• B
• cmb
In summary, the 50:50 conclusion is flawed because it is based on a misdirection: focussing on the 'swap' option when there are two other options that the host can follow.
cmb
I understand that there are other threads on this, (e.g. https://www.physicsforums.com/threads/monty-hall-vs-monty-fall.661985/ which gives a thorough account) but they support the proposition that a 'swap' scenario results in a 2/3 win probability rather than the 'intuitive' 50:50 assumption.

I want to discuss the 50:50 conclusion, which I think is correct and the 2/3 is a fallacy which I will explain.

To summarise the paradox, if you are presented with 3 doors in a competition and there is a car behind one (that you want to win) and a goat behind each of the other two (the booby prize), you pick one door and the chance of you getting the door with the car behind it is obviously 1/3. The host opens one of the other doors to reveal a goat (they know which ones have goats) and asks 'do you want to swap your choice of door'? A cursory inspection of that moment suggests the chance of you now getting the car is 50:50, but a slightly deeper inspection suggests a 2/3 chance if you go with a 'swap' strategy.

One might argue that your chance of picking the door with the car was 1/3, so it therefore remains 1/3. Therefore, all other options (i.e. swapping from that door) must then be 2/3, so it is better to swap, if you want to win.

I have to disagree for the following observation (which I will then back up with a reason the 2/3 is a fallacy); let's say that just at the moment you are about to decide whether to swap doors or not, you go sick and your quiz partner steps in. They are now presented with two doors, which, for sure, one has the car and one has the goat. There is simply no possible way that the probability for them is not 50:50! Yet, somehow the probability for you was different?

This makes absolutely no sense, and the fallacy exists here:- once the host opens one door to a goat, the 1/3 probability that you picked the car in the first 'round' then changes. It is no longer a 1/3 chance that you picked the car, it is now a 1/2 chance. It just is, it is clearly no longer the same probability as before. This is nothing less than Bayesian probability in which the confidence in a given observation changes according to prior observations. In this case, the later observation that there is a goat behind one door increases the probability of your original choice to 50:50. It does not remain the same. This is nothing less than an axiom of science; if you derive a hypothesis that has a very low probability (e.g. your prize car is behind the first door) you then test it, and by making observations that do not disprove that original hypothesis, then the probability of that hypothesis improves.

This is precisely the Monty Hall scenario: The probability your choice of door has a car behind it increases once you get a further non-contradictory observation.

OK, so that deals with the fallacy in words, but this doesn't un-stitch the description of events that lead to a 2/3 'count' of outcomes, if laid out sequentially in some structured matrix. So, where is the fallacy in that?

It is as follows; the 'misdirection' is the focus on the options for what happens when the game player picks a goat, so they could swap and win the car? Right? Well, yes, but what's missing is that there are TWO options that the host can follow if YOU pick the car correctly in the first place! TWO outcomes - they can either pick the 'leftmost' goat or the 'rightmost' goat. These are TWO options, not one, and this is where the 'mathematical' fallacy exists.

See, like this;

 Door 1​ Door 2​ Door 3​ ​ I pick​ Host Opens​ I stick​ I swap​ car​ goat​ goat​ ​ 1​ 2​ win​ lose​ ​ ​ ​ ​ 1​ 3​ win​ lose​ ​ ​ ​ ​ 2​ 3​ lose​ win​ ​ ​ ​ ​ 3​ 2​ lose​ win​

(It doesn't matter what the actual combination is behind the doors, the same would be the case for each combination.)

The point is that if you DO pick the car, then the host has TWO options. One outcome they pick one goat, the other they pick the other one. In a typical explanation of this, this is simply put down as 'the host picks a goat', as if that was one outcome. It isn't, it is two outcomes that look like one. The host can ONLY pick one other option if I pick a goat, but he has TWO options if I pick the car.

There are 4 options for any given combination behind the doors. Two are 'wins' and two are 'loses', whether you fix to one strategy or the other.

