A fascinating mathematical problem

I would think that the chosen door would have the car behind it.

My logic (?) is that the Game Show involves the usual hype and audience tensioning normally seen.

Had the contestant chosen the goat - he would have merely opened the door and revealed the failure - cue adverts - onto next contestant.....

Falsely raising the hopes (by eliminating the goat) of the contestant and then showing the failure can only serve to be negative. The contestant gets upset, the audience won't like it (unless they are warped like us and like to see suffering ) Then there are the TV producers, sponsors and advertisers.........

I am sure that there are more learned and emminent forum members who may differ....................as always

It's a well known problem that has already been discussed on the forum in the Brain Teaser thread. The surprise outcome is that you are better off changing your choice after being shown the door with the goat.
John

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It's a well known problem...

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Not to me.

I still don't see why it's not just a 50/50 chance once the first door has been opened.

I'll have to give it more thought.

Steve B

quote:

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Originally posted by John Channing:
It's a well known problem that has already been discussed on the forum in the Brain Teaser thread.

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I suspected that this may have come up before but a search for "Monty hall" didn't turn anything up. Then I thought that even if it HAD come before it might be of interest to people who don't read the "Brain Teaser" thread because it is a spooky example of how intuition and logic can give different results.

Minky,

Page 2 of the Brain Teaser thread....Yes it really was that long ago.....somewhat "dressed up" by BAM copied here....

Game show

Alright, here's an old one so I hope you haven't heard the answer already. If you haven't it is quite interesting. Enjoy!
Late one cold, rainy night in November after a few pints and a rather dubious curry you flop down onto your favourite listening couch and turn on your Naim system. Ah...that's better. But even the PRaT isn't enough to keep your eyes from closing and you soon fall into a deep, deep slumber and start to dream...

Congratulations! You have been selected as a participant in a very popular national TV gameshow. This is your first time on TV and on such a show and you are overwhelmed with excitement and nervousness. The gameshow host is none other than Paul Stephenson. Dressed in a gold lame suit with sequens, standing centre-stage on a Fraim. The grand prize: a Naim NAP500 and installaton in your home by Roy George himself. Your heart pounds with excitement.

You find the first rounds easy and before long the other contestants have been eliminated and you have made it to the final challenge. The audience go wild! You try hard to look at them through the bright spot lights and recognize them as your chums from the Naim Forum - all clapping and cheering you on. What a night!

Now the final challenge begins. The lights dim and the cheers of the crowd die down. Paul shows you three doors numbered 1, 2 and 3. He says that the NAP500 is hidden behind one of the doors. You must choose which one. If you choose correctly you win the NAP500. If not you win Paul's hideous suit. The tension mounts. The audience can't contain themselves: shouts begin..."door 1"..."no door 2"...etc. You hesitate. Paul urges you to make a choice. Finally you choose door number 3 (after your existing Nait 3 system). A green light comes on over door 3. Oh no...did I choose wisely...

A hush comes over the theatre as Paul walks over to door number 1 and motions to open it. The atmosphere can be cut with a knife. Slowly, Paul opens door 1 to reveal nothing! Phew that was lucky. The crowd bursts into a relieved cheer. But soon Paul has gripped the handle of door 2 and the crowd go silent. You can hear a pin drop. The tension is unbearable...

Suddenly Paul stops turning the handle of door 2 and turns towards you. He says he is in a good mood tonight and wants to help you as much as he can. He offers to do you a favour and allow you to change your selection. You can now stick with your original choice...door 3 or switch to door 2. It is up to you and you have just 10 seconds to decide. The audience erupts! "door 2 go for door 2"..."no, no stick with door 3"...

You have 5 seconds left. You are sweating. Everyone is hanging on your next words. Do you stick with door 3 or switch to door 2?

The answer ?

well you have to change your mind.....

Take a peek at page 2 of the Brain teaser and follow X pages of inconclusive debate....

I agree with Minky, its well worth a repeat.

Cheers

Don

The thing about that problem that surprised me is how everyone seems to think the intuitive answer is to stick with your original choice. It's surely obvious that the one you picked can only be right 1/3 of the time so the other choice must be right 2/3 of the time. It's not like you'd stuggle to make the choice if after picking a door you were offered a chance to choose *both* the other two doors instead.

My mate was asked a variant of that question at a graduate interview for an investment bank in 1989 and got it wrong, but got the job anyay and is now a multi-millionaire. I got the answer right and am a struggling computer programmer in a dead end job. There really is no justice.

Matthew

Change your mind

I use this problem with chairmen and other top executives who have a make your descision and stick with it attitude.

They just can't cope.

