Monty Hall Problem

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Monty Hall Problem

The problem I am about to give is a problem one is usually given in a Probability class:

 

You go to Monty Hall for a game.  The game composes three doors.  One door has a million dollars behind it, the other two are empty.  You get two chances, the first one and then the second one after the host has opened a different door.  You pick a door, and the host opens a different door showing that it is empty.  Assuming you want the million, should you switch, or should you stay with your first choice?  Explain. (I hope I explained this well enough).

While it can be reasoned out without an appeal to Bayes Theorem, if you want a more rigorous defence of your choice, using Bayes Theorem will give you the correct answer. 

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There is an ongoing debate

There is an ongoing debate about this on You Tube. I'm not great with math so I get a little lost.

I would say it makes no difference because I think the paradox skips the option that you picked the correct door to begin with. 


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Ig wrote: There is an

Ig wrote:
There is an ongoing debate about this on You Tube.

Don't listen to the people on youtube.  I have seen some of the comments about things.  A few smart people, a whole bunch of morons.

 

Ig wrote:
I would say it makes no difference because I think the paradox skips the option that you picked the correct door to begin with.

Nope.  The correct answer is to switch.  The reasoning is as follows:  There are three doors, one which has a prize and two that are empty.  Now, the probability that you picked the wrong door is 2/3, while the probability that you picked the right door is obviously 1/3.  Since you have a 2/3's chance of being wrong the first time, you should switch.  Since Monty is going to open an empty door, and since you probably picked an empty door, the prize will be in the unopened door...in all probability. 

Alot of people have the intuition that since there would be only two doors left, the probability is 1/2.  However, this fails to take into consideration the fact that your first choice was probably an empty door. 

"In the high school halls, in the shopping malls, conform or be cast out" ~ Rush, from Subdivisions


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I used to watch the show

I used to watch the show when I was a kid and dont recall. Did he offer the switch all the time?


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Ig wrote: I used to watch

Ig wrote:
I used to watch the show when I was a kid and dont recall. Did he offer the switch all the time?

TV show? 

"In the high school halls, in the shopping malls, conform or be cast out" ~ Rush, from Subdivisions


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Chaoslord2004 wrote:The

Chaoslord2004 wrote:

The problem I am about to give is a problem one is usually given in a Probability class:

 

You go to Monty Hall for a game. The game composes three doors. One door has a million dollars behind it, the other two are empty. You get two chances, the first one and then the second one after the host has opened a different door. You pick a door, and the host opens a different door showing that it is empty. Assuming you want the million, should you switch, or should you stay with your first choice? Explain. (I hope I explained this well enough).

 

I can't even fathom why this is a problem for people. It's obvious that you should switch, as your odds improve from 1/3 to 2/3

I think it helps if you consider that by switching you get "both doors" the one opened, and the other remaining door..... It might also help if people imagined that the door was closed, and that they were getting both doors...... oddly enough, in that case, you know that one MUST be empty...

Or, maybe it would help people if they imagined that there were 100 doors, and 98 of them were opened after your first pick... I think here people would get it... "I'm getting 99 doors by picking the other door, whereas I'm only getting 1/100 if I stick with my choice."

 

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Ig wrote: I used to watch

Ig wrote:
I used to watch the show when I was a kid and dont recall. Did he offer the switch all the time?

 

Lots of times, and oddly enough, the implicit rule was "Never switch!"  It was considered a sign of resolve to stick with your first choice... however, it was often the case that 2 out of the three doors had something good in them.... 

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Chaoslord2004 wrote:

Chaoslord2004 wrote:
TV show?

Monty Hall was the host of a real TV show called Let's Make a Deal. That's the origins of the Monty Hall Problem.

http://www.letsmakeadeal.com/MontyHall.htm

I'm old as fuck. I used to watch it along with most other Americans of my generation. It was very popular.

 Anyways, the reason I asked is because if he always offered the option to switch then the "house" would have bad odds. I would presume they only offered the switch once every 3 contestants or something. I vaguely remember they had 2 or 3 sequences in a show.


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The Monty Hall Problem and

The Monty Hall Problem and the rules of Let's Make a Deal are not the same. You can spend hours reading about the Monty Hall Problem on the internet.

Like the Birthday Problem, the MHP shows how susceptible humans are to fallacious intuitive reasoning. We use 'pretty good' heuristics to guess probabilities, but our natural intuitions on probability are very weak, so we get caught in the tricky problems. These are like intuitive blindspots. Religious texts exploit similar blindspots to convince people of false claims.

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I came across this in the

I came across this in the book about an autistic kid.
Apparently, the 1/3 - 2/3 answer is one a mathematician would give while a logician would insist 50 - 50. The cheeky bugger! Laughing

todangst wrote:
I think it helps if you consider that by switching you get "both doors" the one opened, and the other remaining door..... It might also help if people imagined that the door was closed, and that they were getting both doors......

Good way of putting it.
I'd always wondered how to describe the answer in an intuitive way.
I think this pretty much nails it.


