Can't be sure who the blue eyed person is) or everybody aside from the green eyed person leaves as the rest of them could all have blue eye eyes. And if they are perfect logisticians and make logical decisions instantly, wouldn't they come to the conclusion instantly?

[Supergrunch]

I know originally a lot of people weren't happy with it but still? Really?

Well, generally it's only students and so on, but occasionally more famous people (like Paul Erdős) have got irritated about the proofs.

Surely with no communication either nobody leaves (Can't be sure who the blue eyed person is) or everybody aside from the green eyed person leaves as the rest of them could all have blue eye eyes. And if they are perfect logisticians and make logical decisions instantly, wouldn't they come to the conclusion instantly?

Neither of those answers is correct. And yes, if they can make a deduction, they make it instantly, but that doesn't mean that there aren't deductions that cannot be made until something has happened.

[Will]

I think an important part of this is that the only time they can leave is at a set point each night when a boat comes to get them. You need the time period of the daily boat to work it out.