Thursday, 12 March 2015

Saving Monty Fall

As pointed out by KaySen, there is a treatment of Jeff Rosenthal’s Monty Fall scenario by Christopher Pynes which suggests, amongst other things, that in a Monty Fall scenario the correct answer is still 2/3.  Interestingly enough, Pynes is making precisely the kind of error that I was accusing everyone of making when I was first defending my arguments first raised in The Reverse Monty Hall Problem – that is by treating a single instance, one shot game as if it were part of a series.

I am now persuaded that the intent of this argument was misguided in terms of both the Monty Hall Problem and the Reverse Monty Hall Problem, but I consider it to be entirely valid with respect to the Monty Fall scenario.

The whole concept that Rosenthal is trying to convey is that the host accidentally falls against a door, thereby revealing a goat, rather than deliberately selecting a door to open.  This entails that the accident is a one-off.  There is no implication that the host repeatedly slips and falls against a door, and that in each case a goat is revealed – as implied by Pynes:

In cases where the contestant picks the door with the prize behind it in the first place, Monty never reveals it, but reveals another door: in these cases one wins by staying and loses by switching—1/3 and 2/3 respectively. In the case where the contestant doesn’t initially pick the prize door, Monty’s fall reveals the only door it can that isn’t the prize—win 2/3 of the time by switching. This makes Monty Fall and Monty Hall the same in all instances; they are logically equivalent, and thus must have the same probabilities.

There is no “never”, there are no other instances required to make “all instances” a meaningful concept.  It happens once, whoops, and it’s entirely possible that it never ever happens again.  Pynes, I believe, is the one being seduced by his familiarity with the Monty Hall Problem.

There is, however, a way in which Monty Fall can be saved in so much as it can be reliably repeated.

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Monty de Sade was continuing to cast about for a more devious variant of the three doors, two goats and a car game.  He heard about the use of coins to discuss the Reverse Monty Hall problem and a new idea formed in his evil little mind.

His new game goes like this:

There are three doors, with two goats and a car distributed behind them at random just like before.

The contestant is offered one door to choose from, just like the classic Monty Hall Problem.

Monty then offers a more complex continuation.  If the contestant agrees to proceed, Monty will toss a coin and then open one of the unselected doors on the basis of the result.  If the car is revealed, the contestant will automatically lose.  If the car is not revealed, then the contestant will be offered the opportunity to switch doors.

What is the likelihood of benefitting from a switch?

There are three decisions here on the part of the contestant:

The selection of a door, which we take to be random

Whether to risk losing the car on the basis of a coin toss

Whether to switch

As far as I can tell, noting that I have been wrong before, there is a 1/3 likelihood of selecting the car from the outset.  This means that there is a 2/3 likelihood of the car being behind one of the two unselected doors.  However, when Monty tosses a coin, there is 1/2 likelihood of revealing the car, if the car is behind one of those doors, and a 0/1 likelihood of revealing it if the car is behind the selected door.

This means there is a 1/3 likelihood of having the car already and a 1/3 likelihood that the contestant will lose the car in the coin toss and a 1/3 likelihood that the car won’t be lost in the coin toss, but will be behind the one remaining unselected and unopened door.

This means that the likelihood of benefitting as a result of switching, after surviving the coin toss, would be equal to the likelihood of benefitting as a result of not switching.  In which case, there was no point in going through the charade, was there?

This seems counter-intuitive.  It seems to me that if the contestant bothered to go ahead with the coin toss then, once a goat is revealed, if a goat is revealed, she is essentially committed to switching.  By why, if it’s 50-50?  Some might suggest that it’s not 50-50, but the probability tree below indicates that it is – unless I’ve made an error somewhere.

That said, there are two perspectives one could take, and they seem to have equal weight:

Perspective One:

At the beginning the contestant picks one door, the likelihood of having selected the car is 1/3, so there is a 2/3 likelihood of not having selected the car.  If she has the car, then she loses nothing by having Monty toss the coin and open a door.  If she doesn’t have the car, then she gains nothing by sticking with her selection.  So she should proceed.