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Strafio wrote:todangst

Strafio wrote:


todangst wrote:
I think it helps if you consider that by switching you get "both doors" the one opened, and the other remaining door..... It might also help if people imagined that the door was closed, and that they were getting both doors......

Good way of putting it.
I'd always wondered how to describe the answer in an intuitive way.
I think this pretty much nails it.

Thanks. I think what throws people is that the door is open, so they just don't count that as part of the deal... why would an empty door be helpful? "All that's left is my closed door, and the other closed door, so it's 50/50." So you just discount the closed door altogether, and that's the error.

But if you remember that even given two closed doors, at least one of them must be empty anyway, it helps you see the problem, and that's how I worked it out for myself. (I began as a 50/50 person too.) What cinched it for me was when I began to think "What if there were 100 doors? Then I'd take the other 99 for sure!" Once I saw that, I saw the light, so to speak.

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Strafio wrote: Apparently,

Strafio wrote:
Apparently, the 1/3 - 2/3 answer is one a mathematician would give while a logician would insist 50 - 50.

Im not sure this is entirely true.  I heard about this problem from three logicians...and i myself am a logician in training.  I think this is more a problem for people, educated or not, not trained in probability theory. 

"In the high school halls, in the shopping malls, conform or be cast out" ~ Rush, from Subdivisions


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todangst wrote: Once I saw

todangst wrote:
Once I saw that, I saw the light, so to speak.

See, I never found the answer puzzleing.  I mean, im probably wrong the first time.  If im probably wrong...and I want the money, I should switch.  I don't see it becoming more intuitive once the numbers are increased.  Perhaps thats just me. 

"In the high school halls, in the shopping malls, conform or be cast out" ~ Rush, from Subdivisions


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Chaoslord2004

Chaoslord2004 wrote:

Strafio wrote:
Apparently, the 1/3 - 2/3 answer is one a mathematician would give while a logician would insist 50 - 50.

Im not sure this is entirely true. I heard about this problem from three logicians...and i myself am a logician in training. I think this is more a problem for people, educated or not, not trained in probability theory.

 

My feel is that people discount the value of the open door... they disregard it in their analysis.... in that case, if you just ignore it, then it can only seem 50/50 to you.... once you're able to see how that third door factors into the equation, you're able to see the problem accurately.

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todangst

todangst wrote:
Chaoslord2004 wrote:

Strafio wrote:
Apparently, the 1/3 - 2/3 answer is one a mathematician would give while a logician would insist 50 - 50.

Im not sure this is entirely true. I heard about this problem from three logicians...and i myself am a logician in training. I think this is more a problem for people, educated or not, not trained in probability theory.

 

My feel is that people discount the value of the open door... they disregard it in their analysis.... in that case, if you just ignore it, then it can only seem 50/50 to you.... once you're able to see how that third door factors into the equation, you're able to see the problem accurately.

Sounds like the confirmation bias fallacy.  We discount that which refutes our thesis, while emphasising that which supports it. 

"In the high school halls, in the shopping malls, conform or be cast out" ~ Rush, from Subdivisions


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Chaoslord2004

Chaoslord2004 wrote:
todangst wrote:

 

My feel is that people discount the value of the open door... they disregard it in their analysis.... in that case, if you just ignore it, then it can only seem 50/50 to you.... once you're able to see how that third door factors into the equation, you're able to see the problem accurately.

Sounds like the confirmation bias fallacy. We discount that which refutes our thesis, while emphasising that which supports it.

Interesting, that does seem to be what is going on... the '50/50'  supporter focuses on the fact that there are only 'two options' - so in his view, it's simply absurd to believe switching matters.

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I wouldn't take the

I wouldn't take the statement too seriously.
It was from a fictional book and I doubt the lass who wrote it knew much about logic. She likely had a common sense idea that she was working with rather than understanding proper formal logic.


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Two random comments: 1) My

Two random comments:

1) My head hurts. Sad

2) Why does this remind me of Schrodinger's Cat? 


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Iruka Naminori wrote: Two

Iruka Naminori wrote:

Two random comments:

1) My head hurts. Sad

Aw.

Quote:
 

2) Why does this remind me of Schrodinger's Cat?

I think this is a lot easier than quantum physics.... to me, it just helps to think "when I begin, my odds are 1 out of 3, so is it better to have one door out of three, or two doors out of three?" You can only get 'two doors' by switching.

 

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For those old enough to

For those old enough to remember:

What's probably behind Door Number 2 is a loaf of bread with a hundred dollar bill hidden inside.  However, the person dressed as a rodeo clown, not knowing about the hundred dollar bill, will quickly trade the loaf of bread for the box of Post Toasties with a mood ring inside.  The new car was behind Door Number 1.

Sorry. I just had to do it. This thread has brought back all sorts of images. 

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Ok i set out to prove bayes

Ok i set out to prove bayes therom wrong.

 i wrote a python program that will run though a monty hall game 1000 times in about a second.

This program has 2 variables,  contestant switch or not, and host always pick empty doors or pick a random door.

