The Monty Hall Problem
Threads - Monty Hall Problem
On
16/1/2004, Paul Williams posted:
I realise that many Science-Matters members may be familiar with this
conundrum (below)
Nevertheless, I believe it needs to aired out again on ocassions.
I have still a carton of beer to collect from a good friend regarding
his (incorrect) answer to this question. (He has a MSc and ocassionally
advises me in mathematics - which I'm sadly deficient in)
O.K. Imagine that you are the contestant...
Should you switch doors or should you stay with your first choice - or
does it make any difference?
Zero Sum replied:
We have had it out many times. The best option is to change your
choice.
> THE MONTY HALL PROBLEM
>
> "This story is true, and comes from an American tv game show. Here
is
> the situation. Finalists in a tv game show are invited up onto the
> stage, where there are three closed doors. The host explains that
behind
> one of the doors is the star prize - a car. Behind each of the
other two
> doors is just a goat.
>
> Obviously the contestant wants to win the car, but does not know
which
> door conceals the car.
> The host invites the contestant to choose one of the three doors.
Let us
> suppose that our contestant chooses door number 3. Now, the host
does
> not initially open the door chosen by the contestant. Instead he
opens
> one of the other doors - let us say it is door number 1. The door
that
> the host opens will always reveal a goat. Remember the host knows
what
> is behind every door!
>
> The contestant is now asked if they want to stick with their
original
> choice, or if they want to change their mind, and choose the other
> remaining door that has not yet been opened. In this case number
2. The
> studio audience shout suggestions. What is the best strategy for
the
> contestant? Does it make any difference whether they change their
mind
> or stick with the original choice? "
Peter Macinnis
posted:
And as has been
said before, if you can't see it, do a complete tree diagram.
John
Winckle answered:
I
have seen pages of statistical proof for this and someone actually ran
trials with thousands of examples being run.
I
think there is a simple logical proof that it easier to
understand.
When
you make your initial choice you divide the three doors into two groups
or sets. The 'set of those doors that you chose' (one member) and
'the set of those you did not choose', (two members).
All
doors have an equal chance of having the prize namely 33%
So
'the set of ones you chose' has 33% chance of winning. and 'the set of
those you did not choose' has 67% chance of winning.
When
the game show host eliminates one member of the 'set of those you did
not choose' the odds of both sets remain the same.
The
odds belong to the set not the individual members in it.
However
now that there is only one member in 'the set of those you did not
choose' if you now choose that one your odds change from 33%
chance to 67% chance, or double your chances.
There
are some assumptions that are unstated. One of these is that the
game show host knows where the prize is.
How
does that look?
Mijochelle (Jon
Pearce) wrote:
Having given
some thought to this (and I am no expert in maths or stats), it seems
there are two distinct probability "events" here;
Firstly, when
making the first choice, there is the 33% probability of getting it
right. (One correct from 3 choices).
In the second
event, the choices are reduced to 2, one right, one wrong. Making
a random choice at this point means a 50% chance of being correct.
So there is no "better" choice unless you possess ESP (if you believe
in that sort of thing).
HOWEVER, if the
host is ALWAYS going to reveal (and therefore eliminate) a wrong choice
after your first choice, then the probability of eventually choosing the
right door would always have been 50%, because as a linked event, one
of the "wrong" choices will be eliminated, and you will always be left
with a one in two choice.
To answer the
original question, is there a benefit in changing your original choice,
.. no, I don't think so.
Paul Williams
responded:
The remaining 2 doors
have 3 possibilities - Goat/Goat : Car/Goat : Goat/Car.
If you were to
change doors now, with no help from the host, whichever door you changed
to would have a 1 in 3 chance of winning.
>
HOWEVER, if the host is ALWAYS going to reveal (and therefore eliminate)
> a wrong choice after your first choice, then the probability of
> eventually choosing the right door would always have been 50%,
because
> as a linked event, one of the "wrong" choices will be eliminated,
and
> you will always be left with a one in two choice.
Your first
choice can *never change* from 1 in 3 odds.
The remaining
doors *never change* from 2 in 3 odds.
So if you were
given the choice of swapping your door for the other 2 doors it would
double your odds of winning.
One of these
doors, at least, will always hide a goat - the host eliminates one of
these doors (hiding a goat) but he is still
effectivelygiving you the choice
of swapping your one door for 2 doors.
>
To answer the original question, is there a benefit in changing your
> original choice, .. no, I don't think so.
