Mr. Stikker is quite redundant, no doubt about that. I may only guess that his intention was to preempt a possible question, "And what if you choose the middle door? Or even the right one?"

There is no difference of course which door you start with. This may be obvious to you, me and to many others. Some may even insist that those who do not see this simple truth have no business solving puzzles.

I reserve a judgement. Would you like me to add your comment to the letter of Mr. Stikker?

Thank you,
Alex

I think I see it now.

Essentially the flaw with my counting outcomes approach is that with the Always Switch or Always Stay strategies, what the host does in picking the door to open has no impact on the outcome--hence my two outcomes under case A should really be collapsed into one.

By the way, I don't see how there's 6 possible outcomes. Under the rules as originally stated, the game host will never open the door with the prize. He does have a choice under Case A, where you select the door with the prize, but as I stated above, this choice does not affect the outcome.

Thanks for clearing this up for me.

Don Link

Car Goat1 Goat2
Car Goat2 Goat1

Goat1 Car Goat2
Goat2 Car Goat1

Goat1 Goat2 Car
Goat2 Goat1 Car

(Same reasoning as why chances of getting a head and a tail with a toss of 2 coins is 50% and not 33.3%--HH HT TH TT)

The above 6 starting positons clearly show that staying with the first choice (Column 1)loses 2/6 times and win 4/6 times.

I was reading it where you think of it in terms of game theory. The Contestant wants to win the car. The Host may: want the Contestant to win, want the Contestant to lose, or not care.

The Host, though, only gets to choose whether or not to offer the Contestant a switch.

If the Host wants the most possible wins, he'll only show a goat and offer a switch whne the player is on a goat.

If the Host wants the most possible losses, he'll only show a goat and offer a switch when the player is on the car.

What about the other cases?

The problem - You have 3 doors and choose 1. Then Monty Hall removes
one of the wrong answers from the problem. So you have TWO possible
outcomes - either stay or switch. One of these is the right answer
so you either get it right or you get it wrong. 50% You don't have
6 possible outcomes, or 4, or 3 or whatever you're all saying. You
either get it right or wrong (stay or switch) 2 possible outcomes with 1 correct answer, that says 50%.

This illustrates a common fallacy in analysis of probabilities.
The fact that an experiment can be described as having two possible outcomes does not mean that both are equally likely.

To convince you that the conclusion of equal probabilities is wrong in the Monty Hall problem, consider the situation with a million doors - one car, the rest goats. Monty says Pick one and promises to show you 999998 goats. You have a 1 in a million chance of picking the car to start with. If you didnt choose the car to start with switching will give it to you for sure. If you don't agree that switching is the better strategy I'd like to meet you in Las Vegas.

dave

His first column shows the prize behind door A (1/3 probability); he chooses door A (1/3); the host opens door B (1/2) - but then what? Peter assumes that we MUST switch to door C, and does not provide for sticking with door A.

If we redraw his tree to show the sticking option in all cases, suddenly the probability of winning becomes 1/2, not 1/3. When the prize is behind door A, the eight possible outcomes are split 4 win and 4 lose. They are not all equally likely, but do fall into pairs of equal probability - so the win/lose outcomes are equally likely.

Will that be the end of it? I doubt it!
:-)

(PS: I tried to include the image in this posting - but couldn't get it to work. Sorry.)

If someone would show me an error in this tree, I am willing to be convinced that I am wrong.

:D

I side with the the many doors solution as I 'solved' it in my own mind that way. The inital chance of choosing the correct door is tiny, however after the bad doors are revealed your chance of now getting the right answer are far better. Throughout the entire sequence of events it appears that switching is the way to go. After all the entire sequence is what generates the outcome not just the final switch:

First choice = 1 in 3
Second choice 1 in 2 (remember the first choice was 1 in 3 this means when you switch you are trading a 'bad' choice of 1 in 3 for a better one of 1 in 2, you were less likely to choose the right door on your first choice, it's a way of looking at it.)

-Ian

"Peter assumes that we MUST switch to door C, and does not provide for sticking with door A."

Sorry for the omitting the variant of sticking, however:
"If we redraw his tree to show the sticking option in all cases, suddenly the probability of winning becomes 1/2, not 1/3."

Is completely nonsense and not true!!!!! Draw it again and you'll see.

The only critics about the general statement might be that often in the story two things are left out:
a) The game host knows where the price is
b) He will always first open a door that is not the original chosen one and does not contain the price.

