Here it is:

Three tennis players. Two are equally-matched amateurs; the third is a pro who will beat either of the amateurs, always.

You blindly guess that Player A is the pro; the other two then play.

Player B beats Player C. Do you want to stick with Player A in a Player A vs. Player B match-up, or do you want to switch?

And what’s the probability that Player A will beat Player B in this match-up?

And here’s the background.

It started when Josh Miller proposed this alternative formulation of the Monty Hall problem:

Three boxers. Two are equally matched; the other will beat either them, always.

You blindly guess that Boxer 1 is the best; the other two fight.

Boxer 2 beats Boxer 3. Do you want to stick with Boxer 1 in a Boxer 1 vs. Boxer 2 match-up, or do you want to switch?

I liked the formulation in terms of boxers (of course, and see data-based followup here), but Josh’s particular framing above bothered me.

My first thought was confusion about how this relates to the Monty Hall problem. In that problem, Monty opens a door, he doesn’t compare two doors (in his case, comparing 2 boxers). There’s no “Monty” in the boxers problem.

Then Josh explained:

When Monty chooses between the items you can think of it as a “fight.” The car will run over the goat, and Monty reveals the goat. With two goats, they are evenly matched, so the unlucky one gets gored and is revealed.

And I pieced it together. But I was still bothered:

Now I see it. The math is the same (although I think it’s a bit ambiguous in your example). Pr(boxer B beats boxer C) = 1 if B is better than C, or 1/2 if B is equal in ability to C. Similarly, Pr(Monty doesn’t rule out door B) = 1 if B has the car and C has the goat, or 1/2 if B and C both have goats.

It took me awhile to understand this because I had to process what information is given in “Boxer 2 beats Boxer 3.” My first inclination is that if 2 beats 3, then 2 is better than 3, but your model is that there are only two possible bouts: good vs. bad (with deterministic outcome) or bad vs. bad (with purely random outcome).

My guess is that the intuition on the boxers problem is apparently so clear to people because they’re misunderstanding the outcome, “Boxer 2 beats Boxer 3.” My guess is that they think “Boxer 2 beats Boxer 3” implies that boxer 2 is better than boxer 3. (Aside: I prefer calling them A, B, C so we don’t have to say things like “2 > 3”.)

To put it another way, yes, in your form of the problem, people easily pick the correct “door.” But my guess is that they will get the probability of the next bout wrong. What is Pr(B>A), given the information supplied to us so far? Intuitively from your description, Pr(B>A) is something close to 1. But the answer you want to get is 2/3.

My problem with the boxers framing is that the information “B beats C” feels so strong that it overwhelms everything else. Maybe also the issue is that our intuition is that boxers are in a continuous range, which is different than car >> goat.

I then suggested switching to tennis players, framing as “two amateurs who are evenly matched.” The point is that boxing evokes this image of a knockout, so once you hear that B beat C, you think of B as the powerhouse. With tennis, it seems more clear somehow that you can win and just be evenly matched.

Josh and I went back and forth on this for awhile and we came up with the tennis version given above. I still think the formulation of “You blindly guess that Player 1 is the pro” is a bit awkward, but maybe something like that is needed to draw the connection to the Monty Hall problem.

Ummm, here’s an alternative:

You’re betting on a tennis tournament involving three players. Two are equally-matched amateurs; the third is a pro who will beat either of the amateurs, always.

You have no idea who is the pro, and you randomly place your bet on Player A.

The first match is B vs. C. Player B wins.

Players A and B then compete. Do you want to keep your bet on Player A, or do you want to switch? And what’s the probability that Player A will beat Player B in this match-up?

This seems cleaner to me, but maybe it’s too far away from the Monty Hall problem. Remember, the point here is not to create a new probability problem; it’s to demystify Monty Hall. Which means that the problem formulation, the correct solution, and the isomorphism to Monty Hall should be as transparent as possible.

**P.S.** Josh noted that the story was also discussed by Alex Tabarrok, and a similar form of the problem was studied by Bruce Burns and Marieke Wieth in 2004.

I find the following the most intuitive — keeping the logical structure and framing, but changing the numbers. (From the MR comments.)

There are 100 doors and 1 has the car, picked at random. You pick one. Then Monty opens 98 doors of which he knows they have no prize. You get an offer to switch to the one other remaining door. Do you want to switch?

