B The Monty Hall paradox/conundrum

[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.

 

[phyzguy]

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.

 

[mfb]

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.

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.

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.

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?

 

[Dale]

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.

 

[jedishrfu]

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

https://abcnews.go.com/blogs/headlines/2012/05/goat-takes-over-mans-car-eats-seats

 

[Orodruin]

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.

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.

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.

 

[Orodruin]

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."

 

[jedishrfu]

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.

 

[jedishrfu]

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.

 

[PeroK]

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.

 

[sysprog]

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:

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.

 

[PeroK]

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.

 

[PeroK]

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.

 

[Dale]

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.

 

[gmax137]

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.

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]

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.

 

[DaveC426913]

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.

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​

 

[BWV]

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

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

 

[DaveC426913]

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?

 

[phyzguy]

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.

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​

 

[DaveC426913]

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.

 

[BWV]

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.

 

[phyzguy]

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.

 

[BWV]

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

 

[DaveC426913]

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.

 

[DaveC426913]

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.

 

[BWV]

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.

 

[Dale]

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

 

[DaveC426913]

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.

 

[mfb]

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.

 

[marcusl]

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

 

[phinds]

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

 

[Dale]

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.

which leaves us back where we started.

Except with the probabilities calculated.

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.

 

[Ibix]

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.)

A suggestion - get a piece of lined paper and draw a table eighteen lines high, plus a header row. Divide it into four columns headed "Car behind door...", "You choose door...", "Host opens door...", and "You win if you...".

Divide the first column into three equal sized boxes labelled 1, 2, 3. The heights of the boxes represent the equal 1/3 probabilities of the three outcomes.

Copy the divisions in the first column into the second, then subdivide each one for the three options you have at this stage. You should end up with nine two-line high boxes, representing the nine 1/9th probabilities of each choice.

Copy the divisions in the second column into the third and subdivide according to the options the host has. Sometimes this will be one two-line high box and sometimes two one-line high boxes.

Finally, copy the divisions from the third column into the fourth and label each box stick or switch.

This is just the tree mfb drew, but with probability represented by height. I think it's probably the same as one of Dale's tables, but with some cells merged.

If you count the stick and switch entries, you'll find they are 50:50. But you would be making the same mistake cmb made - forgetting to weight by the probability of the outcome. If you count stick and switch entries multiplied by height (i.e. by the probability of the outcome) you'll get the expected 33.3:66.6 ratio.

Note also that your choice is completely constrained by the first and second column. The third one is actually irrelevant, which is to say that "sticking is only right if you guessed right first time" - trivially 1/3.

I'm sorry I can't draw this at the moment - my LaTeX table-fu is not strong enough on a tiny phone screen.

 

[sysprog]

I'm sorry I can't draw this at the moment - my LaTeX table-fu is not strong enough on a tiny phone screen.

Not to worry, @Ibix; @stevendaryl presented a full matrix in this (now closed) thread in his 02-17-2019 post #79 therein:

The correct analysis is that there are 9 possibilities for the situation (x,y) where x is the location of the car (behind doors 1, 2, or 3) and y is the door you pick. All 9 possibilities are equally likely (if everybody plays randomly). They are:

​

1. (1,1)​

2. (1,2)​

3. (1,3)​

4. (2,1)​

5. (2,2)​

6. (2,3)​

7. (3,1)​

8. (3,2)​

9. (3,3)​

The strategy of always switching wins in cases 2, 3, 4, 6, 7, 8. It loses in cases 1, 5, 9. That's the correct analysis.

Now, all of those possibilities have probability 1/9. Now, let's take possibility 1. It has two subcases:

1A: Monty opens door 2.​

1B: Monty opens door 3.​

The probabilities for those two subcases have to add up to the probability for case 1. They are equally likely. So we have:

P(1A) = P(1B) = 1/2 P(1)​

Similarly, there are two subcases for 5 and 9.

So the complete probability matrix looks like this:

$\left( \begin{array} \\ choice & car & open & probability \\ 1 & 1 & 2 & \frac{1}{18} \\ 1 & 1 & 3 & \frac{1}{18} \\ 1 & 2 & 3 & \frac{1}{9} \\ 1 & 3 & 2 & \frac{1}{9} \\ 2 & 1 & 3 & \frac{1}{9} \\ 2 & 2 & 1 & \frac{1}{18} \\ 2 & 2 & 3 & \frac{1}{18} \\ 2 & 3 & 1 & \frac{1}{9} \\ 3 & 1 & 2 & \frac{1}{9} \\ 3 & 2 & 1 & \frac{1}{9} \\ 3 & 3 & 1 & \frac{1}{18} \\ 3 & 3 & 2 & \frac{1}{18} \end{array} \right)$

I think that all the probabilities add up to 1.