Kind regards,
Peter Stikker

p.s. Also a computer simulation will show the same thing.

For instance, if door A was the winner, and you picked it, Monty would open either door B or C. If he opened B, and you switched to C, you lose. If he opened C, and you switched to B, you lose. THESE MUST NOT BE COUNTED AS TWO DiFFERENT POSSIBILITIES!!!! It'should be one possibility, you picked, you were shown a loser, and you switched to a loser. That will straighten out your tree.

(your way of thinking may be correct if there were three different prizes and you were asked whatr is the probability of getting each prize.)

Another approach that leads often to the same conclusion is to think about the following:
The probability that my initial chosen door is correct = 1/3
The probability that my initial chosen door is incorrect = 2/3

If my initial chosen door is correct and I switch I will loose, if my initial chosen door is incorrect and I switch I will win. Hence probability of winning with switching = 2/3, loosing 1/3.

Not switching will actually 'switch' these probabilities around.

As mentioned earlier, the full rules are often left out in order to keep things clear. 'The rules' I mean as in that the host knows behind what door the price is, that he will always first open a door that does neither contain the price, neither the original chosen door etc.

If all of this still doesn't convince you, either create a software programe yourself and run it a few times (you could actually do this in Excel even). :-)

Suppose there are 1,000,000 doors with a car behind one and goats behind the others. Select one! What is the probability the car is behind the one you selected? What is the probability the car is behind one of the other doors? You know that behind at least 999,998 of the other doors are goats. You know Monty knows where the car is. So now he opens 999,998 doors and you see all goats. Do you know any more than you did before he opened these doors? Has the probability that the car is behind your door changed? Has the probability that the car is behind one of the other doors changed? Do you want to switch?

Suppose you are playing a seven door version of the game. You choose three doors. Monty now opens three of the remaining doors to show you that there is no prize behind it. He then says, "Would you like to stick with the three doors you have chosen, or would you prefer to swap them for the one other door I have not opened?" What do you do? Do you stick with your three doors or do you make the 3 for 1 swap he is offering?

(see https://www.maa.org/devlin/devlin_07_03.html)

either
1. If I stick with my 3 doors and do not act on the new information, the information is irelevant. The probability of my success (that the prize is behind one of my original 3 doors) is still 3/7. The probability of the opposite event (that the prize is behind the one remaining door I did not choose and Monty did not show) is 1 - 3/7 = 4/7, because the probability of the prize being somewhere is 1 - certainty.

or
2. I ask myself: What is the probability that the one remaining door I did not pick is empty? The same as the probability that the 4 doors I did not pick were all empty to begin with, that is 3/7. Probability of the opposite event (that the prize is behind the one remaining door) is 1 - 3/7 = 4/7, again, because the probability of the prize being somewhere is 1 - certainty.

Always works. I make the 3 for 1 swap to increase my chances from 3/7 to 4/7.

Most people who claim 50% seem to downplay the effect that Monty has when he shows you the goat.

Think of it this way. You make an initial guess with 1/3 probability of being right. Let's say that Monty DOES NOT open any doors, and he asks you if you want to change your guess. If this is the case, then you have a 2/3 chance that the car is behind the door you didn't pick. This 2/3 chance is spread evenly across the two doors you didn't pick, leaving 1/3 chance for each door. This is normal, right?

So why don't you switch? It's because you don't know which door to switch to, so your chances don't become any better.

Now, when Monty shows a goat, he's actually helping you out a lot. He's saying "You'd like to switch but you don't which door to switch to? Well, DON'T switch here! Now, instead of spreading your 2/3 probability over two doors, it concentrates on one door."

If this doesn't please you, here's a reasonable analysis. You can draw a tree for these possibilit's. Without loss of generality, let's say that your initial guess is Door 1.
1) Let's say that Door 1 is in fact the car. This occurs with a 1/3 chance. Monty picks one of the other two doors, each with 1/2 probability (so the chance that of both you being right and him picking Door 2 is 1/3*1/2=1/6; same for Door 3). In each of these cases, you should stay. Thus, you should stay for 1/6+1/6=1/3 of the games.
2) Let's say that Door 1 is NOT the car. This happens 2/3 of the time. Monty's choice of door is forced. You should switch 2/3 of the time.

And by the way, don't get so damn annoyed at other people's arguments. I used to think as you do, but then I realized that my two apparent choices were actually not equal.