 I have found that if the host picks a random door, it does not matter if the contestant switchs or not.(and the host gets the million 33% of the time)

if the host does always open an empty door, the contestant will score about 33% not switching.

if our contestant does switch, he will then get 66% correct!

 So bayes theorm is correct, as long as the host is choosing the empty door, with no contents.  

 

Since we wouldn't know if the host was choosing randomly or not, it would be beneficial to switch(per the average of 33% and 66% = 49.5%)

 

If you ask me, ill send you the program so you can see for yourself!

hope this has been helpful! 


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BodaciousB wrote: Ok i set

BodaciousB wrote:
Ok i set out to prove bayes therom wrong.

That will be logically impossible, as long as you accept these three axioms of formalized probability theory:

!.  The probability of any proposition falls between 1 and 0.

2.  Certian propositions have a probability of 1

3.  When there is no overlap, P(P or Q) = P(P) + P(Q) 

and the definition of conditional probability:

P(P/Q) = P(P & Q)/P(Q)

if you accept these, you must accept Bayes Theorem.  It follows logically from the axioms. 

BodaciousB wrote:
So bayes theorm is correct, as long as the host is choosing the empty door

Obviously he is choosing an empty door... 

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You loose me at number 3, I

You loose me at number 3, I dont know what Q is.

 

Its ok though I believe you now after writing this program.

you can get it here: http://pastebin.ca/408783

just save it as a .py and run it with IDLE(found at python.org)

 

Just switch the 2 variables around (switch and hswitch)

Fundamental flaw in the program:

    the remaindoor() function picks the NEXT door that has no contents and the contestant did not pick, as opposed to a random door that has no contents, and contestant didn't pick. But, It really doesn't matter in this instance any way.


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BodaciousBrian wrote: You

BodaciousBrian wrote:
You loose me at number 3, I dont know what Q is.

 Q is any proposition.  It says, the probability of the disjunction P or Q, when there is no overlap, is the probability P plus the probability of Q.

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I love this problem. I

I love this problem. I tortured my entire family with it at Christmas a couple years ago.

 Straifo, I got the problem from the same book: The Curious Case of the Dog in the Night Time. What a great story, recommend to everyone.

 

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So, may I just expand the

So, may I just expand the problem a bit?

Suppose you're on "Deal or No Deal" and Howie Mandell gives you the option throughout the game of switching the case you initially picked with remaining cases on stage. When would it be to your advantage to start switching cases? Only when your case and one other remains (here I'll introduce that one case contains the $1,000,000 prize and the other contains $0.05)? Is there a point in between the original 25 cases and one remaining case?

How does the Monty Hall Problem change when being payed by multiple people? Suppose there are two people playing with three doors or three people playing with four doors?

Please discuss...

 

I find the problem quite fascinating because it does seem to show a disconnect between logic and what many would consider "common sense". I think about it every time I hear our fabulous President mention the catch phrase "common sense solutions"...

 

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todangst wrote: I think it

todangst wrote:
I think it helps if you consider that by switching you get "both doors" the one opened, and the other remaining door..... It might also help if people imagined that the door was closed, and that they were getting both doors...... oddly enough, in that case, you know that one MUST be empty...

Or, maybe it would help people if they imagined that there were 100 doors, and 98 of them were opened after your first pick... I think here people would get it... "I'm getting 99 doors by picking the other door, whereas I'm only getting 1/100 if I stick with my choice.

 It started clearing up for me when I started thinking of the unchosen door and the open as one collective door that had a  2/3 chance of being correct. I'm not sure if that makes any sense...

"The obedient always think of themselves as virtuous rather than cowardly."
-- Robert Anton Wilson (1932-2007; R.I.S.)

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Chaoslord2004

Chaoslord2004 wrote:

BodaciousB wrote:
Ok i set out to prove bayes therom wrong.

That will be logically impossible, as long as you accept these three axioms of formalized probability theory:

!. The probability of any proposition falls between 1 and 0.

2. Certian propositions have a probability of 1

3. When there is no overlap, P(P or Q) = P(P) + P(Q)

and the definition of conditional probability:

P(P/Q) = P(P & Q)/P(Q)

if you accept these, you must accept Bayes Theorem. It follows logically from the axioms.

That's a nice, concise way of presenting it. I will add that to my page on induction and bayesian theory. 

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todangst

todangst wrote:
Chaoslord2004 wrote:

BodaciousB wrote:
Ok i set out to prove bayes therom wrong.

That will be logically impossible, as long as you accept these three axioms of formalized probability theory:

!. The probability of any proposition falls between 1 and 0.

2. Certian propositions have a probability of 1

3. When there is no overlap, P(P or Q) = P(P) + P(Q)

and the definition of conditional probability:

P(P/Q) = P(P & Q)/P(Q)

if you accept these, you must accept Bayes Theorem. It follows logically from the axioms.

That's a nice, concise way of presenting it. I will add that to my page on induction and bayesian theory.

Thanks.  Question:  I know that statisticians use probability alot, but how much probability theory is utalized by scientists? 

"In the high school halls, in the shopping malls, conform or be cast out" ~ Rush, from Subdivisions