And indeed they do sum to 1, and furthermore (just as LaPlace would say, when omitting a deriviation, "il est aisé à voia que"), it is easy to see that:

the probability sum of the 50% of the rows on which the first 2 row-column cells are equal is 1/3 and represents the probability that sticking wins,​

​

and that

​

the probability sum of the 50% of the rows on which the first 2 row-column cells are unequal is 2/3 and represents the probability that switching wins.​

Here's the code (from rosettacode.org) for a Python simulation:

from random import randrange

doors, iterations = 3,100000  # could try 100,1000

def monty_hall(choice, switch=False, doorCount=doors):
  # Set up doors
  door = [False]*doorCount
  # One door with prize
  door[randrange(doorCount)] = True

  chosen = door[choice]

  unpicked = door
  del unpicked[choice]

  # Out of those unpicked, the alternative is either:
  #   the prize door, or
  #   an empty door if the initial choice is actually the prize.
  alternative = True in unpicked

  if switch:
    return alternative
  else:
    return chosen

print "\nMonty Hall problem simulation:"
print doors, "doors,", iterations, "iterations.\n"

print "Not switching allows you to win",
print sum(monty_hall(randrange(3), switch=False)
          for x in range(iterations)),
print "out of", iterations, "times."
print "Switching allows you to win",
print sum(monty_hall(randrange(3), switch=True)
          for x in range(iterations)),
print "out of", iterations, "times.\n"

You can see it run on codepad at http://codepad.org/E7HfuR4s

 

[Dadface]

My problem here is I can't see where there is a problem. I might be overlooking something.

1. If your original choice is the car (1/3 chance) and you switch, you will will win either goat A or goat B.

2. If your original choice is goat A (1/3 chance) and you switch, you will win the car.

3. If your original choice is goat B (1/3 chance) and you switch, you will win the car.

So by revealing a goat the host is giving a 2/3 chance of winning the car.

 

[sysprog]

My problem here is I can't see where there is a problem. I might be overlooking something.

1. If your original choice is the car (1/3 chance) and you switch, you will will win either goat A or goat B.

2. If your original choice is goat A (1/3 chance) and you switch, you will win the car.

3. If your original choice is goat B (1/3 chance) and you switch, you will win the car.

So by revealing a goat the host is giving a 2/3 chance of winning the car.

The problem that most people who experience a problem experience comes from incomplete reasoning: at the point at which they're offed the option to stick or to switch, they see the facts of there being two doors (chosen and not chosen), two options (stick or switch), and two unknowns (both doors conceal something), as meaning that the two possibilities are equiprobable.

Your not seeing a problem indicates that you didn't disregard the 1/3 original probably of having chosen the car, or the fact that sticking means keeping that probability, or the fact that switching means trading that 1/3 for the other 2/3, and not missing the fact if the car is behind one of the doors that you didn't originally pick, showing you which one it isn't behind doesn't change the fact that it's behind one of them.

It's easy for some people to get distracted by the main non-pivotal fact of the procedure. The act of opening of the only non-car door available to him, or of randomly choosing and opening one of two non-car doors available to him, has no effect on the chance of winning if sticking versus the chance of winning if switching. If the host didn't open a door, but simply asked the contestant whether he'd like to keep the door he's chosen, or switch for the other 2 doors, probably almost no-one would opt to keep the originally chosen door.

 

[Ibix]

Not to worry, @Ibix; @stevendaryl presented a full matrix in this (now closed) thread in his 02-17-2019 post #79 therein:

Almost - he didn't make the 1/9th row twice the height of the 1/18th rows, nor merge cells.