Let's
say you always pick Door #1:
Door
# 1 | Door# 2 & Door # 3
Goat
| Goat & Car
Car
| Goat & Goat
Goat
| Car & Goat
It
can be clearly seen that staying with Door #1 will win the Car 1 in 3
times.
As
a Door with a Goat is always eliminated from your remaining choices, it
can be seen that changing your choice to the remaining door will win the
car 2 in 3 times.
John
Winckle explains this well (see below)
There
is a set of 1 Door (your first choice) with a 33.3*% chance of winning.
There
is a set of 2 Doors remaining, with a combined 66.6* chance of winning.
The
second set is reduced to 1 Door which has a 66.6* chance of winning.
(The
host always eliminates a door with a goat)
If
there were 100 doors with 99 goats and one car:
Choosing
one door gives you an *invariable* 1 in 100 chance of picking the car.
If
you could change to the remaining doors, 99 times out of 100 you would
win.
If
the host eliminates 98 doors with goats, should you change doors?
David Maddern
queried:
I can not handle
how you come to the conclusion that "The remaining 2 doors have 3
possibilities - Goat/Goat : Car/Goat :
> Goat/Car."
When the host
explains there is a car behind one of the doors and then opens one door
to reveal a goat.
The resultant
possibilities are (car / goat) and (goat / car) and since there is only
one car there is a 50:50 chance of choosing the car Which one you
choose does not change the physical entities in front of you.
Which one you point at does not affect what is behind the doors.
In that they are symmetrical. Both remaining doors are
equal. It is not affected by you poining at it.
Also you say
that "Your first choice can *never change* from 1 in 3 odds"
Surely it does
when one door is opened.
As your degrees
of freedom reduce from three to two the probability of your getting it
right increases.
Same phenomenon
as having got 4 lottery numbers marked off increases your chance of
having won the big prize. Of course you have already won or lost
it as far as the lottery commission is concerned.
Your contention
is that one door being revealed negatively changes the chance of the one
already nominated negatively defies logic.
So what have I
missed. Is there some double meaning somewhere?
Zero Sum answered:
>
you. Which one you point at does not affect what is behind the doors. In
> that they are symettrical. Both remaining doors are equal. It is
not
> effected by you poining at it.
No, but you 'point to' before the
event. The doors are not equal. The chance is no 50:50.
>
Also you say that "Your first choice can *never change* from 1 in 3
> odds" Surely it does when one door is opened.
Nope. Nothing has changed about
your frst guess, its chances are still one in three.
>
As your degrees of freedom reduce from three to two the probability of
> your getting it right increases.
And the choice was made before the
any alteration in degrees of freedom so how can the probablility of you
'getting it right' change?
>
Same phenomenon as having got 4 lottery numbers marked off increases
> your chance of having won the big prize. Of course you have
already won
> or lost it as far as the lottery commission is concerned.
No. This is not the same case at all.
>
Your contention is that one door being revealed negatively changes the
> chance of the one already nominated negatively defies logic.
Your contention that one door being
revealed negativly can affect a
decision made in the past defies
logic.
>
So what have I missed. Is there some double meaning someqwhere?
Contingncy
and he existence of time.
Draw
a table.
Anthony Morton
posted:
I suspect what's
going on here is that most treatments of probability, even at advanced
level, fail to stress the vital connection with information.
Instead, probability is taught as though it springs magically from the
definition of an 'event space'. As a result people are easily
lulled into erroneous reasoning about problems like this one.
If you have a
set of similar 'events' andyou have no
information that could imply that one event is more likely than another,
then
and only then should you assign
equal probability to each event, equal to the reciprocal of the number
of 'degrees of freedom'. This is what Laplace rather quaintly
called the Principle of Insufficient Reason, and what modern authors
call the 'principle of indifference'.
Unfortunately,
the principle of indifference is invoked so frequently in applications
that it's easy to imagine that it automatically applies in all
situations, and much teaching of probability fails to point out that
this is a fallacy.
So let's look at
the Monty Hall problem again. At the start you have three doors,
one with a car inside and the other two with goats. You don't have
any reason to think one's more likely to contain the car than any
other. So when you choose one door, the probability you've picked
the car is one-third. No problem here.
Now Monty the
host opens one of the other doors to reveal a goat. Subsequent
reasoning hinges on three crucial (but veiled) assumptions:
1. Unlike you,
Monty knows where the car is. So Monty doesn't have to work with
probabilities; for him, the car is behind one door with probability one
and behind the others with probability zero. Yours and Monty's
probabilities are different because you have different information.
