Monty Hall Problem... For Newbies

[bigred]

As I understand it (explained by Derren Brown) You start with a car and two goats (1:3 car - 2:3 goat) you pick one, let's say door 1 and are left holding your choice (1:3) while MH opens door 3 and shows a goat. If you maintain your original choice, door 1 (1:3) your odds are still as they were from the outset. However, if you swap, you are now picking from a 1:2 or fifty - fifty option, thereby improving your chances of winning the car.

I think!

That's the logic, but I think it's flawed. The odds DO change. Once he reveals a goat, the odds of you having the winning door are now 50%. If you change, your odds are also 50%. For those who disagree, let me put it this way: based on the prevailing logic, if he shows the OTHER door and it's also a goat, you're saying your chances are still 1 in 3. ie the odds don't change. But clearly they do.

PS Monty always knew he wasn't going to show the car. Duh. At least I've never seen or heard him "randomly" pick another door and boom there's the car.

 

[jiggeryqua]

Let's work through the possibilities. There are 3 doors, A B and C. The car is behind door A. If you pick door A then Monty can open either door B or door C.

A B C
Car Goat Goat<snip>

This means that changing gives you a 2/3 chance of winning, while sticking gives you a 1/3 chance of winning.

Well that looks like a better approach than any of the repetitious posts, which don't address why the odds 'shift' to only one of two available, indistinguishable doors....because deciding in your mind which door you want isn't affecting the material world.

Of course it matters that Monty knows, but it doesn't matter that he always does it. The game is controlled - that's TV. Of course we know that he's showing a safe goat. But there will always be one, whichever door you choose, whether there's a car or a goat behind it, there will always be a safe goat door. Monty opens it, because he always knows where it is. Whichever door you pick, he knows where there's a safe goat. We know there's a safe goat (there's probably a mathematical proof, but who needs one? There's a safe goat.) He shows us the thing that must, 100%, exist. A certainty, and lo, there it certainly is, behind one of the doors that you have already not picked.

So far, it's looking good, you're on the right track, you have mad skillz, you rejected one wrong door, it must have improved the chances you picked the car ...of course it hasn't, it hasn't changed what's behind your door and I'm not falling for some Derren Brownish mind-game: seeing the guaranteed goat does not alter my thinking, whether he's been doing it all season or not is fatuously irrelevant here. He's not going to show you a car, whichever door you picked, and he can always show you a goat, whichever door you picked. Showing you a goat tells you nothing about the door you picked.

Back to your example. You start by deciding where the car is, which I'm not sure is helpful? For a probability to be useful to me here it has to relate to what I do know, not what I don't. In the absence of useful information, we have three doors (X, Y & Z, so we can have C,G & G behind them)

X Y Z

(CGG, GCG or GGC)​

I pick a door. Whichever door I pick, one of the other doors has a goat behind it. I'm confident of that. 100%. Isn't this just the one about the change and where did the other £2 go? This 'spare' 1/3 probability should surely be shared equally between the remaining doors, which are indistinguishable to the blind eye of chance. There are 3 possible set-ups and I've chosen one. Showing me that something I already knew to be true is true doesn't change much. We're down to two options, certainly, but why would it be more likely to be the other one? It will always be possible to eliminate one option after you pick, whether you're right or wrong. It's neither harder nor easier to find a safe goat, whatever is behind your chosen door. We see the goat, and we now have a 50/50 chance of a car - because both doors can equally hide one or other of two options, just as they always did.

I made my own wee table, paper and pen, wollery - I shan't faff about here. Five colums: your pick; where the car is; which door is opened; result of change; result of stick. 12 rows, equal numbers of wins and losses. When you collapsed some of your options, was that related to it apparantly not mattering to Monty which safe goat he shows you if you have picked the car? I include both possibilities, because we can never know if he had a choice of goat. Soapy Sam certainly seems to think it matters, but there are (in that framework) at least four possible scenarios. I have 12 rows, they all seem like valid rows...