It is a 50:50 chance once the host opens a goat door. The question is whether he has opened 'the other' goat door or 'one of' the goat doors. These things are not equal.

This closes the circle between the 'obvious' fact that any 'new' contestant who comes into the competition just as the host opens a goat door is, clearly and obviously, presented with a 50:50 chance, yet the maths didn't say this. The reason is the 'two choice' option the host had is 'hidden' within the definition of the question.

I would welcome a confirmation or rebuttal on this, I have been mulling this over for a couple of weeks and I simply could not reconcile the 'new contestant' scenario with the apparent numbers from the mathematical description. But once you realize the host is actually making one of two choices (in effect, they are pre-selecting two of the contestant's options and reducing them to one) then you realize it is a 50:50 outcome after all. At least, I think this is the case, because the alternative strays so far from intuition it begs us to look for the fallacy in the maths, and I believe this is it.

There have been so many threads on this that I hesitate to respond to this one. But I will tell you what made it clear to me. Suppose, instead of 3 doors, that there are 1,000,000 doors, with 1 car and 999,999 goats. You choose a door, so your odds of choosing the car are clearly 1/1,000,000. Then the host opens 999,998 doors which all contain goats. Do you really believe that the odds that your door has the car has increased to 50:50? The car didn't move, so how could your odds have increased? Think about it from this standpoint.

FactChecker, DrClaude, BWV and 2 others
If you think 50:50 is intuitive then you should work on your intuition. Not everything that has two options is 50:50 and the two options are clearly different, there is no reason to assume they are the same.
cmb said:
which I think is correct
This is just silly. It's like claiming "I think 4+5=10". Even if you might find some argument why it would be you should realize that it is obviously wrong.
cmb said:
let's say that just at the moment you are about to decide whether to swap doors or not, you go sick and your quiz partner steps in. They are now presented with two doors, which, for sure, one has the car and one has the goat. There is simply no possible way that the probability for them is not 50:50! Yet, somehow the probability for you was different?
If your quiz partner knows which door you picked originally then changing the person didn't change anything, they should still switch to the other door. If your quiz partner doesn't know, then they do not have the knowledge needed to distinguish the doors, and they have to resort to random guessing. You, still watching, (should) know that one door is better than the other.
You have additional knowledge. You know that the door you picked originally couldn't have been opened, even if it had a goat.
cmb said:
once the host opens one door to a goat, the 1/3 probability that you picked the car in the first 'round' then changes. It is no longer a 1/3 chance that you picked the car, it is now a 1/2 chance.
That makes no sense. The host will always open a goat that is not your door - that event is guaranteed to happen, a Bayesian makes no update on a guaranteed event because it doesn't provide additional information (about the door you picked).
If sticking to your door would win a price with 50% probability then you must pick the right door initially half of the time. Despite there being three doors.

The four options in your table are not equally likely. If they would it would mean you pick the correct door initially with 50% probability. And it should be clear that you do not.

------

Play the game with pen and paper. Or let a computer play it 100,000 times. You'll see that switching the door gives you a 2/3 chance to win.

You are repeating the same old misconceptions that have been discussed over and over again in previous threads. I don't see anything new here. Do you really think continuing this discussion is useful?

scottdave, Vanadium 50, BWV and 1 other person
cmb said:
I would welcome a confirmation or rebuttal on this
First rebuttal is experiment. This has been tested and the standard analysis is correct. The switching strategy does in fact measurably increase your odds in the amount predicted.

The second rebuttal is that your analysis is incomplete. You need to include the cases where the goat is behind 2 and where the goat is behind 3. And look at the probabilities for each case.