Until you point out two things

1. Get an expert in to help identify the options so that you can make an informed choice (or become that expert yourself)

2. The RIGHT answer doesn't involve changing your mind, it only involves following a different (and better) course of preplanned actions, which for some reason isn't obvious to many people (like the rest of life!)

Most of them seem able to come to terms with item 2

Cheers

Don

It's surely obvious that the one you picked can only be right 1/3 of the time so the other choice must be right 2/3 of the time. It's not like you'd stuggle to make the choice if after picking a door you were offered a chance to choose *both* the other two doors instead.

Matthew,
That is probably the best and simplest explanation I have ever read of the problem!
John

Try looking at it like this

If you get to choose NO doors your chances are 0
If you get to choose ONE door your chances are 0.333 (ie 1/3)
If you get to choose TWO doors your chances are 0.667 (ie 2/6)
If you get to choose THREE doors your chances are 1 (ie a dead cert)

You get to choose ONE door by choosing which door to open and sticking with that choice.

You get to choose TWO doors by deciding which WHICH door (the third one) you WON'T open.

You CHOOSE (say) doors A & B and FORCE the game show host to open them both, simply by saying "I choose door C". The host is thus FORCED to open either A or B (lets say B) and you then simply FORCE him to open the other (A) by "changing your mind!".

The technique works whether the host knows where the car is or doesn't know where the car is.

Cheers

Don

I agree with John.

Matthew

so dies it affect the probability that after you change your mind that he then walks over to door number 3 and starts to open that one?

I think that the way you have come up with the 2/3 probability of door 2 is flawed. I don't think it matters if you change you mind at all, but if you change your mind or not you now have a 1/2 chance you will win.

Why must the other choice be right 2/3 of the time? What if rather than hovering by door 2 he is hovering by door 3 when asking you the question? or what if he opened door 2 first and now gives you the choice of changing?

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Page 2 of the Brain Teaser thread....Yes it really was that long ago.....somewhat "dressed up" by BAM copied here....

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Don, do you know off the top of your head all the questions in that thread?

I still think its a 50/50 as to which one it is behind after he has opened the first door...
The door you picked originally, will either have the car, or a goat. That isnt going to change, nor is the probability of you picking the car if you change your mind.
Like winning the lottery... No matter what numbers you choose, the probability of you winning is still the same...

Don, do you know off the top of your head all the questions in that thread?

more or less...yes

really sad, isn't it!

Cheers

Don

Two-sheds, Patrick, Adam

Matthew is right. (for once)

One step at a time.

Three doors. One car, two goats.

You get to pick only ONE door. probability of picking the car? 0.333

You get to pick TWO doors. Probability of picking the car? 0.667

These are fact. They won't change....ever.

Can we all agree so far?

Cheers

Don

Two-sheds

copied from Page 3 of the Brain Teaser

"posted Fri 23 November 01 16:03
Slamming Doors

I am trying to create an excel spreadsheet to simulate the game show described by Bam. (this is getting bloody sad isn't it ?)
So far I have created two sheets, one where the contestant 'sticks' with his original choice, the other where he always changes his mind. In both cases, Mr S always opens an empty box. (sometimes Mr S has a choice about this box, sometimes he doesn't). In other words, Mr S KNOWS where the prize is.

Running this simulation 1,000 times always gives a probability close to .333 if you don't change your mind and .667 if you always do change your mind. (Excel hides the boxes, picks the first choice, does Mr S's thinking for him - BRILLIANT - etc, etc)

The results of 10 runs of each spreadsheet are given below

.336; .322; .342; .355; .346; .344; .330; .360; .332; .363; Avg = .343
.682; .672; .704; .614; .648; .684; .669; .679; .685; .664; Avg= .670

I intend to extend the spreadsheet to investigate the results assuming Mr S DOESN'T know where the prize is, ie he opens box 1 or 2 at random. This means that from time to time he will reveal the prize - much to the contestant's dismay! What I'm not sure about, is whether the contestant is still better off changing his mind (when he gets this chance). At present, I am inclined towards...................???

Cheers

Don"


Cheers

Don

A proof for those who insist on seeing numbers rather than relying on one's intuition:

1. Pick a door (A). Chance of being correct is 1/3

2. Chances of it being one of the other two doors (B, C) is 2/3.

[I assume we all agree to this point]

3. Host eliminates door B which, since it never has the car, has a car probablity of 0.

[This is the bit people get wrong and assume this has a probability of 1/3 -- omitting the fact that the host knows how to pick the one without the car]

4. Probability of it being behind door C is 2/3 - 0 = 2/3. Door A's probability remains unchanged at 1/3.

Matthew

Patrick,
You are wrong. Do a search on the internet for "The Monty Hall problem" if you need more convincing.
JOhn

Here are all the possible outcomes, highlighted in bold is the door the gameshow host will open.