Grab a volunteer, three plastic cups and a cotton ball. Have the cotton ball represent the car, and the other empty cups the goats. Have your volunteer turn away while you place the cotton ball under a cup. Have the volunteer first switch all the time, for maybe 20-30 trials, and then stay every time for 20-30 trials. I did a science fair project on the subject once and my results were as follows:

Switching for 1000 trials: 664 wins, 336 losses

Staying for 1000 trials: 332 wins, 668 losses

jman_red

Anyone who wants to convince themselves of the correct answer should draw a decision tree, like Peter Stikker (www.cut-the-knot.com/peter.shtml). But once it's drawn, create a table of the probability for winning and losing against sticking or switching doors. I found it convinced my head - though not my heart. It'still FEELS wrong, but now I know although that it really is twice as likely that you'll win if you switch.

Stick & win: (1/36)*6 = 1/6
Stick & lose: (1/18)*6 = 1/3
Switch & win: (1/18)*6 = 1/3
Switch & lose: (1/36)*6 = 1/6

Thank you to all the contributors to this forum, whose postings have finally led me to the truth.

:7

What interest me most is the fact that there are so many approaches - many of which are expressed in mathematical lingo that I do not understand. So I just ignore those, and merely mentally register the fact that many "more learned than I" persons disagree with my "novice" viewpoint.

This applies to the most recent posting (02 Aug 02), wherein a referenced math author declared the Vos Savant view as wrong.

It'seems to me that a simpler review is important -easy for anyone to understand.

Let's look at the problem from the start. The contestant is standing in front of three doors, behind one of which is a prize - a car. The other two doors each shield a "booby" prize - a goat. The contestant chooses a door, and the host then opens one of the other doors displaying a goat. Ths host then allows tha contestant a second choice: stay with the initial choice or switch to the other door. Which is the best winning strategy? My review follows.

The probability of the contestant initially picking the "car" door is 1/3 - and the probability or his/her picking a "goat" door is 2/3. I think no one will disagree with that.

If he/she picks the "car" door, regardless of which door the host opens, the contestant loses if he/she decides to switch. This means that in 1/3 of the initial choices, a later switch decision results in a loss - and a win if a "stay" decision is made.

If he/she initially picks one of the "goat" doors (2/3 probability) the host must choose a a "goat" door, and if the contestant switches, he/she wins - and loses if a decision is made to switch.

In other words, not knowing what is behind the door of the initial contestant choice, once that decision is made, and the host performs his chore, the odds of switching - to win the car - are 2/3.

No probability "trees" or equations are necessary - just simple, easy to understand, every day logical reasoning.

Assume door 1 is chosen
Another way to look at it is, there are really two games being
played. One when the car is beyond is behind door 1 and one when
the car is behind door 2 or door 3

The winning percentage for each door for the second game is
100% 1/3 of the time. The winning percentage for each door
for the first game is 0% 1/6 of the time (if Monty treats
each door the same).

The total winning percentage for either door is then
100% *1/3 + 0% 1/6
-------------------- = 1/3 / 3/6 = 6/9 = 2/3
1/3 + 1/6

1/3 + 1/6

And one more thing for those counting the possible outcomes, you can't distinguish the 2 goats from each other even if they have different colors because it's just either you win or you loose. In the 3 doors, 1 door is the prize and the other 2 doors are nothing.

Look at how the problem is worded.
Step 1: Choose a door.
Step 2: The host will open a door, revealing it to be a goat, not the prize
Step 3: You may either switch to the last remaining door or stay at your own.

If you initailly choose wrong, which is 2/3 probability, then the host will eliminate the other door that is incorrect and you will switch to the correct door. That is true both statistically and in real life. Try it yourself.

It'seems to me, however, that you simply missed step 2.

jman_red

1) change
2) do not change

let us analyse the game for each of two players.

Firstly, the easy one. Somebody who does not change. He has already decided not to change and can therefore 'skip' any door being revealed. Therefore the decision is random selection out of three.

P(choosing the car with initial door selection)= 1/3
therefore
p(win car)=1/3

I am about to demonstrate that the alternative strategy, changing, gives a higher probability of winning. 'Not changing' is now NOT an option.

As before, there is a 1/3 probability of choosing the car first. However, for this individual, choosing the car would lead to finally picking a goat after swapping. EVERY OTHER initial choice of door leads to winning as Monty will reveal the other goat, leading the player straight to the car. The only way of not being led to the car is to choose the car the first time!!