 

[Ibix]

...and that python code is ugly. I'd use:

import random

# Specify how many doors and how many times we play
doors = 3
trials = 1000

# Choose where the car is each game... 
car = [random.randint(1,doors) for i in range(trials)]
# ...and the door you choose each game
youChoose = [random.randint(1,doors) for i in range(trials)]

# Define the rules the host follows... 
def hostLeavesClosed(y, c):
    if y == c:
        # You chose the car. Host chooses a random door to leave shut
        hostsChoices = [d for d in range(1, doors + 1) if d != y]
        return random.choice(hostsChoices)
    else:
        # You did not choose the car. Host leaves that door shut
        return c
# ...and work out what he does each game
yourOtherChoice = [hostLeavesClosed(y, c) for (y, c) in zip(youChoose, car)]

# Calculate how many times you'd have won by sticking... 
stickWins = sum([1 for (c, y) in zip(car, youChoose) if c == y])
# ...and by switching
switchWins = sum([1 for (c, y) in zip(car, yourOtherChoice) if c == y])

# Display the results
print("With " + str(doors) + " you would have won " + str(stickWins) + " times by sticking and " + str(switchWins) +" times by switching out of " + str(trials) + " trials.")
print("Theory predicts " + str(trials / doors) + " wins by sticking and " + str(trials - trials / doors) + " wins by switching")

 

[sysprog]

Almost - he didn't make the 1/9th row twice the height of the 1/18th rows, nor merge cells.

In his $\LaTeX$ matrix he had 12 rows, 4 for each door instead of 3, with 2 of each 4 showing 1/18 instead of 1/9, and that's functionally equivalent to, and perhaps more perspicuous than, your device of double height to reflect the bifurcation of the those 9ths into 18ths.

 

[Ibix]

In his $\LaTeX$ matrix he had 12 rows, 4 for each door instead of 3, with 2 of each 4 showing 1/18 instead of 1/9, and that's functionally equivalent to, and perhaps more perspicuous than, your device of double height to reflect the bifurcation of the those 9ths into 18ths.

But the whole point of my construction is that the height of each row is proportional to the probability of the outcome. I chose to do it that way because Dave wanted - I think - a visual representation of the probabilities. Steven's version doesn't do that - it just lists the probabilities at the end of the table, which Dave already said wasn't what he wanted.

 

[sysprog]

But the whole point of my construction is that the height of each row is proportional to the probability of the outcome. I chose to do it that way because Dave wanted - I think - a visual representation of the probabilities. Steven's version doesn't do that - it just lists the probabilities at the end of the table, which Dave already said wasn't what he wanted.

To me the difference doesn't matter; however, I do see your point.

 

[hutchphd]

perspicuous

Excellent new word for my old brain. Thanks.

 

[DaveC426913]

Divide the first column into three equal sized boxes labelled 1, 2, 3. The heights of the boxes represent the equal 1/3 probabilities of the three outcomes.

No.

In the case of experimental results, probabilities don't appear.

All you have is the initial conditions, and the results. And the results will literally show two 'switch' wins for every one 'stick' win.

 

[DaveC426913]

OK, rather than concentrating on probabilities, I'm focusing on results - played games - i.e. the probabilities have already played out - and we are left with the results (after all, that is the only thing that ultimately matters).

To be clear: I am certain of the proof that switching is the winning strategy; I'm now simply trying to (succinctly) demonstrate it as fact rather than theory.

Here is my table showing 36 games. (And yes, I know it is effectively identical to the OP's table.)

Game #​	Setup​	I pick​	Host opens​	I choose to​	to Door #​	Outcome​	Stick​	Switch​
1​	Car, Goat, Goat​	1​	2​	Stick​	1​	Win​	1​	​
2​	Car, Goat, Goat​	1​	3​	Stick​	1​	Win​	1​	​
3​	Car, Goat, Goat​	1​	2​	Switch​	3​	Lose​	​	​
4​	Car, Goat, Goat​	1​	3​	Switch​	2​	Lose​	​	​
5​	Car, Goat, Goat​	2​	1​	Stick​	2​	Lose​	​	​
6​	Car, Goat, Goat​	2​	3​	Stick​	2​	Lose​	​	​
7​	Car, Goat, Goat​	2​	1​	Switch​	3​	Lose​	​	​
8​	Car, Goat, Goat​	2​	3​	Switch​	1​	Win​	​	1​
9​	Car, Goat, Goat​	3​	1​	Stick​	3​	Lose​	​	​
10​	Car, Goat, Goat​	3​	2​	Stick​	3​	Lose​	​	​
11​	Car, Goat, Goat​	3​	1​	Switch​	2​	Lose​	​	​
12​	Car, Goat, Goat​	3​	2​	Switch​	1​	Win​	​	1​
​	​	​	​	​	​	​	2​	2​
​	​	​	​	​	​	​	​	​
13-24​	Goat, Car, Goat​	​	​	​	​	​	2​	2​
​	​	​	​	​	​	​	​	​
25-36​	Goat, Goat, Car​	​	​	​	​	​	2​	2​

Obviously, the car being behind door 2 or 3 gives identical results, so whatever happens in those scenarios is identical to the first 12.