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scottdave and jedishrfu
Lastly, what do you do if the goat is found sitting in the car eating the seats?

sbrothy, kith, atyy and 3 others
You are, quite bluntly, wrong about this. Your main fallacy is in your interpretation of what a probability means and what a conditional probability is. In the case of the Monty Hall problem, the probability can either be viewed as a frequency of the outcomes when repeated many times (frequentist interpretation) or as a confidence in which door the car hides behind (Bayesian interpretation). Both interpretations will give the same result in this case, you will have a 2/3 probability of selecting the door of the car if you switch doors.

cmb said:
once the host opens one door to a goat, the 1/3 probability that you picked the car in the first 'round' then changes. It is no longer a 1/3 chance that you picked the car, it is now a 1/2 chance. It just is, it is clearly no longer the same probability as before. This is nothing less than Bayesian probability in which the confidence in a given observation changes according to prior observations. In this case, the later observation that there is a goat behind one door increases the probability of your original choice to 50:50. It does not remain the same.
This is wrong. Assuming an equal prior for all doors, if you just opened a door with a goat, you would have increased each of the other doors to 0.5. However, this is not what happened. What happened was that the host opened a door he knew did not have a car among the ones you did not pick. This only redistributes the probability between the doors you did not pick because you effectively exempted your door from being opened.

Let us say you pick door A and that the host opens door C. Call the event that the car is behind your door A as well and the event that the host opens door C c. Then P(A|c) P(c) = P(c|A) P(A) by Bayes' theorem. Now, P(c|A) = 1/2 because if A is true the door is chosen randomly. Furthermore, P(c) = P(c|A)P(A) + P(c|B) P(B) + P(c|C) P(C), with B and C being the events that the car is behind door B/C, respectively. We have P(c|A) = 1/2 as discussed, but also P(c|B) = 1 and P(c|C) = 0 as the host will open C for certain if B is the car and not open C if the car is there. Assuming equal prior probabilities of 1/3, this leads to P(c) = 1/2 = P(c|A). Consequently, P(A|c) = P(A) = 1/3.

Now, for event B the situation looks different. We have (by the same argument) P(B|c) P(c) = P(c|B) P(B). Here P(c) is still 1/2, but P(c|B) = 1 as already discussed. It follows that
P(B|c) = P(c|B) P(B) / P(c) = 2 P(B) = 2/3.

cmb said:
I have to disagree for the following observation (which I will then back up with a reason the 2/3 is a fallacy); let's say that just at the moment you are about to decide whether to swap doors or not, you go sick and your quiz partner steps in. They are now presented with two doors, which, for sure, one has the car and one has the goat. There is simply no possible way that the probability for them is not 50:50! Yet, somehow the probability for you was different?

Yes, their probability is different unless they have been told what occurred. If you just present them with two doors, they will pick correctly in 50% of the cases because they will not know the prehistory. However, if you repeat the setup a million times, you will find that your friend gets the car 1/3 of the time when they pick the same door you did and 2/3 of the time when they pick the other door. Of course, they will pick your door 50% of the time, resulting in that they actually get the car with probability 0.5*(2/3 + 1/3) = 0.5.

Edit: If you insist on the 50-50 probability I have a nice betting opportunity to offer you ...

jedishrfu
The Monty Hall problem is a good example of how someone who understands conditional probability (the host) can hoodwink someone who does not (contestant).

It took me some time to understand it and the 1000+ doors argument won me over making it fairly obvious.

jedishrfu said:
It took me some time to understand it and the 1000+ doors argument won me over making it fairly obvious.

Speaking as someone who accepted the standard argument with conditional probabilities, can you put your finger on what exactly in the 1000+ doors argument that convinced you? Just on the face of it I would have guessed that it was prone to the same sort of fallacies as the 3 door case. Is it just the absurdity of having chosen the right door from the beginning?

To quote John Oliver on LHC being the end of the world: "It either happens or it doesn't, so 50-50."

The 1000+ door argument showed how likely you've chosen the wrong door. Of course a person could have chosen the right door but we all know how unlikely that is. I remember struggling with the concepts behind the three door version and the Marilyn Vos Savant controversy. However, convincing people is much harder since our sense of probability is so skewed.

https://en.wikipedia.org/wiki/Monty_Hall_problem

I still struggle today with Baye's Theorem attempting to understand it well enough to teach. For me, these problems sometimes appear as dyslexic thinking where I have to sort through my thoughts carefully before I come to the right realization.