P(selecting car first time)=1/3
and
P(being shown one of the two incorrect doors)=1

thus
P(swapping from incorrect door to correct door)=2/3

Hence this is the better strategy as it doubles the expectation of winning.

This result becomes clear as soon as you consider new rules as if swapping was required, or the converse.

The way this should be defined is as follows:

Decision tree 1:

C = Car
G1 = Goat 1
G2 = Goat 2

Possibilities for Decision Tree 1:

C G1 G2
G1 C G2
G1 G2 C

The chances of picking the correct door at random are 1 in 3.

The contestant picks door number 1. They have a 1 in 3 chance of having picked the correct door.

Monty opens door X and shows the contestant a goat. Monty then offers to let the contestant choose another door.

This creates a NEW decision tree, where the possibilities are:

C G1
G1 C

This is an original decision and is not linked in any way other than chronology and prize to be won to the previous decision.

Thusly, there is a 50% chance to select the correct door.

In this case, staying and changing are equally possible, and are equally beneficial.

Note that when considered as a separate decision, the odds of matching are different.

Compare this to matching dice. The odds of rolling two six sided dice simultaneously and coming up with the same number are far different than the odds of rolling a single six sided die, recording the first number, and then rolling a second six sided die hoping to match that first number. In this case, the second option is far more likely (1 in 6 rather than 1 in 36).

Subbie

It's easier to see with more doors also. Someone used an example with one million doors but his logic was flawed because you would be dead before you could open them all.

It's easier to see the ansewer if you ten doors with 9 monkeys and a car behind them. It's a little absurd to think that after you have made your 10% chance pick and then the host shows you 8 monkeys that your odds increase to 50%. The answer is more obvious even without using math if you look at two things: the host didn't randomly pick 8 doors and monkeys are more fun than goats.

The guy picks a card. The probability of the card being the ace is 1/3. You show him one of the others being a king.

1. If he does nothing, you might have shown the king to pigeons (and not to him) with the same effect, the probability of his success (that his initial choice is the ace) is still 1/3. The probability of the opposite event (that the ace is the card he did not pick and you did not show) is 1 - 1/3 = 2/3, because the ace surely is somewhere.

2. He asks himself: What is the probability that the other card I did not pick is also a king? The same as the probability that the 2 cards I did not pick were both kings to begin with, that is 1/3. Probability of the opposite event (that the other card is the ace) is 1 - 1/3 = 2/3, again, because the ace surely is somewhere.

If you want just to convince the guys by an experiment, do it with something that does not cost them much, say nuts, pennies, etc.

In the simplified Quiz Show example

Doors 1 2 3

Contestant chooses a door, host shows what’s behind one of the other doors, which instantly eliminates this door from the probability equation.

I am not a mathematician but a wee bit of common sense is all that is required.

This theory cleary shows that even after one of the doors is opened, it is still part of the equation. If the contestant is silly enough to still choose this door then the theory is correct, else, well I think you know the answer

When you are presented with three doors, you choose one which you do not wish Monty to discard. It doesn't matter what's behind it. You're not trying to pick the prize door.

Monty opens a door, which has no prize behind it. You are now presented with a simple choice - and here's what you know:

One door has the car, one door doesn't. That statement is 100% absolutely certain.

NOW, you make the choice - which is it? Door A or Door B? It's a 50:50 choice...

If you think about it, whether or not it is good idea to switch depends on whether or not the contestant initially chose the door with the prize.

If the contestant initially chose the door with the prize, then he must stick with that choice in order to win the prize. That is a no-brainer.

If the contestant initially chose a door without the prize, then if he changes his selection, he is certain of winning the prize. Why? Because Monty has opened the only other door that does not conceal the prize.

What is the probability that the initial selection was the correct selection? 1/3, yes? And the probability the initial selection was incorrect? 2/3, yes?

If you work that through, you will come to the conclusion that the contestant does indeed double his chances of winning by changing his selection. That is backed up by possibly hundreds of experiments.

Repeat after me.

Dont' open the door then place the car.
Place the car then open the door.
Place, then open, don't open then place.
If you first place the car in door number one you will
see that only half that time is door two opened.
When the car is placed in door3 all the time door 2 is
then opened.

Place then open. Place then open. Don't open and place

2. In the Monty Hall problem, the contestant chooses on two occasions. In the first, unconstrained, he stands a one third probability of choosing the door hiding the car, two thirds of not doing so. In the second, unconstrained, he has a probability of one half of picking the car, one of the two other options having been eliminated.