 

[ddt]

It doesn't matter which of those options it is. As long as we know he won't show the car accidentally, the problem remains the same.

Actually, it doesn't matter - for the probabilities - if Monty Hall knows where the car is or not. The probabilities stay the same if he randomly opens another door.

In the canonical setup, you precisely get the car in case your initial pick is a goat-door. That remains the case if Monty picks a random other door. Suppose your initial pick is a goat-door and Monty randomly opens another door. Then either he picks the other goat-door, and you switch to the third door - which has the car - or he picks the car-door and you obviously switch to the (now opened) car door.

 

[Crossbow]

I have heard about the 'Monty Hall Problem' for some months now, and I think that one of the major elements of the problem is being omitted; and it is this omission of this one element that it the source of the confusion.

Specifically, if the contestant initally picks the correct door, then Monty Hall shows the contestant that they won the big prize soon thereafter.

However, if after the inital pick Monty Hall shows an incorrect door, then Monty Hall is essentially giving a strong clue to the contestant the his initial selection was incorrect; because if the contestant picked the correct door, then Monty Hall would be telling him that his inital selection was the correct selection.

Therefore, when Monty Hall shows the contestant an incorrect door and offers them a chance to change their selection, then Monty Hall is non-explictly telling the contestant that their original selection was incorrect.

I mapped it out using the above assumptions and it always worked out that when the contestant changed his initial selection, then the contestant would always win the big prize provided that Monty Hall would let the contestant know immediately if his initial choice was correct or not.

However, it has been many years since I have seen the show, therefore can someone who is more familiar with the show please inform me (and the rest of us) one detail: If the contestant picks the right door, then does Monty Hall inform them of the fact right away?

Thanks.

 

[ddt]

That's the logic, but I think it's flawed. The odds DO change. Once he reveals a goat, the odds of you having the winning door are now 50%. If you change, your odds are also 50%. For those who disagree, let me put it this way: based on the prevailing logic, if he shows the OTHER door and it's also a goat, you're saying your chances are still 1 in 3. ie the odds don't change. But clearly they do.

No, the odds don't change solely by Monty opening a door. Monty can always open a goat door, so the sole fact he opens a door does not change the probability of 1/3 that your initial choice was the car. And that is fundamental to the problem: switching essentially means that you choose both other doors.

 

[ddt]

I have heard about the 'Monty Hall Problem' for some months now, and I think that one of the major elements of the problem is being omitted; and it is this omission of this one element that it the source of the confusion.

Specifically, if the contestant initally picks the correct door, then Monty Hall shows the contestant that they won the big prize soon thereafter.

However, if after the inital pick Monty Hall shows an incorrect door, then Monty Hall is essentially giving a strong clue to the contestant the his initial selection was incorrect; because if the contestant picked the correct door, then Monty Hall would be telling him that his inital selection was the correct selection.

Therefore, when Monty Hall shows the contestant an incorrect door and offers them a chance to change their selection, then Monty Hall is non-explictly telling the contestant that their original selection was incorrect.

I mapped it out using the above assumptions and it always worked out that when the contestant changed his initial selection, then the contestant would always win the big prize provided that Monty Hall would let the contestant know immediately if his initial choice was correct or not.

However, it has been many years since I have seen the show, therefore can someone who is more familiar with the show please inform me (and the rest of us) one detail: If the contestant picks the right door, then does Monty Hall inform them of the fact right away?

Thanks.

And this is why it is essential in phrasing the problem that Monty always opens a goat door after your initial pick, whether your initial pick was a goat or the car.

 

[phunk]

The people who doubt this should just go ahead and try it. Get some way to randomly pick a number 1-3, for example roll a die and subtract 3 if you get 4, 5, or 6. Use that to pick which door will have the car behind it, and then use it again to pick the contestant's guess, then figure out if you would win if you stay or switch. Repeat until your denial goes away.