+1

Sandro-

Yeah, that is my favorite too. It has a limitation: people have a hard time generalizing from that explanation back to the 3 door, they return to the paradox.

The goal here was to make switching in the three door version feel like the intuitive right answer. The open question is why it feels intuitive. Some discussion here.

+1 too.

+1

“Then Monty opens 98 doors of *which he knows they have no prize*.” This is critical and is often left out of the statement of the problem. For a game to be well defined, you have to know Monty’s strategy as well. Does Monty only open goat doors, or does he randomly open doors, sometimes revealing the car? If it is the latter, then the player gets no useful information from the doors Monty opens (I think).

Exactly. And Monty have to do anything at all? (Might he just have opened your choice immediately if he knew you had lost?)

In the tennis example, was the tournament order (i.e that those you didn’t chose play first, then the winner playing your choice) something that was driven by your choice, or did it just work out to be that way? And if the latter, was the tournament set up oblivious to who was the pro? The answer to make this most cleanly like MH is that the tournament order was determined based on your choice as described – which is unnatural; you’d normally think of your self as bystander independently speculating on the outcome of the tournament. (On the other hand, in MH you are explicitly the center of attention, and the game show rules are obviously contrived for drama’s sake – so it makes more sense that the strategy is focal on your initial choice.)

Any restatement of MH which continues to obscure (or hides even deeper) the need to explicitly outline “his” strategy is, IMO, unfit as an explication/clarification.

Once you explicitly state Monty’s strategy, the solution becomes pretty obvious because the game is easy to model.

So, it seems that the popularity of the MH problem and it’s reputation for subtlety (and quite frankly the fun of it) is due to poor exposition. You get fierce arguments because people are making different assumptions and those assumptions are kind of hidden.

Terry:

There are two interesting aspects of the Monty problem. One interesting aspect is the ambiguity that you note, which is interesting in part because the problem at first appears to be clearly defined; it is only upon careful reflection that it becomes clear that the probability depends on aspects of Monte’s decision process (the “likelihood”) which were not specified in the problem statement. The other interesting aspect of the problem is the non-intuitiveness of the answer, even after the ambiguity has been cleared up. The purpose of the tennis reformulation is to address this second issue.

exactly.

Agree.

Plus, once you are explicit about Monty’s strategy it is straightforward to see if the tennis restatement above is isomorphic. You just check to see if the math is the same. Is the probability of a goat in the player’s first choice door mathematically the same after Monty does what he does?

Agreed!

But then what is the purpose of proposing variants (100 doors? Tennis?) that push our intuitions to a different place, but leave the underlying problem in place (i.e. strategy vagueness about what is actually going on).

In this light, the boxing/tennis variants seem to be misguided and pointless. They are trying to find a story, with conceptually the same fatal flaw of vagueness as MH has, but are framed so that normal intuitions come out “right”? Ok, suppose you succeed at that – you haven’t educated anyone!

Totes agree.

Bxg:

Consider a visual illusion, for example the image that includes two gray squares that are the same color but which look much different because of their differing background images. You can put a mask over the image that shows just the two squares and reveals that the two colors are identical. The mask makes the true colors apparent and intuitive, which in turn allows us to get more of a gut feeling about the illusion.

The purpose of the tennis or boxing reformulation is the same thing: it’s supposed to be a sort of mask that makes the correct probabilities clear, so that we can better see what it was about the original problem that gave us wrong intutions.

I think it makes the correct choice clear, I don’t think it makes the correct probabilities clear. In that sense I partially agree with Bxg, but I don’t see how Bxg can justify the claim that this exercise is “misguided and pointless.”

Let me explain:

First the goal here was just a quick response trying to fit the desiderata of this quoted tweet.

The thing that the this framing highlights is a problem with the original Monty Hall problem. The original problem involves sleight of hand: it hides the fact that the two unchosen doors interact, so to speak. This new formulation reveals that interaction, clearly. The conjecture here is that people can now intuitively grasp that information has been revealed, that the pro is more likely to be Boxer B. I can see a reason to justify your skepticism. In the boxer framing it is possible that people are just over-generalizing the typically useful feeling that someone is better because they won, i.e. they think Boxer B is better than what they thought before, not that Boxer B is more likely to be the pro. But this hypothesis, and the conjecture can be tested. Competition isn’t the only way to have the other two doors interact.

