Postage Stamp Chat Board & Stamp Bulletin Board Forum
 

Here is the Monty Hall problem in its classic form:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1 [but the door is not opened], and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?"

Is it to your advantage to switch your choice?

Vote, and reasons please.

Norm

_________________
Geelong, VFA Premiers 1878, 1879, 1800, 1882, 1883, 1884, 1886, AFL Premiers 1925, 1931, 1937, 1951, 1952, 1963, 2007, 2009, 2011, .

I'd stay with my original choice. If I'm gonna be a goat, then I'm gonna be a goat

_________________
FORESTS OLD, PASTURES NEW
(An expert is one who knows more and more, about less and less)

Yes, according to statistics, it IS to your advantage to switch and here's why:

When you made your initial choice, you had a one in three chance (33.3%) of selecting the correct door. Now that one door has been revealed, you have the opportunity to choose the other door. Now your probability is one out of two or 50%, that the prize is behind the other door. Logically, a 50% chance is greater than 33.3%, so accordingly, you should change your choice of doors.

_________________
"Making money is a hobby that will complement any other hobbies you have, beautifully." - Scott Alexander
I collect Souvenir Sheets postally used on cover, WW used stamps up to 1957 except everything of Brasil and Argentina.

Surely then there's also a 50% of selecting the correct door by sticking with the one you picked in the first place?

_________________
Nigel

Stay with original choice
He (the host) ain't going to trick me to change

_________________

Collecting - Stamps, poster stamps and covers that feature Kangaroos.
Wanted: Illustrated Tourist cover of "Waratah Park", Sydney, NSW. Australia

IF I had a brand new car to my liking and the car behind the door isn't what I'm looking for, then I would switch otherwise I'd stick with the car door.

Kloster

Carte d'or?

Sorry Norm, I couldn't resist...going now

_________________
FORESTS OLD, PASTURES NEW
(An expert is one who knows more and more, about less and less)

This becomes a circle.

You could think "I picked the right door, so the host is pushing me to change my mind, so I'll get the goat." But then you would think "The host thinks I'm thinking that, so I should stick to the first choice." Then, "The host thinks I'll think he thinks I'm thinking I'm right, so I should change." And on and on, until you have yourself totally confused and the host isn't sure if your insane babbling is "stay or change" final answer.

_________________
Collecting Mongolia; Thailand; Indo-China; Mourning Covers; OHMS.
My online 'store': http://stampsfromaethelwulf.blogspot.com/

Chance of being behind the other door is actually 66.666666 etc % as they must add up to 100%. PS Did you see this on Mythbusters a few weeks back,
the psychology was very interesting!

Andrew

What if you wanted to win a goat instead of a car?

It had to be said....

Did not see that on Mythbusters...but,

When you have 3 choices, yes, 66.6% chance it is behind another door.

Now that one door is out of the equation, you have two doors to choose from, hence 50% chance. That's the pure statistical math behind it.

_________________
"Making money is a hobby that will complement any other hobbies you have, beautifully." - Scott Alexander
I collect Souvenir Sheets postally used on cover, WW used stamps up to 1957 except everything of Brasil and Argentina.

Do you get to keep the goat?

_________________
Nigel

I know what is argued that the answer must be, but I can't get my head around taking 1 door away and leaving 2 regardless of which I chose first, gives me a 50:50 chance whichever door I then chose.

My gut would stay with my first choice as changing my mind and losing would feel worse.

To me if you start with 3 doors and thus a 33.33% chance for each but then remove a door, the equation resets and you now have a 50% chance.

There is a further argument in how the puzzle is actually worded.

There is also the argument of which country you are in - the car may help you get around but the goats can feed you. So what's the most beneficial prize?

_________________
Greg - Looking for Goulburn Australia Cancels and Grangemouth Scotland Cancels and Covers
Member of the S.T.A.M.P Club for Slightly Twisted And Mad Philatelists - Motto: "Bring back the lick!"

Yes ... and keep you company in some Eastern Federations.

I want the goat

_________________
FORESTS OLD, PASTURES NEW
(An expert is one who knows more and more, about less and less)

I think I'll pass this round and try again later

_________________
My blog, if you're interested: petitmondemoi.blogspot.com

I have a 1/3 chance of picking the car first up, and 2/3 chance of not picking it.

So if I switched:

If I originally picked the car (probability 1/3) and switched, then I lose.

If I originally picked the goat (probability 2/3) and switched, then I win.

And if I didn't switch:

If I originally picked the car (probability 1/3) and didn't switch then I win.