Then Monty tosses the coin and, phew, doesn’t reveal the car.

Given that there was 1/3 likelihood of the car being behind the selected door and a 2/3 likelihood of it being behind one of the other two doors, it’s more likely that the car is behind the remaining unselected and unopened door and the contestant should therefore switch.

Perspective Two:

The contestant selects a door.  If she has selected the door with the car, then she loses nothing by having Monty toss the coin and open a door.  If she hasn’t selected the car, then she gains nothing by sticking with her selection.  So she should proceed.

Monty tosses the coin and, phew, reveals a goat.  The likelihood of Monty revealing a goat when there are two goats behind the unselected doors is twice that of revealing a goat when there is only one goat that could be revealed.

Given that Monty revealed a goat, it’s more likely that the car is behind the selected door and the contestant should therefore stay.

So, I throw the question out there: what is the best strategy for the contestant?

Note that this time I have no firm answer of my own beyond a gut feeling that the rational contestant should risk a coin toss to make use of the 2/3 likelihood of not having selected the car and switch if a goat is not revealed, having dodged an automatic loss entirely by chance.  I do recognise that the probability calculations don’t seem to support this gut feeling but perhaps someone has some better probability calculations.

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And how, precisely, does this save Monty Fall?  If my 50-50 solution is correct, then it is possible to run the scenario over and over again, as many times as you like, and then extract one iteration of the game in which Monty reveals a goat.  This then represents the single instance involved in the Monty Fall scenario, it’s just that the “luck” is now based on a coin toss rather than a humorously placed banana peel.

1. As far as I can tell, the computations in your table are perfectly correct. Which means that the question :

> what is the best strategy for the contestant?

should be easy to answer . You just have to look at the table :

Strategy 1, not taking the "risk" : 1/3 chances of winning.

Strategy 2, taking the risk and switching if possible : 1/3 chances of winning

Strategy 3, taking the risk and staying if possible : 1/3 chances of winning

All three strategies are completely equivalent, probabilitiy-wise.

> This means that the likelihood of benefitting as a result of switching, after surviving the coin toss, would be equal to the likelihood of benefitting as a result of not switching. In which case, there was no point in going through the charade, was there?

Exactly.

Now concerning Perspective one and two, they are in a nutshell exactly the same kind of mistake you did before. Namely that you take into account only part of the situation, and you forget to take into account that probabilities change when something is revealed (that's the point of conditional probabilities).

In perspective one, you only look at the choice of the contestant. You say that apriori there is a 1/3 chance that the car is behind the selected door. That's true. But once a door is opened revealing a goat, this probability goes to 1/2. This is just a simple computation of conditional probabilities :

P(selected door contains car | open door contains a goat) =(1/3)/(2/3) = 1/2.

In perspective two, you only look at the placement of the car. You say that it's twice more likely to reveal a goat if the car is behind the selected door. That's true. But it's twice more likely to have a goat behind the selected door, so the two things cancel out. And you can use Bayes theorem to compute the same probability :

P(selected door contains car | open door contains a goat) = P(open door contains a goat | selected door contains car) * P(selected door contains car) / P(open door contains a goat ) = 1*(1/3) / (2/3) = 1/2

> I do recognise that the probability calculations don’t seem to support this gut feeling but perhaps someone has some better probability calculations.

As your calculations are right, it means that your gut feeling is wrong. I don't have better probability calculations to offer, as your computations are already correct and complete.

1. In a sense, the perspectives are sort of right though - even if the final conclusion of each is wrong. Winning as a consequence of both switching and staying is more likely, but only to the extent that both options move from 1/3 to 1/2 in likelihood. I think we tend to fill in the gaps with what makes more sense to us, making a statement more precise even if by doing so we make it less accurate. In this case we turn what I wrote:

"it’s more likely that the car is behind the remaining unselected and unopened door"

into

"there’s a greater likelihood of the car being behind the remaining unselected and unopened door than of it being behind the selected door (2/3 vs 1/3)"

rather than

"the likelihood of the car being behind the remaining unselected and unopened door in now greater than it was originally (1/2 rather than 1/3)"

Perhaps this is where the gut feeling comes from, especially when we are familiar with Monty Hall.