In my case, the only contests I've ever won and there have been several are ones where only I entered.
- my dog won best of breed at a small 100+ breed dog show in Texas because he was the only Groenendael present

https://en.wikipedia.org/wiki/Groenendael_dog

- a sales contest at work to win a matchbox sized drive by providing contacts for the matchbox team. I provided TI as a possible candidate for the drives for their calculator products (didn't pan out) but since I was the only one who submitted anything I got the drive.

- two programming contests at work where I got a small award for my program given by mgmt to the top 5 who entered. Only 5 entered both times. One program was an infinite T3 game using embedding technology. The other was a music staff and piano part. Embedding the piano and you could tap out a song. Embedding the piano inside a staff part and you could write sheet music. They were used to sell document embedding technology.

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Another way to scale it is to have 333 cars 666 goats and 999 doors. The opens 333 doors one at a time showing no cars and only goats. However, this may still confuse some folks.

cmb said:
I understand that there are other threads on this, (e.g. https://www.physicsforums.com/threads/monty-hall-vs-monty-fall.661985/ which gives a thorough account) but they support the proposition that a 'swap' scenario results in a 2/3 win probability rather than the 'intuitive' 50:50 assumption.

Why not play the game and see? You could simulate it on your own, with a friend or by writing a computer programme. A friend of mine, many years ago, didn't believe me until he sat down to write a computer simulation of the game.

Take three playing cards. Let's say two Jacks for the goats and and Ace for the car. Mix them up.

First, you test the "stick" strategy.

You pick a card (and you have to stick with it). Then, you play the role of Monty and look at the cards and turn over a Jack. Then you turn over your card to see whether it's an Ace.

You count how often you win with the stick strategy.

Then, you test the "switch" strategy.

You pick a card. You play the role of Monty and reveal a Jack. Then, you switch to the other card and turn that over.

You count how often you win with the switch strategy.

There are exactly 3 doors, and sticking loses 2 out of 3 times, so there's only 1 out 3 chances to lose left for switching.

Here's a sample simulation program from https://rosettacode.org/wiki/Monty_Hall_problem:
Code:
Sinclair ZX81 BASIC
Works with 1k of RAM.

This program could certainly be made more efficient. What is really going on, after all, is

if initial guess = car then
sticker wins
else
switcher wins;

but I take it that the point is to demonstrate the outcome to people who may not see that that's what is going on. I have therefore written the program in a deliberately naïve style, not assuming anything.

10 PRINT "     WINS IF YOU"
20 PRINT "STICK","SWITCH"
30 LET STICK=0
40 LET SWITCH=0
50 FOR I=1 TO 1000
60 LET CAR=INT (RND*3)
70 LET GUESS=INT (RND*3)
80 LET SHOW=INT (RND*3)
90 IF SHOW=GUESS OR SHOW=CAR THEN GOTO 80
100 LET NEWGUESS=INT (RND*3)
110 IF NEWGUESS=GUESS OR NEWGUESS=SHOW THEN GOTO 100
120 IF GUESS=CAR THEN LET STICK=STICK+1
130 IF NEWGUESS=CAR THEN LET SWITCH=SWITCH+1
140 NEXT I
150 PRINT AT 2,0;STICK,SWITCH

Output:
WINS IF YOU
STICK           SWITCH
341             659
That page has simulations in dozens of computer languages, and they all illustrate the same conclusion: sticking wins 1/3 and switching wins 2/3.

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WWGD, DaveC426913 and PeroK
cmb said:
See, like this;

 Door 1​ Door 2​ Door 3​ ​ I pick​ Host Opens​ I stick​ I swap​ car​ goat​ goat​ ​ 1​ 2​ win​ lose​ ​ ​ ​ ​ 1​ 3​ win​ lose​ ​ ​ ​ ​ 2​ 3​ lose​ win​ ​ ​ ​ ​ 3​ 2​ lose​ win​

Let's analyse this table. The car is behind door 1. In four games you pick door 1 twice and doors 2 and 3 only once each!