If I originally picked the goat (probability 2/3) and didn't switch, then I lose.

So the best probability of winning involves switching.

_________________
John

Looking for used pre-decimal OS/NSW and G/NSW state perfins, and NWPI KGV overprints.

Technically, yes ... since you were more likely to have selected one of the goats in the first round.

The Monty Hall Problem. Which door would you pick?

1. I would stay with my original choice 56% [ 9 ]
2. I would switch to the other closed door. 44% [ 7 ] x

Total votes : 16

Norm

_________________
Geelong, VFA Premiers 1878, 1879, 1800, 1882, 1883, 1884, 1886, AFL Premiers 1925, 1931, 1937, 1951, 1952, 1963, 2007, 2009, 2011, .

Here is an elegant solution, taken from Wiki. There anly three possible combinations of car and goats. You pick door One (although the same solution works regardless of which door you pick).

1 .....2..... 3
Car Goat Goat
Goat Car Goat
Goat Goat Car

A player who stays with Door 1 wins in only one out of three of these equally likely possibilities, while a player who switches wins in two out of three. The probability of winning by staying with the initial choice is therefore 1/3, while the probability of winning by switching is 2/3.

Norm

_________________
Geelong, VFA Premiers 1878, 1879, 1800, 1882, 1883, 1884, 1886, AFL Premiers 1925, 1931, 1937, 1951, 1952, 1963, 2007, 2009, 2011, .

Q. Does the car door ever get opened from the remaining two doors?

_________________
I AM ALWAYS IN THE MIDDLE
Post your KGV faults on the stampboard wiki.

Yes it does, and a third Goat jumps out.

Norm

_________________
Geelong, VFA Premiers 1878, 1879, 1800, 1882, 1883, 1884, 1886, AFL Premiers 1925, 1931, 1937, 1951, 1952, 1963, 2007, 2009, 2011, .

Rephrase has the prize door ever been opened leaving the contestant with the choice of two doors which both contain a goat?

_________________
I AM ALWAYS IN THE MIDDLE
Post your KGV faults on the stampboard wiki.

No. Monty always knew which door contained the car, and always opened a goat door.

Norm

_________________
Geelong, VFA Premiers 1878, 1879, 1800, 1882, 1883, 1884, 1886, AFL Premiers 1925, 1931, 1937, 1951, 1952, 1963, 2007, 2009, 2011, .

There are two events here.

Select one from three.

Then select one from two.

In selections of this type, the outcome of the first event defines what's left over for the second event, but has no influence on the final outcome. The proability of each outcome for the second choice is still 50%.

Toss a fair coin 10 times and get 10 heads in a row. The probability of a head on the next toss is still exactly 50%.

Statistically, whether you stick or change makes no difference whatever.

The various logics presented above that claim otherwise are flawed.

Any discussion should be on the basis of whether Monty wants you to change your selection, and whether he really wants you to have the car or the goat. That is the only thing that should influence your choice ... and that needs a psychologist, not a statistician.

Aethelwulf got it right ... I think.

_________________
A man might as well marry ... if he finds a good wife he will be happy ... if not, he will become a philosopher.
Collecting Greater New Guinea & Macropods (Kangaroos & Wallabies).

I beg to differ Muruk.

As I've shown above, I have a 2/3 chance of winning if I switch, and a 1/3 chance of winning if I don't.

_________________
John

Looking for used pre-decimal OS/NSW and G/NSW state perfins, and NWPI KGV overprints.

But the probability of the first event is only 33% and the probability of the second event changes to 50%. Thus changing doors improves the odds.

If you had 100,000 doors and one car, the probablitiy of the first event would be roughly 100,000 to one. Eliminate 99,998 options and one still has the car, the probability of the first event is not relevant. The odds change from 100,000 to One to Two to one. 50%. As opposed to 100,000 to 1.

Norm

_________________
Geelong, VFA Premiers 1878, 1879, 1800, 1882, 1883, 1884, 1886, AFL Premiers 1925, 1931, 1937, 1951, 1952, 1963, 2007, 2009, 2011, .

Statistically, you are comparing two separate events.

First event your chances are one in three.

Second event your chances are one in two.

Your improvement in odds is because you have moved from event one to event two ... NOT because you have switched choices (or not).

_________________
A man might as well marry ... if he finds a good wife he will be happy ... if not, he will become a philosopher.
Collecting Greater New Guinea & Macropods (Kangaroos & Wallabies).

Your logic is flawed.