Stick | Switch
Goat | Goat Car ->>> Switch wins
Car | Goat Goat ->>> Stick wins
Goat | Car Goat ->>> Switch wins

Convinced?
John

"The point is that Matthew's, Don's etc's logic assumes you know how the game works in advance. This is not stated in the 'problem'"

Not true -- the question is quite clear and nothing in my answers requires other knowlege. Some might argue that it's not stated that the host knows where the car is but this is rather implicit in the question given that he has a 100% success rate in opening a non-car door.

(Can't speak for Don's answers as frankly I don't understand them -- presumably because I am not a Chairman or other top executive).

"If you are not convinced, try an experiment for yourself and see if you can 'win' the car 66.6% of the time rather than 50% of the time"

I bet you a million pounds it doesn't.

"The bit you're getting wrong is that there are not actually 3 possibilities; in fact there are only 2, because a 'choice' with a probability of zero is always going to be eliminated by the host"

You are assuming that the host picks his choice before you pick and sticks with that. What actually happens is the host picks after your choice so, in the 2/3 of cases where one of his two doors has the car, he opens the one that doesn't have the car -- effectively revealing more information about the third door.

"(I only ever scored 100% in one degree exam and it was Probability Theory)"

Good job they didn't have this problem then

Matthew

I'm with Patrick on this one...

The way I see it is this.

The problem as presented to me is I have a choice of 1 door in 3, so I pick a door. I have a 33.333... percent change of guessing correctly.

If I have chosen ( guessed ) "wisely", I can now play again.

Now I have a choice of 1 of 2 doors, or a 50 percent chance of being correct. The fact that there were once 3 doors and only one choice has absolutly no bearing on the choice I now have.

What the host does, does not know, says or does make no difference to the problem as stated above, unless the prize is placed after I make a choice.

So where have I gone wrong in my reasoning?

Yours in thickness....

regards,

michael

1. You choose A which has a 1/3 chance

2. B and C also have 1/3 chance each -- meaning B and C together have a chance of 2/3.

[This is 100% non-debatable I assume]

If at this point you were allowed to choose to B and C you therefore have a 2/3 chance of being correct and this choice must be preferable to sticking with A.

[Again I assume that is obvious]

The only problem you have is that with your new choice of B and C you don't know which one has the car. If you are allowed to open both you clearly have 2/3 chance, yes?

So if you pick B+C and then someone tells you which one of these doors definitely doesn't have a car your remaining door must still have 2/3 of a chance, yes? If not, and it's now 50:50, how did the extra probability "transfer" back to door A?

In essence by changing you are getting to open two doors instead of one.

Matthew

Patrick - I initially agreed with you but now I realise the following with the two options below -

Car | Goat Goat ->>> Stick wins
Car | Goat Goat ->> Stick wins

If you are picking A, then the two scenarios above have a combined probability of 1/3, not 1/4 each.
The reason being is that 1/3 of the time the car is behind A then 1/2 the time the host reveals a goat behind B and 1/2 the time he reveals a goat behind C.
1/3 x 1/2 = 1/6

So each scenario happens 1/6 of the time.

Whereas for the other two scenarios -

Goat | Goat Car ->>> Switch wins
Goat | Car Goat ->>> Switch wins

The probabilities are -
1. Goat is behind A = 2/3
2. After that, the chance of a goat being behind B or C is a 1/2.
= 2/3 x 1/2 = 1/3
So each of the scenarios above have a 1/3 chance of happening. Since each of those scenarios result in you winning if you switch then the probability of you winning if you switch is 2/3.

Does that make sense? I am a bit vague on probability formula these days, but I think it goes something like that.

Charlotte

Erm no, but my girlfriend seems to have discovered the Naim message board!

Patrick,

The Monty Hall Show ran for quite a while. So most, if not all the contestants knew the process. It was always the same.

Contestant picks a door
Host opens one of the OTHER doors and ALWAYS shows a goat
Contestant gets to change his mind or stick

For those who suggested an "experimental" run, well, that what my computer simulation did.....and sure enough, STICK and you win 0.33 of the games. CHANGE you mind and you win 0.67 of the games.

NOW....if part way through the game, just after Monty has opened the door to reveal a goat, an alien drops in, unaware of the rules and what has just happened.....well, he has two doors to choose from, one with a goat and one with a car. So HIS chances of choosing the car at this point ARE 50/50

Cheers

Don

But Minky, what did you think of "The curious incident of the dog in the night-time"?

I read it a few weeks ago, just interested what you though..... Apart from the intriguing mathematical bit

Computers are supposed to work on 1's and 0's - in other words "Yes" or "No" - why does mine frequently say "Maybe"?......