The data you present here assumes you pick the winning door 50% of the time. You can ignore all the door opening by Monty. That's irrelevant. You've picked the right door 50% of the time whatever happens.

Dale, sysprog and Orodruin
cmb said:
I have to disagree for the following observation (which I will then back up with a reason the 2/3 is a fallacy); let's say that just at the moment you are about to decide whether to swap doors or not, you go sick and your quiz partner steps in. They are now presented with two doors, which, for sure, one has the car and one has the goat. There is simply no possible way that the probability for them is not 50:50! Yet, somehow the probability for you was different?

Let's analyse this. First, let's assume that Monty knows where the car is. Let's say it's behind door 2 and you pick door 1. He opens door 3.

For Monty, there are no probabilities here. He knows where the car is. He knows that if you stick you lose and if you switch you win.

You don't know where the car is, but you do know the rules of the game. You can calculate that the probability of winning is 1/3 if you stick and 2/3 if you switch.

Your friend then takes over. Let's assume he doesn't know the rules of the game. Then he has even less knowledge than you. He just sees two doors and - based on his limited knowledge - he would compute a different probability.

But, if you explain to your friend the rules, so that he has the same knowledge as you, then he should calculate it's better to switch.

And, if Monty tells your friend where the car is (so your friend has the same information as Monty), then he can win 100% of the time.

A good example of this is the following experiment.

You have a pack of cards and two friends outside the room. You look at the top card, which is the Queen of Spades. Your first friend comes in and you tell her the top card is a spade. Your second friend then comes in but you tell him nothing. The two friends then have to try to guess the top card.

You're first friend has a much better chance (1/13) of getting it right. The second friend only has a 1/52 chance of getting it right. And you can calculate all this. The extra knowldege gives your first friend a better chance of winning.

In the Monty Hall problem if the contestant uses all the knowldege at his/her disposal, then he/she should switch and win 2/3 of the time. But, if the contestant doesn't use all the knowledge available and sticks, then they only win 1/3 of the time. And, if the contestant tosses a coin to see whether to stick or switch, then they would win 1/2 of the time.

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sysprog
Another way to look at it is as follows: the stick strategy wins if you initially pick the car. The switch strategy wins if you initially pick a goat. The probability of initially picking the car is 1/3 and the probability of initially picking a goat is 2/3.

Motore, etotheipi, DrClaude and 6 others
jedishrfu said:
appear as dyslexic thinking where I have to sort through my thoughts carefully before I come to the right realization
Me too. This Monty Hall thing has always bugged me, so I have a lot of sympathy for the previous posters who are harshly judged by others as having "bad intuition."

I don't know if this is a true story, but
A particularly interesting exchange occurred between Vazsonyi and his good friend Paul Erdos. Erdos was ―one of the century‘s greatest mathematicians, who posed and solved thorny problems in number theory and other areas and founded the field of discrete mathematics, which is the foundation of computer science. He was also one of the most prolific mathematicians in history, with more than 1,500 papers to his name. Vazsonyi relates how in 1995, after relating the goats and Cadillac problem and the answer (always switch), Erdos responded ―"No. That is impossible..."
https://files.eric.ed.gov/fulltext/EJ1060344.pdf

I wouldn't criticize anyone too harshly for following Erdos' intuition.

That's not to say the "don't switch" intuition isn't wrong, it is. Wrong. So maybe that Erdos story isn't accurate.

I finally made myself a spreadsheet to run a bunch of cases and with 50,000 trials I "win" 33% of them following the "don't switch" approach - and win 66% by switching. Creating the spreadsheet is illuminating.

I think the problem is interesting for a number of non-mathematical reasons. For the TV show contestants I think there's more than math to their intuition. First, Monty is a fast-talking con man who doesn't want to give away his car. This is of course false, if too many people lose the show won't work. Plus, it isn't his car.