 

[Jrrarglblarg]

Mythbusters managed to concoct a form of blind test. Jamie kept his choice every time and Adam changed it every time. The results were startling -- Adam had something like 80% hit rate and Jamie had less than 20%. (numbers made up from memory)

Me? I'd change but cringe while doing so. It seems we have an internal bias towards our first choice in these things.

 

[Crossbow]

And this is why it is essential in phrasing the problem that Monty always opens a goat door after your initial pick, whether your initial pick was a goat or the car.

OK thanks! That is what I was thinking as well, but since it had been so long since I had seen the show, I was not sure.

Therefore, when explaining the 'Monty Hall Problem', I think that it would really help to also provide the important data that Monty Hall will let the contestant know right away if he selected the right door or not; and so far (except here at the start of this thread), I have not seen that bit of data provided.

 

[Dessi]

My young nephews struggled with the 3-door Monty Hall problem, but it seemed to make a little more sense to them when I explained the scenario with 100 doors instead:

Imagine there are 100 doors, 99 with goats and 1 with a car. You select a door at random; it could be a goat, or it could be a car. Monty Hall then opens 98 doors at random, showing 98 goats. Only two doors remain: do you switch? Hell yes you do.

 

[Soapy Sam]

I made my own wee table, paper and pen, wollery - I shan't faff about here. Five colums: your pick; where the car is; which door is opened; result of change; result of stick. 12 rows, equal numbers of wins and losses. When you collapsed some of your options, was that related to it apparantly not mattering to Monty which safe goat he shows you if you have picked the car? I include both possibilities, because we can never know if he had a choice of goat. Soapy Sam certainly seems to think it matters, but there are (in that framework) at least four possible scenarios. I have 12 rows, they all seem like valid rows...

No, I don't believe it matters. And I only see the three possible scenarios, two of which are functionally identical from Monty's POV.
If you pick the car, he can open either remaining door.
If you pick a goat, he can only open the remaining goat door, but it makes no difference to anything which door that is.

If the car is behind C and you pick either A or B, Monty opens B or A accordingly. If you stick, in either case, you lose.
2 of three possibilities result in failure if you switch.

If you pick C, ie you pick the car, Monty opens either A or B, it makes no difference which.
If you switch, you lose, because you already chose the car.

That really is all there is to it.

 

[jiggeryqua]

No, I don't believe it matters. And I only see the three possible scenarios, two of which are functionally identical from Monty's POV.

And I don't see why I should care about Monty's point of view. The possibilities, from which we derive the probability, allow for both scenarios when you have already chosen the car behind door X. Monty opens Y, or Monty opens Z. All I know is I see the inevitable goat. That they are functionally identical for Monty does not make them functionally identical to probability.

 

[wollery]

Back to your example. You start by deciding where the car is, which I'm not sure is helpful? For a probability to be useful to me here it has to relate to what I do know, not what I don't.

No, it doesn't matter.

You can work through it using all the various possibilities if you wish, having the car behind door A, then have it behind door B, then door C, but by doing that you're just multiplying the outcomes by 3.

Car behind A
Pick A. Change. Lose.
Pick B. Change (to A). Win.
Pick C. Change (to A). Win.

Car behind B
Pick A. Change (to B). Win
Pick B. Change. Lose.
Pick C. Change (to B). Win.

Car behind C
Pick A. Change (to C). Win.
Pick B. Change (to C). Win.
Pick C. Change. Lose.

You could do the same for sticking and see that you still get a 1/3 chance of winning three different ways.

 

[Soapy Sam]

And I don't see why I should care about Monty's point of view.

You don't need to. He does, in case he opens the wrong door and shows you the car.
What counts is that you can only choose the car one way, but a goat two ways. Either way you choose one goat, Monty reveals the other goat. If you stick in either of those cases, you get the second goat.
The only other possibility is that you choose the car.If you do, Monty can open either other door , showing a goat. If you switch IN THIS 1 CASE, you will lose, because you chose right first time. In either of the other two cases, a switch wins.