The choice between two doors makes a 50/50 probability ... it cannot be anything else no matter how you try to explain it.

Whatever you chose at first has no bearing whatever on the choice you have now.

Anyone who has a robust method of providing a greater than 50% probability on either of the choices here should extend and use their method to win a fortune at Tattslotto. You could predict which numbers will win based on what didn't win last week. Good luck.

_________________
A man might as well marry ... if he finds a good wife he will be happy ... if not, he will become a philosopher.
Collecting Greater New Guinea & Macropods (Kangaroos & Wallabies).

And that is why I asked my two questions

50/50 Is the only answer to the odds of winning

The Monty Hall Problem.

Q.Which door would you pick?

A.Either of the two doors that are left the third is irrelevant, change your mind your choice.

My answer was stay with the door I 1st picked

_________________
I AM ALWAYS IN THE MIDDLE
Post your KGV faults on the stampboard wiki.

While I picked the option to switch to the other closed door, the problem is not well defined.

Monty knows what is behind the doors. Does he always give the option to switch, or just sometimes? (I remember watching the show in my youth, and I don't think he always gave the option.)

If he only gives the option if you have already picked the car, you should never switch.

If he only gives the option if you have already picked a goat, then you must switch.

If he always gives the option, it is better to switch. A good way to see this is to just try it. Let a friend play Monty. He rolls a die to determine which door has the car. You pick a door, and he tells you another door which has a goat. Keep track of whether you win by staying or switching. Do this a large number of times.

I've done this, and it doesn't take too long to realize that you win over half the time by switching.

No, your initial selection gave you a One in Three chance of winning.

Door 1 will contain

Car
Goat
Goat

You have chosen Door One, and Monty has exposed a goat in Door 2 or 3. If you keep door One, you still only have a one in three chance of winning, not a one in two, since Monty knows where the car is and has selected a goat door. If you change doors, you increase that chance to one in two.

Norm

_________________
Geelong, VFA Premiers 1878, 1879, 1800, 1882, 1883, 1884, 1886, AFL Premiers 1925, 1931, 1937, 1951, 1952, 1963, 2007, 2009, 2011, .

Yes ... Norm is correct.

If you don't change your selection, then your odds of winning are 1 in 3, since you made the selection while there were still three doors.

Wrong.

Your chances are one in two if you switch, and one in two if you don't.

Your probabilities do NOT carry forward from when you had three choices, they MUST be recalculated from the new position.

Frank_King is correct ... the choice must be based on psychology, not probability.

_________________
A man might as well marry ... if he finds a good wife he will be happy ... if not, he will become a philosopher.
Collecting Greater New Guinea & Macropods (Kangaroos & Wallabies).

My head's buzzing

_________________
My blog, if you're interested: petitmondemoi.blogspot.com

There's a bee caught in your hair.

_________________
A man might as well marry ... if he finds a good wife he will be happy ... if not, he will become a philosopher.
Collecting Greater New Guinea & Macropods (Kangaroos & Wallabies).

This simulator is a good illustration:
http://www.grand-illusions.com/simulator/montysim.htm

If you run it for 1000 times, and keep the original choice, then you win about a third of the time, and lose two thirds of the time. Repeating the simulation, but swapping the doors, then you win about two-thirds of the time, and lose one-third of the time.

_________________
Chris de Haer
Author of 'A Closer Look at Modern Australian Stamps' in Stamp News Australia.
Latest news on my blog http://stamps.chrisdehaer.com.au Follow me on Facebook twitter.com/stampchris

This is also the logic used by the simulator ... it is flawed.

It assumes that after the first choice you are left with:

1 .....other
Car Goat
Goat Car
Goat Car

Therefore 2 chances of Car by switching to choice number 2 ... looks great.

BUT

If you have chosen the car (first choice in column 1) then Monty has a choice of two goats to reveal and can reveal either one.

So the second choice is really:

1 .....other
Car Goat1
Car Goat2
Goat Car
Goat Car

Still 50/50 as expected.

The four statistical possibilities written out longhand are ... if you chose number one at first:

You have the car ... goat 1 remains hidden (goat 2 revealed)
You have the car ... goat 2 remains hidden (goat 1 revealed)
You have goat 1 ... car remains hidden (goat 2 revealed)
You have goat 2 ... car remains hidden (goat 1 revealed)

Rework for door 2 and/or 3 and the outcome remains the same.

_________________
A man might as well marry ... if he finds a good wife he will be happy ... if not, he will become a philosopher.
Collecting Greater New Guinea & Macropods (Kangaroos & Wallabies).