Question: Are all the games 1-36 equally likely to have happened? i.e. Does every row have the same probability of showing up in the table? If not, what rows am I missing?

 

[PeroK]

OK, rather than concentrating on probabilities, I'm focusing on results - played games - i.e. the probabilities have already played out - and we are left with the results (after all, that is the only thing that ultimately matters).

To be clear: I am certain of the proof that switching is the winning strategy; I'm now simply trying to (succinctly) demonstrate it as fact rather than theory.

Here is my table showing 36 games.

Game #​	Setup​	I pick​	Host opens​	I choose to​	to Door #​	Outcome​	Stick​	Switch​
1​	Car, Goat, Goat​	1​	2​	Stick​	1​	Win​	1​	0​
2​	Car, Goat, Goat​	1​	3​	Stick​	1​	Win​	1​	0​
3​	Car, Goat, Goat​	1​	2​	Switch​	3​	Lose​	0​	0​
4​	Car, Goat, Goat​	1​	3​	Switch​	2​	Lose​	0​	0​
5​	Car, Goat, Goat​	2​	1​	Stick​	2​	Lose​	0​	0​
6​	Car, Goat, Goat​	2​	3​	Stick​	2​	Lose​	0​	0​
7​	Car, Goat, Goat​	2​	1​	Switch​	3​	Lose​	0​	0​
8​	Car, Goat, Goat​	2​	3​	Switch​	1​	Win​	0​	1​
9​	Car, Goat, Goat​	3​	1​	Stick​	3​	Lose​	0​	0​
10​	Car, Goat, Goat​	3​	2​	Stick​	3​	Lose​	0​	0​
11​	Car, Goat, Goat​	3​	1​	Switch​	2​	Lose​	0​	0​
12​	Car, Goat, Goat​	3​	2​	Switch​	1​	Win​	0​	1​
​	​	​	​	​	​	​	2​	2​
​	​	​	​	​	​	​	​	​
13-24​	Goat, Car, Goat​	​	​	​	​	​	2​	2​
​	​	​	​	​	​	​	​	​
25-36​	Goat, Goat, Car​	​	​	​	​	​	2​	2​

Obviously, the car being behind door 2 or 3 gives identical results, so whatever happens in those scenarios is identical.

(And yes, I know it is effectively identical to the OP's table.)
Question: Are all the games 1-36 equally likely to have happened? i.e. Does every row have the same probability of showing up in the table? If not, what rows am I missing?

I suggest you have confused "equally likely events" with your strategy. Your 12 scenarios are really only 6, in each case you either win by sticking or win by switching.

And, yes, the other 12/24 cases are just the same by with the car in a different position.

 

[DaveC426913]

I suggest you have confused "equally likely events" with your strategy. Your 12 scenarios are really only 6, in each case you either win by sticking or win by switching.

No. The above are results. Actual games played. There is no 'either/or'.

'I stuck' is an outcome of a game. 'I switched' is an outcome of another game. The sum total is statistical win for 'Switch'.

If I were to take the results of the actual programmatic experiments, I would see 10,000 results. Many of those rows are identical, but there has to be a minimum number of results that, when sorted for readability, will show the Switch option winning twice as often as the Stick option.

That is what the above table is intended to be. So which rows am I missing?

 

[PeroK]

No. The above are results. Actual games played. There is no 'either/or'.

'I stuck' is an outcome of a game. 'I switched' is an outcome of another game. The sum total is statistical win for 'Switch'.

If I were to take the results of the actual programmatic experiments, I would see 10,000 results. Many of those rows are identical, but there has to be a minimum number of results that, when sorted for readability, will show the Switch option winning twice as often as the Stick option.

That is what the above table is intended to be. So which rows am I missing?

Ah, I just realized you have the bogey games in there. Games 5, 7, 9 and 11 cannot happen. You have the host opening door #1 to reveal the car! He never does that. Which in a way is the source of the paradox.

That's actually the Monty fall problem, where one game in three is spoiled by the host.