Second, I think the contestants have a strong feeling about their initial choice. They'd rather stick and lose, than switch and lose. If they switch and lose, that means they had the car in their hand and gave it away. Kind of like getting voted off Survivor, with an immunity necklace in your pocket.

Dale said:
Another way to look at it is as follows: the stick strategy wins if you initially pick the car. The switch strategy wins if you initially pick a goat. The probability of initially picking the car is 1/3 and the probability of initially picking a goat is 2/3.

This really helped me "get it."

I had another thought - suppose the zonk prizes really are goats, and further, that you'd rather have a goat than a car (maybe an extreme Amish farmer?). So imagine your objective is to win a goat. Your initial pick gives you a 2/3 chance. Would you switch? No way!

Dale and jedishrfu
gmax137 said:
I have a lot of sympathy for the previous posters who are harshly judged by others as having "bad intuition."
I have a lot of sympathy for other posters who are sincerely and humbly confused and don't understand the resolution. I have very little sympathy for posters like the OP who come in telling everyone else that the well-known standard resolution is wrong and that everyone else has committed a fallacy to get the wrong answer. The harshness is probably not a reaction to the bad intuition but to the arrogance.

phinds, gmax137 and mfb
I'd love to see this table fleshed out to show the correct outcomes.
I would do it myself (as I have done in the past) but I keep tripping over the same logical errors.

I can even see where the OP made the error. S/he is equating possibilities with probabilities: (2 rows each "must" mean 50:50 probability - which is wrong*) but I can't figure out how to correctly represent it in the table.

cmb said:
 Door 1​ Door 2​ Door 3​ ​ I pick​ Host Opens​ I stick​ I swap​ car​ goat​ goat​ ​ 1​ 2​ win​ lose​ ​ ​ ​ ​ 1​ 3​ win​ lose​ ​ ​ ​ ​ 2​ 3​ lose​ win​ ​ ​ ​ ​ 3​ 2​ lose​ win​

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Its actually easy to see from the table - if you always swap, you win 2/3rds of the time. If you always stick you win 1/3 of the time

gmax137 said:
Second, I think the contestants have a strong feeling about their initial choice. They'd rather stick and lose, than switch and lose. If they switch and lose, that means they had the car in their hand and gave it away.
This is an important point - and an example of the well-documented Loss Aversion effect

BWV said:
Its actually easy to see from the table - if you always swap, you win 2/3rds of the time. If you always stick you win 1/3 of the time
I'm afraid it's not easy for me to see from the table. I see two wins for 'stick' and two wins for 'swap'. What do you see?

Well, you need to add a probablity entry to the table. Each door pick has a probablility of 1/3, so when you split the pick of door 1, each of the first two entries only have a probability of 1/6. Then it all works out. With "I stick" you win 1/3 of the time, and with "I switch" you win 2/3 of the time.

cmb said:
 Door 1​ Door 2​ Door 3​ Prob​ I pick​ Host Opens​ I stick​ I swap​ car​ goat​ goat​ 1/6​ 1​ 2​ win​ lose​ ​ ​ ​ 1/6​ 1​ 3​ win​ lose​ ​ ​ ​ 1/3​ 2​ 3​ lose​ win​ ​ ​ ​ 1/3​ 3​ 2​ lose​ win​

BWV
phyzguy said:
Well, you need to add a probablity entry to the table. Each door pick has a probablility of 1/3, so when you split the pick of door 1, each of the first two entries only have a probability of 1/6. Then it all works out. With "I stick" you win 1/3 of the time, and with "I switch" you win 2/3 of the time.
That's where I was going with it, but it is not sufficient to just stick another column on the table and label it with 1/3 and/or 1/6 - it should be visualized in a self-explanatory fashion.

There should be six rows, or a multiple of six.

The ideal solution might be to show the first n games, with the lucky (contrived) happenstance that every game followed a different possible path.