I disagree - the simulator is not flawed.



Quickly looking at this, I think the problem is that you are discounting:
You have goat 1 ... goat 2 hidden (car is revealed)
You have goat 2 ... goat 1 hidden (car is revealed)

That makes six possible outcomes, but only four of them are 'winning' options, hence the 2/3 above.

Take a look at http://www.youtube.com/watch?v=mhlc7peGlGg

_________________
Chris de Haer
Author of 'A Closer Look at Modern Australian Stamps' in Stamp News Australia.
Latest news on my blog http://stamps.chrisdehaer.com.au Follow me on Facebook twitter.com/stampchris

The car is never revealed, so if you insist on including those outcomes, you must assign them a probability of zero.

If the car is revealed you have already lost, so a stay or switch option would only choose between goats.

My analysis does not change.

_________________
A man might as well marry ... if he finds a good wife he will be happy ... if not, he will become a philosopher.
Collecting Greater New Guinea & Macropods (Kangaroos & Wallabies).

I completely agree with Muruk on this. I wouldn't change my choice unless I could phone a friend

Sorry John, now I have to disagree with you.

I've had more of a think and have convinced myself to reverse my position.

If choosing from scratch after one door is opened, the choice is clearly 50/50.

But the choice of staying or switching does not produce the same outcome.

After your first selection:

You have the car ... next choice is car (stay) or goat (switch):

You have goat 1 ... next choice is goat 1 (stay) or car (switch).

You have goat 2 ... next choice is goat 2 (stay) or car (switch).

So, stay = 1 chance in 3 ... switch = 2 chances in 3.

In this case the first selection does have bearing on the second, and a tree analysis is appropriate ... not straight single event probability.

_________________
A man might as well marry ... if he finds a good wife he will be happy ... if not, he will become a philosopher.
Collecting Greater New Guinea & Macropods (Kangaroos & Wallabies).

After your first selection:

A goat is revealed as asked in an earlier post and that door will always contain a goat

You have the car ... next choice is car (stay) or goat (switch):

You have goat 1 or 2 ... next choice is goat 1 or 2 (stay) or car (switch).

So, stay = 1 chance in 2 ... switch = 1 chances in 2.

_________________
I AM ALWAYS IN THE MIDDLE
Post your KGV faults on the stampboard wiki.

Yes if it was a New Zealand contestant that becomes a major factor.

The problem is that the door opened is conditional on whether you have chosen the car or not.

If you have chosen the car and keep that door the host can pick any door, and you still win. If you have picked a goat, the host can only pick the door with the goat behind it. This adds an extra level of complexity as Muruk says.

The simulator shows Muruk's conclusion is correct. If you think the simulator is flawed, have a play with it. You'll find it isn't flawed.

_________________
Chris de Haer
Author of 'A Closer Look at Modern Australian Stamps' in Stamp News Australia.
Latest news on my blog http://stamps.chrisdehaer.com.au Follow me on Facebook twitter.com/stampchris

To me it still is a 50/50 chance once the third door is exposed. You are only second guessing by changing doors.

I disagree. I'm with Muruk on this one - and it's not often I can say that

In the initial analysis, I choose Door 1:

1 .....2..... 3
Car Goat Goat
Goat Car Goat
Goat Goat Car

After the host reveals that Door 3 contains a goat, we are left with:

1 .....2..... 3
Car Goat XXX
Goat Car XXX
XXX XXX XXX

So, this is now a COMPLETELY DIFFERENT event, and the probability of the car being behind Door 1 is exactly 50%.

The error in the previous calculations is that they LINK the two events to a single set of probabilities.

Now, if the host opened a RANDOM door, not knowing where the car is, then the probabilities would indeed differ if he revealed a goat.

BUT that is not the case here - to re-iterate, statistically we are looking at two DIFFERENT events.

Lies, damned lies and statistics....

Me, I'd stick with my original choice.

This is a game with a choice of three doors, you pick one door, you do not know what is behind that door only Monty does.

Of course you will only have one of two things behind that door and that is the car or the goat again you still don't know.

Monty picks a door with a goat behind it because he has to or he has a choice of two you still don't know, but he does.

He asks the question stay or swap?

Stay or swap your choice there can only be one or the other behind both doors.

It is a game of chance with a choice of two door at the end and the chance of you picking the correct door is 50/50

_________________
I AM ALWAYS IN THE MIDDLE
Post your KGV faults on the stampboard wiki.