 

[GlennB]

OK thanks! That is what I was thinking as well, but since it had been so long since I had seen the show, I was not sure.

Therefore, when explaining the 'Monty Hall Problem', I think that it would really help to also provide the important data that Monty Hall will let the contestant know right away if he selected the right door or not; and so far (except here at the start of this thread), I have not seen that bit of data provided.

No. He never does that. The contestant always has a dilemma.

 

[wollery]

I made my own wee table, paper and pen, wollery - I shan't faff about here. Five colums: your pick; where the car is; which door is opened; result of change; result of stick. 12 rows, equal numbers of wins and losses. When you collapsed some of your options, was that related to it apparantly not mattering to Monty which safe goat he shows you if you have picked the car? I include both possibilities, because we can never know if he had a choice of goat. Soapy Sam certainly seems to think it matters, but there are (in that framework) at least four possible scenarios. I have 12 rows, they all seem like valid rows...

Look at it this way, there are only three actions -

1. You pick a door.
2. Monty opens one of the other two to show a goat.
3. You decide whether to change or stick.

If you choose the door with the car and change then you lose.
If you choose the door with the car and stick you win.
If you choose the first door with the goat and change then you win.
If you choose the first door with the goat and stick then you lose.
If you choose the second door with the goat and change then you win.
If you choose the second door with the goat and stick then you lose.

Changing gives you two ways to win and one way to lose.

Sticking gives you one way to win and two ways to lose.

There are equal numbers of wins and losses overall, 3 ways to win and 3 ways to lose, but two of the ways to win come from changing doors.

 

[dlorde]

That's the logic, but I think it's flawed. The odds DO change. Once he reveals a goat, the odds of you having the winning door are now 50%. If you change, your odds are also 50%. For those who disagree, let me put it this way: based on the prevailing logic, if he shows the OTHER door and it's also a goat, you're saying your chances are still 1 in 3. ie the odds don't change. But clearly they do.

If he shows the other door and it's also a goat, it means you got lucky - your 1/3 chance paid off, as it will do, one time in three. What actually happens in any one run of the game isn't a guide to calculating the odds.

When you make your choice, the odds of you choosing the car are 1/3. The odds of the car not being your choice are 2/3. When the goat door is opened, the odds of the car not being your choice are still 2/3, but you now know the door to pick to take that 2/3 chance.

 

[wollery]

OK thanks! That is what I was thinking as well, but since it had been so long since I had seen the show, I was not sure.

Therefore, when explaining the 'Monty Hall Problem', I think that it would really help to also provide the important data that Monty Hall will let the contestant know right away if he selected the right door or not; and so far (except here at the start of this thread), I have not seen that bit of data provided.

If he did that there would be no dilemma at all.

Think about it, if you get it right he tells you right away, you win.
If he doesn't tell you right away then you've got one of the wrong doors, he then opens one of the other doors revealing a goat, which means that you know the two wrong doors and the last door must be the car. So you change doors and win.

In that scenario there'd be no way to lose! They might as well just give every finalist a car and be done with it.

 

[GlennB]

And I don't see why I should care about Monty's point of view. The possibilities, from which we derive the probability, allow for both scenarios when you have already chosen the car behind door X. Monty opens Y, or Monty opens Z. All I know is I see the inevitable goat. That they are functionally identical for Monty does not make them functionally identical to probability.

Monty doesn't have a pov. His choices are not 'functionally identical'.

In your illustration you're only considering the case where you made the right choice. Then, and then only, you'll see a goat at random. That wasn't a 50% shot, it was 1/3.

Otherwise, 2/3 times, he's obliged to open one specific door, the only one with the goat.

 

[Dr. Keith]

I think it is important to note that Monty Hall hates the Monty Hall Problem.

Why? Because he never did this on Let's Make a Deal. The Problem was made up by mathematicians to argue about, not Monty Hall or his show.