PeroK
DaveC426913 said:
I'm afraid it's not easy for me to see from the table. I see two wins for 'stick' and two wins for 'swap'. What do you see?
There should only be one win for stick, not two - it does not matter which door the host opens if your original pick was the car.

DaveC426913 said:
There should be six rows, or a multiple of six.
But there aren't six possibilities. If you pick the car, the host has two choices of which door to open, but if you pick one of the goats, the host has only one choice. So there aren't six options, there are only four. The fallacy is in thinking these four options have equal probability when they actually don't.

Stephen Tashi
 door 1 door 2 door 3 I pick I stick I Swap car goat goat 1 W L 2 L W 3 L W

PeroK, jedishrfu, hutchphd and 1 other person
phyzguy said:
But there aren't six possibilities. If you pick the car, the host has two choices of which door to open, but if you pick one of the goats, the host has only one choice. So there aren't six options, there are only four. The fallacy is in thinking these four options have equal probability when they actually don't.
Right.

But - again - the table doesn't show probabilities. And simply adding another column for probability won't clarify why those probabilities are what they are.

BWV said:
 door 1 door 2 door 3 I pick I stick I Swap car goat goat 1 W L 2 L W 3 L W
You have eliminated the host's involvement. If the host's involvement is not relevant then you are not representing the Monty Hall problem at all.

DaveC426913 said:
You have eliminated the host's involvement. If the host's involvement is not relevant then you are not representing the Monty Hall problem at all.
No, the host's action is captured with the switch column - the host always offers a switch after opening a door with a goat - which is why if you did not originally pick the door with the car, you always win on a switch. In 2 and 3 if the host did not reveal a goat, a switch would be a 50/50 chance of winning

mfb
Numbers on lines are relative probabilities, numbers on the right are absolute probabilities.

Edit: Now with 6 cases for symmetry.

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DaveC426913, sysprog, phinds and 1 other person
The probability tree or decision tree shown by @mfb is the most efficient tool for analyzing this scenario. However, if you insist on a table, here is the full table.

 Car Location P(A) My Guess P(C|A) Monte Reveals P(E|A,C) P(A,C,E) Stay Wins E(Stay) Switch Wins E(Switch) 1 0.3333 1 0.3333 1 0 0 1 0 0 0 1 0.3333 1 0.3333 2 0.5 0.05556 1 0.0556 0 0 1 0.3333 1 0.3333 3 0.5 0.05556 1 0.0556 0 0 1 0.3333 2 0.3333 1 0 0 0 0 1 0 1 0.3333 2 0.3333 2 0 0 0 0 1 0 1 0.3333 2 0.3333 3 1 0.11111 0 0 1 0.111111 1 0.3333 3 0.3333 1 0 0 0 0 1 0 1 0.3333 3 0.3333 2 1 0.11111 0 0 1 0.111111 1 0.3333 3 0.3333 3 0 0 0 0 1 0 2 0.3333 1 0.3333 1 0 0 0 0 1 0 2 0.3333 1 0.3333 2 0 0 0 0 1 0 2 0.3333 1 0.3333 3 1 0.11111 0 0 1 0.111111 2 0.3333 2 0.3333 1 0.5 0.05556 1 0.0556 0 0 2 0.3333 2 0.3333 2 0 0 1 0 0 0 2 0.3333 2 0.3333 3 0.5 0.05556 1 0.0556 0 0 2 0.3333 3 0.3333 1 1 0.11111 0 0 1 0.111111 2 0.3333 3 0.3333 2 0 0 0 0 1 0 2 0.3333 3 0.3333 3 0 0 0 0 1 0 3 0.3333 1 0.3333 1 0 0 0 0 1 0 3 0.3333 1 0.3333 2 1 0.11111 0 0 1 0.111111 3 0.3333 1 0.3333 3 0 0 0 0 1 0 3 0.3333 2 0.3333 1 1 0.11111 0 0 1 0.111111 3 0.3333 2 0.3333 2 0 0 0 0 1 0 3 0.3333 2 0.3333 3 0 0 0 0 1 0 3 0.3333 3 0.3333 1 0.5 0.05556 1 0.0556 0 0 3 0.3333 3 0.3333 2 0.5 0.05556 1 0.0556 0 0 3 0.3333 3 0.3333 3 0 0 1 0 0 0 0.3333 0.666667