I know that you are a mathematician and you have worked through this dispassionately, however, are you really entirely free of gut-feelings? Do you not have any intuition as to what would be the best option, despite knowing what probabilities tell you?

Finally, I think that the sense that there are two equal but incompatible likelihoods is a very good hint that something has been left out in one's considerations. It's reassuring that, now at least, I've got it right with my computations.

2. In a game where it is possible for Monty to reveal a car then conditional probability applies and the chances are 50/50. Where there is no possibility he can reveal a car then the 2/3 applies on a switch. In Monty Fall he always falls on a goat, so it is the same as classic Monty. However your new game is 50/50 because he can reveal a car. It does not matter what strategy the contestant takes, switching or staying yields a 50/50 chance. Not being a math person, this is how I work out the probability. first move is to list he possible outcomes;

Where the contestant always picks door 1. ( ) represents which door Monty opens being 2 or 3 and ** is the location of the car

a: *1* (2) 3
b. *1* 2 (3)
c: 1 (*2*) 3
d: 1 *2* (3)
e: 1 (2) *3*
f: 1 2 (*3*)

A. and B. become compressed due to Monty being able o pick either one and only reveal a goat.

In a game where it is possible for Monty to reveal a car then conditional probability applies and the chances are 50/50. Where there is no possibility he can reveal a car then the 2/3 applies on a switch. In Monty Fall he always falls on a goat, so it is the same as classic Monty. However your new game is 50/50 because he can reveal a car. It does not matter what strategy the contestant takes, switching or staying yields a 50/50 chance. Not being a math person, this is how I work out the probability. first move is to list he possible outcomes;

Where the contestant always picks door 1. ( ) represents which door Monty opens being 2 or 3 and ** is the location of the car

a: *1* (2) (3)
c: 1 (*2*) 3
d: 1 *2* (3)
e: 1 (2) *3*
f: 1 2 (*3*)

So now to play it out. Monty randomly picks door 2 and there is a goat. So eliminate all possibilities where he picks door 3 and also the possibility there is a car behind door 2. his is what is left.

In a game where it is possible for Monty to reveal a car then conditional probability applies and the chances are 50/50. Where there is no possibility he can reveal a car then the 2/3 applies on a switch. In Monty Fall he always falls on a goat, so it is the same as classic Monty. However your new game is 50/50 because he can reveal a car. It does not matter what strategy the contestant takes, switching or staying yields a 50/50 chance. Not being a math person, this is how I work out the probability. first move is to list he possible outcomes;

Where the contestant always picks door 1. ( ) represents which door Monty opens being 2 or 3 and ** is the location of the car

a: *1* (2) (3)
e: 1 (2) *3*

You have a 1 in 2 chance of having the car and Monty has a 1 in 2 chance of having he car.

Just thought I would list how I work it out, it may give it another conceptual outlook.

3. Sorry can't edit where I accidently pasted the first paragraph to the middle :(

1. With respect to your first paragraph, I draw your attention to Vietoris' comment - he argues that Pynes simply got it wrong.

2. If you ran an actual simulation on Monty fall, it will show the exact same results as Monty Hall. Because Monty always falls on a goat door. If in some simulations he can fall on a car door, then it will be 50/50. Run the simulation yourself and see the results where monty falls on a goat door every time.

3. Apparently, you understand that there is a huge fundamental difference between the two following scenarios :

A - I consider one possible situation where Monty opens a goat door, in a Universe where the rules forces Monty to open a goat door.
B - I consider one possible situation where Monty opens a goat door, in a Universe where the opening of the door was random.

As you said, in scenario A, the odds are 2/3 - 1/3. And in scenario B, the odds are 1/2 - 1/2.

The fact that we are in a situation where Monty did open a goat door, does not mean that in the Universe (in the mathematical sense) it is impossible for Monty to open the car door. In other words, just looking at the situation, we cannot deduce in what scenario we are.