3. If the contestant constrains his second choice, this last probability may change. If he decides always to stick to his first choice, the second ‘choice' is now forced. He has essentially eliminated the probability of one half of winning on the second choice; he wins only if his first choice was correct Thus the probability of winning overall remains one third.
If he decides always to switch, then he has anulled the one third probability of being right first time. He wins if his first choice was wrong with a probability of two thirds, since Monty has eliminated the other possible wrong first choice. If his first choice was right - with a probability of one third - he has given it up, and loses. (In this case, it might appear that he can lose in two ways, depending on Monty's choice of door, but neither of these possible events alter the contestant's choices, and hence lie within the one in three chance of the his first choice being wrong.)

4. If the contestant decides before his first choice to switch at his second chance, he converts his two chances of choosing wrongly to winning choices, since Monty has to eliminate the other wrong choice. He does this at the expense of losing his one in three chances of having been right first time.

5. So decide to switch every time, and never change your mind at the second choice. If you do, half the time you'll be wrong.

Yup, if Monty places the car AFTER the door is opened
but not before.

>2. In the Monty Hall problem, the contestant chooses on two
>occasions. In the first, unconstrained, he stands a one
>third probability of choosing the door hiding the car, two
>thirds of not doing so. In the second, unconstrained, he has
>a probability of one half of picking the car, one of the two
>other options having been eliminated.

No, only one half if the car is placed AFTER the door is
opened. The car is distriuted BEFORE the door
is opened. You are opening then placing. You cannot do
that! Place the car THEN open the doors in ALL combinations.
Place then open. Do not open then place! Do you understand?,
you are driving me mad.
>

>3. If the contestant constrains his second choice, this last
>probability may change. If he decides always to stick to his
>first choice, the second ‘choice' is now forced.

No sir, the probabilty on the second door is 2/3 even if the
choice is made after one door is eleminated. Run a simulation
and see, but make sure to label the same doors with the same number
each round.

He has
>essentially eliminated the probability of one half of
>winning on the second choice; he wins only if his first
>choice was correct Thus the probability of winning overall
>remains one third.
> If he decides always to switch, then he has anulled the
>one third probability of being right first time. He wins if
>his first choice was wrong with a probability of two thirds,
>since Monty has eliminated the other possible wrong first
>choice. If his first choice was right - with a probability
>of one third - he has given it up, and loses. (In this case,
>it might appear that he can lose in two ways, depending on
>Monty's choice of door, but neither of these possible events
>alter the contestant's choices, and hence lie within the one
>in three chance of the his first choice being wrong.)
>
>4. If the contestant decides before his first choice to
>switch at his second chance, he converts his two chances of
>choosing wrongly to winning choices, since Monty has to
>eliminate the other wrong choice. He does this at the
>expense of losing his one in three chances of having been
>right first time.
>
>5. So decide to switch every time, and never change your
>mind at the second choice. If you do, half the time you'll
>be wrong.

NO, That is wrong, wrong, wrong. If the door would have
been 2/3 by "keeping" your switch choice after the door was open,
then not switching is 1/2? 1/2+2/3=1? You are crazy

The name's Keith Devlin

>7 chance of winning. See Kevin Devin's column at
>https://www.maa.org/devin/devin_07_03.html

The correct reference is

https://www.maa.org/devlin/devlin_07_03.html

Actually his explanation is wrong. He writ's:

1. The probability that the prize is behind door B or C (i.e., not behind door A) is 2/3.

2. The prize is not behind door C.

Combining these two pieces of information yields the conclusion that the probability that the prize is behind door B is 2/3. Endquote

In fact if the contestant chooses door A and Monty never opens
C when the car is behind A the chances on B are 100%

That is correct but one third of the time, i.e. when the car is behind door A, Monty also opens door C (i.e., one of the wrong doors). Thus Keith Devlin's column is correct.
Have a Good Day
KJ Ramsey

Read what I said again. I said that Monty (chooses) never
to open Door C when the car is behind A

"Monty never opens C when the car is behind A"

you then said:

"but one third of the time , when the car is behind A
Monty also opens door C"

Wrong, I JUST said Monty chooses NEVER to open door
C when the car is behind A. He opens door B instead!