The table can be simplified somewhat by simply identifying the door that the car is behind as 1 and removing the other lines of the table.
 Car Location P(A) My Guess P(C|A) Monte Reveals P(E|A,C) P(A,C,E) Stay Wins E(Stay) Switch Wins E(Switch) 1 1 1 0.3333 1 0 0 1 0 0 0 1 1 1 0.3333 2 0.5 0.16667 1 0.1667 0 0 1 1 1 0.3333 3 0.5 0.16667 1 0.1667 0 0 1 1 2 0.3333 1 0 0 0 0 1 0 1 1 2 0.3333 2 0 0 0 0 1 0 1 1 2 0.3333 3 1 0.33333 0 0 1 0.333333 1 1 3 0.3333 1 0 0 0 0 1 0 1 1 3 0.3333 2 1 0.33333 0 0 1 0.333333 1 1 3 0.3333 3 0 0 0 0 1 0 0.3333 0.666667

Forgive me for appearing pedantic in pursuing this.Since we don't have to deal with scenarios that the rules forbid (such as Monty opening the door with the car - probability zero), we get this:

 Car Location​ P(A)​ My Guess​ P(C|A)​ Monte Reveals​ P(E|A,C)​ P(A,C,E)​ Stay Wins​ E(Stay)​ Switch Wins​ E(Switch)​ 1​ 1​ 1​ 0.3333​ 2​ 0.5​ 0.16667​ 1​ 0.1667​ 0​ 0​ 1​ 1​ 1​ 0.3333​ 3​ 0.5​ 0.16667​ 1​ 0.1667​ 0​ 0​ 1​ 1​ 2​ 0.3333​ 3​ 1​ 0.33333​ 0​ 0​ 1​ 0.333333​ 1​ 1​ 3​ 0.3333​ 2​ 1​ 0.33333​ 0​ 0​ 1​ 0.333333​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ 0.3333​ ​ 0.666667​

which leaves us back where we started.

But, as I said, all this has done is add a column (or two) that provides the probability value.
You've added the probability equations P(E|A,C) and P(A,C,E), but that's not going to help a layperson (like me) see why those probabilities are what they are. (eg. Think of showing this to my family over the dinner table.)

I think what I would like (and may end up producing myself), is the actual first n results of the experiment -except not random. The first n results of an actual game played out, will not list probabilities, simply list the final outcomes of n games - and the table will literally have two 'switch' wins for every one 'stick' win.

Just multiply the probabilities by 100 and you can treat them as games out of 100, after rounding them suitably.

I still think the simplest approach is "if you pick the right door initially then you win when keeping it, if you do not you win when switching". What is the probability to pick the right door initially? 1/3, obviously.

Dale
Well **that** OP sure released a flood of responses. I see that cmb is long gone, however...

PeroK
marcusl said:
Well **that** OP sure released a flood of responses. I see that cmb is long gone, however...
He's probably too embarrassed to come back.

Be a good time to close this thread. We have more than enough on the Monty Hall problem

DaveC426913 said:
Since we don't have to deal with scenarios that the rules forbid (such as Monty opening the door with the car - probability zero),
Sure, we don’t have to, but by including them in the table it is clear that we also need to explicitly calculate the probabilities and we can check that those go to zero. When you are making a table like this it is best to be systematic and include everything.
DaveC426913 said:
which leaves us back where we started.
Except with the probabilities calculated.
DaveC426913 said:
that's not going to help a layperson (like me) see why those probabilities are what they are
Why not? Which don’t you see. I can explain any step.

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