> but I don’t see how Bxg can justify the claim that this exercise is “misguided and pointless.”

Ok, that was coming on a bit strong, but nevertheless don’t see it as being that interesting to reformulate the problem while keeping the same TYPE of ambiguity/impression that the original MH does. Sure, you’ve found a way or restating things (albeit with its own imprecision) where most intuitions do indeed align with the “right” answer.

Who cares?

I think it comes down to this: does someone who suddenly sees the ‘right’ answer after seeing the tennis version

thereby gain a real appreciation of why their intuitions were led astray in the original MH? I’m dubious that the answer is often ‘yes’, but perhaps I’m very wrong. Either way though, I do wish that any intution-aid to MH would explicitly confront the strategy-choice failing in the original rather than hide it in similar obscurity to the original.

Bxg:

I wasn’t sure what you meant by the following 3 criticisms:

(1) “keeping the same TYPE of ambiguity/impression that the original MH does”

(2) “albeit with its own imprecision” [though I did improve precision here]

(3) “I do wish that any intution-aid to MH would explicitly confront the strategy-choice failing in the original rather than hide it in similar obscurity to the original.”

I read your comments here.

Perhaps I don’t get what you mean by “strategy-choice failing” but what I can say is that for the purposes of getting across the idea that you have learned something when a fighter loses, it doesn’t matter whether the tournament order was determined by the contestants choice, or if two boxers were chosen to fight at random and it happened to not be the contestants choice. The only thing that matters is that only two boxers (tennis players) can fight at once. Of course if you added a story about how before the show Monty chose at random which door he would not “inspect,” then it may feel different, even if the idea is the same in the contingency he decides to inspect the two doors that you didn’t choose and reveal a goat.

The one thing you wrote that I did understand was this: “Does someone who suddenly sees the ‘right’ answer after seeing the tennis version thereby gain a real appreciation of why their intuitions were led astray in the original MH?”

I think the answer here will be a clear no, but I don’t see why that should be the goal here. It certainly wasn’t mine. People’s ability to solve a problem is dependent on how it is framed to them. For example, people do a poor job with the card framing of the Wason Selection, but they do well with a social rule framing. The same is true in Monty Hall, where people have trouble generalizing the 100 door intuition to 3 doors. People don’t do well generalizing the insight from an intuitive/familiar framing to other less familiar (or abstract) framings.

The focus here was on how colliding/comparative aspect of Monty’s choice-rule is hidden in the original Monty Hall problem. My goal was to create a framing of the problem with equivalent structure that made that transparent. That is what people get hung up on. I believe that was achieved. I don’t believe it was hidden “in similar obscurity to the original,” as you say. That said, I have no idea why people’s intuition works here.

Bxg:

My reply disappeared. It is being moderated and should appear above this comment when it is released.

Joshua B:

“Perhaps I don’t get what you mean by “strategy-choice failing””

You are not alone in not seeing this. It is subtle, and some statements of the problem do not have the failing, so it can depend on how you were first told the story.

In short, the statement of the problem often does not tell us what other things Monty could have done, does not tell us why Monty did what he did. For instance, if Monty is trying to make the contestant lose, the game is very different. Below, I explained this:

The wikipedia page on the MH problem is an example of a statement of the problem with this failing (not all statements have this failing):

Monty might have other options or decision rules than the one we usually assume. He could (1) open the other door showing the car, (2) randomly choose one of the other doors to open, (3) do not open another door at all, (4) only open the goat door when the contestant chooses the car door, or something else.

gotcha, the malevolent Monty Hall problem. Yeah, missed your exchange and I did not get that that was what Bxg was referring to.

I can see the issue of putting human characters in a problem and expecting your readers to naturally assume (or accept!) robotic rule-based behavior from them.

When presented well and I don’t think that is the key intuitive block for people when facing the MH problem.

I also don’t think it is hidden in the version I presented; seems unnatural to think that there is a hidden malevolent opponent (or nature) that only shows you the outcomes of matches when you have chosen the pro.

Even if Monty had opened 98 goat doors by chance, I would still change my choice. It’s just very unlikely that he would have managed that. But given that he did, it seems that it changes the probability of me having picked the correct door.

Monty’s knowledge of where the car is makes the story more reasonable, in that you could see that this could happen again and again, but it doesn’t seem critical to me for a single play. Am I missing something? I like the 100 door formulation though.