Delvins column is wrong. In fact not enough information
was provided to determine the odds on door B which could
vary between 50%-100%. All that can be said with certainty
is that if the contestant always switches to both door B
and Door C the overall odds are 2/3. That is a hellavu lot
different than saying the odds on door B are 2/3

Remember in the (2/3) cases when the car is behind door B or C
the chance of winning is 100%. Your reasoning is that 2/3
of the time (when the car is behind B or C) the chances of winning is 2/3.

The error in your reasoning is to associate a probablity of losing
with "those instances when the car is not behind A".

Indeed 2/3 of the time the contestant will win 100% of the
time. Half the time being the cases when Monty opens door C.
Considering the 2/3 of the time the car is behind B or C,
those cases in which door C is opened do nothing to alter
the winning percentage. It's still 100%.

To figure out the overaall winning percentage for Door C you have to
figure out how many times door C lOSES. The losing cases come
from when the car is behind door A. Here Monty has a choice,
he can open door B or C. If he always opens door B he shifts
all the losing cases to door B and door C always wins.

half of that (2/3) or 1/2(2/3)=1/3 Monty opens door C
and the car is behind B so we know that the contestant
wins 1/3 of the time.

All we know then is how many times the contestant wins.
What fraction of the time Does

>Remember in the (2/3) cases when the car is behind door B or
>C
....
> The losing cases come from when the car is behind door A. Here Monty has a choice, he can open door B or C. If he always opens door B he shifts ...

Once again you are not using the lettering system used by Devlin where the letters are not applied until after the choices are made. Monty does indeed have a choice between two doors but the labels "B" and "C" are not applied until after the contestant choses door "A" and Monty next opens a door. Regardless of which door Monty opens that door is labeled "C" and the remaining door is labeled "B". It is easy to see from this system of labels that the probability that the car is behind door "A" is indeed 1/3 since the contestant has no idea where the car is and that the probability that the car is behind door "C" is zero since Monty never next opens a door with a car behind it. It follows that the probability that the car is behind door "B" is equal to 1 - 1/3 - 0 = 2/3. This agrees with the post by Tattertale where it was explained that the chances of winning are doubled if the contestant switches. However, the contestant would never switch to door "C" because that is the door that Monty opened.

Have a Good Day
KJ Ramsey

That is not true here is what Delvin said:

"Suppose the doors are labeled A, B, and C. Let's assume the contestant initially picks door A. The probability that the prize is behind door A is 1/3. That means that the probability it is behind one of the other two doors (B or C) is 2/3. Monty now opens one of the doors B and C to reveal that there is no prize there. Let's suppose he opens door C."

Delvin did not say that Monty picks one of the two remaining doors
and then labels that door C.

>Monty does indeed have a choice between
>two doors but the labels "B" and "C" are not applied until
>after the contestant choses door "A" and Monty next opens a
>door. Regardless of which door Monty opens that door is
>labeled "C" and the remaining door is labeled "B".

I agree that such a labeling system changes the problem
but that is not the spirit of the orginial question in which
three doors each have distinct identities given by A B and C
or if you prefer 1,2,3. People can judge for themselves what
they think Delvin really meant.

The two labeling systems are completely different.
Your system corresponnds to always switch for both doors
which averages out to 2/3. You can't relabel door C door
B and say its the same!

Suppose the contestant chooses A and Monty never chooses
C with a car behind A under "my" system. What are the odds
on B and C? Door C can never lose....think about it...

>Suppose the contestant chooses A and Monty never chooses
>C with a car behind A under "my" system. What are the odds
>on B and C? Door C can never lose....think about it...

The question is not whether a contestant wins or looses with a particular door but what are the probabilities of the car being behind B given the fact situation. Having Monty open door B is not the fact situation at issue.

Have a Good Day
KJ Ramsey

In the one third
>time that the car is behind A, neither B or C can win
>regardless of whether the contestant switches to them.

But one can lose more than the other depending on which
door Monty picks, suppose he always opens B, switching
to B never loses..

>>Suppose the contestant chooses A and Monty never chooses
>>C with a car behind A under "my" system. What are the odds
>>on B and C? Door C can never lose....think about it...
>
>The question is not whether a contestant wins or looses with
>a particular door but what are the probabilities of the car
>being behind B given the fact situation. Having Monty open
>door B is not the fact situation at issue.
>

Consider the following 6 cases, where Monty never opens Door
C with the car behind A. Initially

X00
X00

0X0
0X0

00X
00X

Now after Monty opens his door given by an M

XM0
XM0

0XM
0XM

0MX
0MX

Notice that in 4 of the cases door B is opened with
a win rate of 50% for switch.