So to understand in which scenario (Universe) we are, we have to read carefully the formulation of the Monty Fall problem :

> the host slips on a banana peel and accidentally pushes open another door, which just happens not to contain the car.

So, for me, this is an unambiguous statement that the fact that there was a goat behind the door was the result of luck, and not of the rules of the game. It means that Monty could also have opened the car door accidentally. So, to me it's perfectly clear that Monty Fall corresponds to scenario B. I can see no reasonable interpretation of the word "accidentally" that would somehow indicate that the "choice" of which door is open is not uniformly random.

Apparently you (and also Pynes) disagree with this analysis of the sentence. But I don't understand why ... What makes you think "Monty always falls on a goat door" in the Monty Fall scenario ? I don't understand how the phrasing of the problem could make you (and Pynes) think that.

4. KaySen, I draw your attention to xor7486's comment. The simulation has already been done.

5. I was not aware of this statement in Monty fall;
"the host slips on a banana peel and accidentally pushes open another door, which just happens not to contain the car"

This does make a difference if he intended to open one door but accidently opened another, because that means the other goat IS behind the door he intended to open in the first place. This means he either opened the contestants door or the other door of the remaining 2 unpicked doors.In either case he has now revealed the location of the car in general terms on his side or Monty's side), if the contestant knows that Monty does a reveal in the game. If the contestant is unaware that Monty was going to perform a reveal on one of the two remain unpicked doors and reveal a goat, only then from not having knowledge are the choices 50/50. There is a significant difference from "falling on a door other than one of he goat doors" than a random pick. The math is entirely different as far as possibilities go, and far more simple. At this stage I am thinking Monty Fall is really a nothing when trying to understand it conceptually.

6. Perhaps I wasn't clear earlier. My point is that the entire difficulty of these kind of problems is translating the "word problem" into a mathematical model. Once you have a mathematical model, it's piece of cake. So the only thing people should be discussing is how to interpret the sentence.

> This does make a difference if he intended to open one door but accidently opened another, because that means the other goat IS behind the door he intended to open in the first place.

With this interpretation, your analysis is correct.

But I did not interpret the sentence like that (and I still don't really understand how one can interpret it as you did). For me the sentence "accidentally pushes open another door" unambiguously means that a random event (the banana) makes him push open a door. I see the word "accidentally" as a proof that the banana peel was randomly placed in front of one of the doors, and hence it could have been any door (the door that the contestant selected or the door with the car). It turns out this door is "another door" (meaning a door other from the one the contestant already chose), and it also turns out that the door does not contain a car.

Apparently, Pynes interprets the sentence differently. He says that somehow Monty is forced to push open a door that does not contain a car (for me, that's completely absurd). And you just interpreted the sentence in another way, saying that Monty wanted to open a goat door but accidentally open the other door because of the banana peel (that's not absurd, but it's really far-fetched). I would say that this is more an english problem than a mathematical problem ...

Perhaps the fact that I'm used to read similar problems makes me interpret sentences in a very specific way. What I mean is the following :
I interpret Rosenthal's formulation to mean "we are in situation B", because I'm thinking that if the guy actually wanted to specify a situation that is not situation B, he would have formulated the problem very differently.
For example, if Rosenthal wanted to describe a situation where Monty was going to open a goat door but on his way he slips and open another door, then he would say something like :
"on his way to open a door as usual, Monty slips and accidentally pushes open THE OTHER door."
The fact that he did not mention that Monty had the intention to open a door, and the fact that he just says "another door", makes me believe that my interpretation is correct and that is what Rosenthal had in mind.

4. Hey this blog is great. Your analysis is right and I think the table you calculated is correct. We could call this variant of the game Fair Monty Hall. The solution is simply that there are two strategies, and both with a winning likehood of 1/6+1/6 = 1/3, so in this game you have the same winning probability as in the simple three door game (you pick one door and if it contains a car you win). This happens because the AutoLoss columns sum up to 1/6+1/6=1/3 as well.
I think this is the most exact way of saving Monty Fall, which is itself logically equivalent to Monty Hall (in which is stipulated that a car is never revealed).

FB

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