2. Monty knows
which door you chose, and can make his choice of door based on that
knowledge.
3. You are to
assume that Monty deliberately chose a door that he knew would reveal a
goat.
As a result of
(3), Monty has shared some of his information with you.
So you are no
longer in a situation of zero information, and so you should no longer
automatically assign probabilities based on the principle of
indifference.
However, the new
information you have is of a rather limited kind: it does not affect
your degree of uncertainty about your initial choice.
This is because
no matter what your initial choice was, Monty can always open a second
door so as to reveal a goat.
This is a point
better understood by enlarging the event space. If you've blindly
drawn a card from a 52-card deck, Monty can always look through the
remaining 51 cards and show you 50 that are not the Ace of
Spades. But regardless of this, the probability that you
drew the Ace of Spades remains at 1/52: it does not magically jump to
1/2.
In each case,
Monty has provided you with no information about your initial selection,
but on the other hand has provided substantial information about the
remaining alternative. In the card analogy, Monty will have found
the Ace of Spades among the 51 remaining cards with probability 51/52,
and after revealing the other 50 cards will hold the Ace with this
probability. In the original problem, the car is behind one of
the other two doors with probability 2/3, and Monty's opening of one
door that he knew in advance would contain a goat does not change this
probability.
This contrasts
with the very different scenario where Monty opens a second door 'at
random'. What this really means is that he deliberately makes no
use of whatever prior information he has. (To ensure he really
does this and doesn't cheat, it might be better to imagine he asks you,
or an uninformed audience member, to choose the second door.) Of
the three assumptions above, only (2) is operative.
This puts Monty
in exactly the same zero-information situation as you in your initial
choice. With probability 1/3, he will open the door with the car
inside.
If Monty opens a
second door 'at random' and it happens to have a goat inside, the
situation looks superficially the same as in the real Monty Hall
problem. But there is a crucial difference: here it was not
certain that Monty would reveal a goat. In fact, the probability
that the door would have a goat inside now depends on whether your
initial choice was the car or a goat. If you had chosen the car,
the probability would be 1; if a goat, the probability would only be
1/2. This means that the opening of the second door reveals some
information about what's behind your door. Thus, the probability
you've chosen the car now changes from its 'a priori' value of 1/3 to a
new 'a posteriori' value.
This new value
can be determined from probability calculus. Define the following
propositions:
A: The door you
chose has the car inside.
B: A second door
opened at random reveals a goat.
I'll use a tilde
to denote negation. Thus initially, with no information, we have
P(A) = 1/3 and P(~A) = 2/3.
The 'a
posteriori' probability is P(A|B), the probability of A given B is
true. By the product rule of probability, this is equal to
P(A|B) = P(AB) / P(B)
where P(AB) is
the probability that both A and B are true. This joint probability
can be calculated using the product rule again:
P(AB) = P(BA) = P(B|A) P(A) = 1 x 1/3 = 1/3
since if A is
true, Monty could not help but reveal a goat: P(B|A) = 1.
We also need the
probability P(B) that B is true, irrespective of the truth of A.
This is found using the 'rule of total probability' taught in high
school:
P(B) = P(B|A) P(A) + P(B|~A) P(~A).
We noted above
that P(B|A) = 1 while P(B|~A) = 1/2.
Thus
P(B) = 1 x 1/3 + 1/2 x 2/3 = 2/3
and
P(A|B) = P(AB) / P(B) = (1/3) / (2/3) = 1/2.
So here we have
full agreement with intuition: if Monty opens a second door at
random and
it turns out to have a goat inside, the probability that you've chosen
the car jumps from 1/3 to 1/2. (On the other hand, if the second
door turns out to have the car inside, your probability jumps from 1/3
to zero.)
What happens in
the real Monty Hall problem? Here we have not B but a new
proposition C: A second door opened by Monty, who knows in advance there
is a goat inside, reveals a goat.
Mathematically,
what makes the difference here is that P(C|A) = P(C|~A) = 1; your
initial choice doesn't affect Monty's ability to reveal a goat. So
we have
P(AC) = P(C|A) P(A) = 1 x 1/3 = 1/3
P(C) = P(C|A) P(A) + P(C|~A) P(~A) = 1 x 1/3 + 1 x 2/3 = 1
P(A|C) = P(AC) / P(C) = (1/3) / 1 = 1/3.
So in the real
Monty Hall problem, the fact that C is a logical certainty means that
the a priori and a posteriori probabilities of A
are identical.