XM0
XM0

0MX
0MX

In only two of the cases was door C opened with a win
rate of 100%

0XM
OXM

However the average win rate is still 2/3
because (4/6)(50%) + 2/6(100%) = 1/3 + 1/3 = 2/3

So be very careful when talking about the win rate
for a particular door without knowing Monty's actions
in the car behind door A case.

From what I see everyone is right - it is both 2/3 and 1/2. It depends from when you are measuring the probability of an event.

I seem to remember from my university days that in probability we have the phrase 'given that....' and that there is a difference between measuring the probability of an event from end to end (i.e. from the whole of the group of events) and the final or single event (e.g after all but one of the events have occured). Therefore, we need to decide whether we are measuring the probability from the BEGINNING of the whole chain of events or whether we measure the probability of the final event.
So, we have: 'given that someone has chosen a door and the host has shown a goat etc etc' and we are then left with two coices - one is a goat and one is the prize. AT THAT POINT there is a 50/50 chance - whatever has gone before is irrelevent. However, at the begininng of the whole chain of events - i.e. before we have made our first choice - there is a 2/3 choice of winning if we swap as per the example. Both are right....... but it depends whether you measure from the the beginning of the whole game or just the final choice..

Any comments?

If you have two doors, one with a car, another with the goat. You're asked to find the car. The probablilty is 50/50

car goat -> choose.. p = 50/50

(fine)

If you choose one and are given a chance to choose again, the probability is still 1/2

car goat -> choose one, think it over, then choose again. p = 1/2

(so far so good?)

Now let's say the peron is offered a third incorrect choice which he/she knows is incorrect. Since there is no way he/she will choose it, it will not affect the outcome so the probability is still 1/2

car goat known-goat -> choose one knowing that a door you will never pick is the wrong one, think it over, choose again.. p = 1/2

If you look really carefully, this is the Monty Hall problem.

Once Monty opens the one door with a goat, you're presented with a brand new problem which in no way depends on the previous choice. Like economic sunk costs, the previous decisions were become irrelevant.

To visualize this, imagine the person making the "switch, no switch choice" is different. This new person is told the following: one of these doors has a goat, the other one a car. A third door has a goat but really, you don't care about that. One last thing, a guy/girl before you made a guess which you don't know whether it is right or wrong. He/She picked one of the two doors you're given. His/Her guess was harder because I had more doors then. Anyhow he/she guessed door 'x' but I guess you probably don't care since you still don't know if it was right. Now please choose a door." Basically you can see thatn it the Monty Hall probelem isn't one of probabilistic saviness, rather simply one packed with irrelevant information, enough to fool the most ambitious mathematicians.

When Monty opens a door revealing one which is wrong. You pat yourself on the back for not having guessed that particular goat but that still does not change the nature of your new question. Now Monty says "there is a goat and a car behind on the two remaining doors. Which one do you think has the car, the one you chose originally(door A) or the other one(door B)?"

car goat -> choice.. p = 1/2

It doesn't take a lot of intelligence to see through this, it only takes the ability to see beyond your own arrogance.

The root of this problem is that too often, logicians and mathematicians treat randomsness as if it were logical: always following the rules and axioms it is supposed to. In the math world, that's fine. The real world however, where most probability is applied, it's not so simple. Weather predictions are wrong; polls don't tell you who will win the election; having a low city crime rate will not mean you can never get mugged. If people were able to recognize this simple fact, perhaps less time would be wasted fiddling around with numbers and losing money in casinos.

Who needs Monty?

Indeed, assume you (the contestant) are given a chance to point to a door and, after a while, offered to either open this door or the other two, i.e., both of them. Every one knows up front that behind one of these doors there's a goat. Of course, there may be a goat behind the other door as well. You receive the prize provided you open a door behind which it is. What should you do?

We may go further, and do away with the need to make a decision: to switch or not to switch. Point to a door, say magic words, and let all three doors open at once. Your only function is to keep scores: would you win if you switched, would you win if you did not. Monty is around there available to do you a favor and open a door with a goat behind, saving you a physical effort. Do you care for his help?

I believe a professional gambler would have had an easier time with this. Let's hypothesize the gambler will win something if he guesses correctly, but will loose something if he is wrong. Thus, he is more highly motivated than, for example, I. Consider the gambler's reasoning. "I'm being forced to pick one of three doors and only one is the right door. So, I'm probably picking the wrong door. Ah, now Monty is giving me a chance to change my pick and he's shown me that one of the other doors is not the one I want. Since I was thinking that one of the other two doors was likely the correct one, and Monty has eliminated one of them, I'll choose the other. For my money, that's almost like certainty."