The TV show
'Deal Or No Deal' is a slightly more complex instance of the
pseudo-Monty Hall problem. Here the contestant makes an initial
choice out of 24 briefcases and then eliminates alternatives herself.
Unlike the real
Monty Hall situation, if the revealed alternatives turn out not to be
the main $2 million prize, the probability the contestant is holding the
$2 million really does go up each time; by the time there is only one
alternative remaining the probability has increased from 1/24 to
1/2. Clearly things would be very different if the alternatives
were being eliminated by someone with prior information about which
briefcase held the prize.
Mijochelle
responded:
>The
remaining 2 doors have 3 possibilities - Goat/Goat : Car/Goat :
>Goat/Car.
Sorry
Peter, but there are now only TWO remaining possiblilities, Goat/Car -
Car/Goat - The game show host ALWAYS opens the door of a Goat
prize. That "Choice" is logically presumed to be excluded, and
thus eliminates the Goat/Goat option at that point. Given that the
Host ALWAYS opens a Goat door, it is always a choice of one door or the
other.
I
agree that at the start it is a 1 in three chance, but probability
deals with chances, not certainties. Once you introduce the
certainty of an event, then the probability of that event is 1 to 1
(100%) (not 33%). The remaining events then make up the
probability.
John Winckle
commented:
It is a puzzle.
If the answer was as you say there would be no point to the puzzle. It
is because it is so counter intuitive that it becomes so interesting.
You are in good company but wrong.
David
Maddern wrote:
Thank
you Anthony for spending the time to explain this.
To
my way of reconciling the world I agree that probability in this case
is about information and is specific to each party (Monty vs observers).
The probability we are talking about is "Of the choice made being
revealed to be a car"
The
way I see it Monty's intention does not change the physical presence
behind the doors.
It
is irrelevant what his intention is to the physical revelation of the
goat to the observer, but there is now more information and half of the
2/3 probability of the two other doors is dispelled and the choice is
now between 2 identical doors, so is 50:50 to the observers that the
choice made be revealed to be a car .
The
logical extension of your explanation is that any of the original doors
has a dual probalility, and whichever one originally chosen will end up
with less probability by virtue of being chosen. Similarly if the
choice is changed in the original three picture , that door then carries
the less probability. Hence each of the doors has a dual
probability.
I
dont think the world works that way, it is not that subject to human
vicissitudes.
About
the cards, if Monty is revealing each of the cards (like he did the
goat) that arent the ace of spades then the final choice is between two
cards, the one not in the pack and the ace, and that is 50:50. The
probability of you originally choosing the ace is 1/52, but as possibles
are discovered, the probability that you now have the ace rises.
Je
vais a dormir
and
Yes,
John but that is like saying someone is guilty because the cops have
him.
Or
God exists cos they wrote a book about him
Zero Sum wrote:
> The way I see it
Monty's intention does not change the physical presence
> behind the
doors.
The way you see
it and the way the universe sees it are not the same. I have a suspicion that
the way the universe sees it is more important.
> It is irrelevant
what his intention is to the physical revelation of the
> goat to the
observer, but there is now more information and half of the
> 2/3
probability of the two other doors is dispelled and the choice is
> now between
2 identical doors, so is 50:50 to the observers that the
> choice made
be revealed to be a car .
The two
remaining doors are not identical any longer, they may be the same size and coulur but
the information associated with the doors has changed.
> The logical
extension of your explanation is that any of the original
> doors has a
dual probalility, and whichever one originally chosen will
> end up with
less probability by virtue of being chosen. Similarly if
> the choice
is changed in the original three picture , that door then
> carries the
less probability. Hence each of the doors has a dual
> probability.
You assert that
this is a logical extension and then asume it as fact. It is neither logical
nor would I call it an extension - a long stretch, maybe.
> I dont think the
world works that way, it is not that subject to human
>
vicissitudes.
The world works
the way it does and having an opinion on it is irrelevant. If you don't accept
the reasoning presented and won't draw a table the only thing I can suggest
is that you perform the experiment and see what
results you
get. That will demonstrate the way the world works.
> About the cards,
if Monty is revealing each of the cards (like he did
> the goat)
that arent the ace of spades then the final choice is between
> two cards,
the one not in the pack and the ace, and that is 50:50. The
> probability
of you originally choosing the ace is 1/52, but as possibles
> are
discovered, the probability that you now have the ace rises.
Please
demonstrate how this occurs. Since it does not, you have Buckley's...
Oh, and I would
love to play cards with you if you are wealthy...