I find many of the other ideas for reasoning to the correct result quite enlightening, but I wonder how much effort went into them. I know my first response upon reading about this problem was that vos Savant was wrong. I had to think about the problem quite a bit to admit'she was right. It took even longer to come up with the "gambler presentation". Perhaps Ms vos Savant and some others immediately saw the correct answer. They have my admiration. However, it seems unfair for some to berate latecomers to this problem for not immediately seeing the correct answer, as well as the "proper way" of getting to that answer.

When I originally thought about the controversy relating to this problem, I began to wonder how many times someone may have given a proof that seemed obviously correct only to find later that it was totally wrong. My original "proof" was like many others - we know one of the other two doors is a wrong guess. When Monty exposes it, we gain no new information. Therefore, there is no reason to change our guess. Equally, there is no reason not to. The flaw, of course, is the assumption we gain no new information. Once we know the correct answer(!), we can easily state and solve the problem using conditional probabilities. Why not use probabability theory in the first place? Because, it was clear we already knew the "correct" answer. Why go through all that formality? Now we know.

I enjoy your web site.

Mike

I am agreeable to sharing the glory.

>I believe a professional gambler would have had an easier
>time with this...

Looks very plausible. To me, I mean.

>However, it'seems
>unfair for some to berate latecomers to this problem for not
>immediately seeing the correct answer, as well as the
>"proper way" of getting to that answer.

I agree, except that in this time and day, I would appreciate the latecomer who did his homework. Such an effort would give an indication of an honest interest to resolve the problem. So much information is now available that it is hard to justify missing or overlooking much of it.

>I enjoy your web site.

Thank you for the kind words.

That's assuming that you KNOW you chose incorrectly from the beginning.

The initial pick, and the choice to stay or switch, are independent events. The first "pick" is a throwaway, unless sometimes he does not offer you the ability to switch. Since that's not the rules of the game, we don't have to worry about that. Think through it like this.

Regular "Let's Make a Deal" (LMAD): three doors, one car, two goats. You make a pick, then Monty throws out a bad door showing you it's a loser. You then determine if you'll switch or stay.

Now let's test out two more versions of the game--
LMAD version 2.0: three doors, one car, two goats. Monty throws out a bad door, and then you select what door you want. You can switch or stay, shouldn't matter either way.

LMAD version 2.1: only two doors, one car, one goat. Pick a door.

Those of you that argue that switching wins 2/3 of the time will still say that regular LMAD has a 66% win rate when switching. You MUST agree that LMAD version 2.0 has a 50% win rate. Same goes for version 2.1-- 50% win rate.

Guess what: all three are the same game. You're just being confused with the extra information in the original game. The game becomes 50/50 when you only have two choices. Don't mix that up with the 1/3 probability of initial success.

Let's analyze this one: LMAD version 2.2. One car, two goats. You are ASSIGNED a door (let's say door A) and told that there is a goat behind it. You can now keep the goat, or switch to one of the other doors. Well, anyone with an IQ over 85 will switch, but now the dilemma is door B or door C. Those are 50% odds, my friend, and this is the same game as the original.

Here's the probability if it helps:
Start of game
Given: doors A, B, and C; prize is behind one door
P(A) = 1/3
P(B) = 1/3
P(C) = 1/3
P(A) + P(B) + P(C) = 1

Assume one door is always the loser, which it is, we just don't know which one. For sake of example, we'll say it's C. Nobody ever chooses door C in our game, and Monty always opens door C. It's a waste of time to even put three doors out there (much as is a person making an initial pick, if you'll see it). Once we get to phase two of our game, here are the new probabilities, and the only ones that matter.
P(A) = 1/2
P(B) = 1/2
P(C) = 0
P(A) + P(B) + P(C) = 1

The fallacy that the 66% crowd believes is that there is not an equal chance of the car being behind doors A and B. They try to draw it up like this...
initial choice is A, P(A) = 1/3
*** THAT'S RIGHT, if there's not a second phase to the game ***
but they leave P(A) at 1/3, and since P(C) is 0, then P(B) must be 2/3, and that's why it pays to switch. In reality, we are guaranteed that P(A) = 1/2 and P(B) = 1/2, which means this is a 50/50 game.