The 100 doors formulation pushes you to assume that Monty only opens goat doors. That is the point of that formulation

But we don’t actually know that. It could just have been by chance that the car was not revealed.

As discussed with bxg above, we need to make explicit Monty’s strategy. Then the whole problem becomes simple (and less interesting).

To formalize this, Monty has 3 choices:

1. Do nothing with probability p.

2. Open a goat door with probability q.

3. Randomly open one of the unopened doors with probability 1-p-q.

Tell me p and q, and the solution is simple. But, if we don’t know p and q, we can’t know the answer.

> But given that he did, it seems that it changes the probability of me having picked the correct door.

Yes and no. Your probability of picking the correct door is 1/3. And once you make your choice Monty opens one of the two other doors at random. At that point there are three possibilities:

With probability 1/3 you picked the correct door and Monty showed a goat

With probability 1/3 you picked an incorrect door and Monty showed a goat

With probability 1/3 you picked an in correct door and Monty showed ther car

(I sent that before finishing the comment, at least it made sense…)

The point is that in the third scenario the game is over. Conditional on the car being still hidden both remaining scenarios (you picked the correct door / you didn’t) are equally likely. To calculate the probability of having picked the right door conditional on Monty showing a goat you have to divide the unconditional probability of picking the right door and Monty showing a goat (1/3) by the probability of Monty showing a goat (2/3).

To be clear, you are right that the probability changes in the “random Monty” case. But that’s why there is no point in switching.

Unlike in the standard “smart Monty” problem, where the probability doesn’t change. That’s why switching is better: the probability of picking the right door is 1/3 (or 1/100) and the probability of the prize being behind the other door(s) is always 2/3 (99/100).

Note that in the Random Monte + you always switch problem, you still win 2/3 of the time if you switch. 1/3 of the time because Monty opened the door with the car, so of course you switch, and 1/3 of the time because he opened the goat but you chose the wrong door in the first place.

In the original, Monty opened all-but-one of the remaining two doors. This generalizes to 98 out of 99 in the variant problem.

But also, and perhaps more simply stated, he opened ONE of the two. This instead generalizes to him opening 1 of the 99 (somehow). If you had sympathy to the “don’t switch” case in MH – and the fact that people do, why we are looking at these variants – is it weaker or stronger in the one-out-of-99 version? What _have_ you learned if MH shows you one specific example of something that you absolutely knew existed either 99 or 98 instances. (I know the answer, but my question is whether it’s useful to contrive a MH variant where our intuition happens to align with the ‘right'(*) answer’.)

What demystified it for me was almost the same as Sandro’s suggestion. I thought of a roulette wheel. The wheel has been spun, but I can’t see where the ball landed but the dealer can. I bet on 8. The dealer then tells me it’s either 8 or 23. Do I want to switch?

I like this one the best

I don’t get it. In MH, you should switch. In this case, you’d just have to be insane to switch. This isn’t how roulette normally works – so why are you being offered this unusual deal?

> In this case, you’d just have to be insane to switch.

No, you should ABSOLUTELY switch, because assuming a fair wheel, and that they haven’t called a house number 0 or 00, there’s only a 1/36 chance that it really is 8, so therefore there’s a 35/36 chance it’s 23.

I’m confused. If B is greater than C then B must be greater than A because there is only one professional among the three players and the other two are equal to each other. Or does this formulation entail that B and C are more like identical six-sided dice so that they are “equal” only in the long run? If so then they are distributions that never overlap that of the pro. It’s as if C’s abilities could be modeled by a single six-sided die but B’s (assumed arguendo to be the pro) require seven of them. Which seems to take you right back to two goats and a Pontiac.

The key for me was understanding that Monty was constrained and had to open one of the doors with a goat behind it. I don’t see how having the goats play tennis against each other helps.

Thanatos:

OK, I guess I didn’t write it clearly enough. “Two are equally-matched amateurs; the third is a pro who will beat either of the amateurs, always. . . .”

Player B winning does not necessarily mean that he is the pro. If B and C are both amateurs, one of them still has to win. They’re equally matched.

I must be dense but I don’t see these as equivalent because they don’t clearly label the probability of ⅔ as located physically; in the original, the probability is within the threading so the ⅔ remains with the original threading. A branching into groups. So the choice is whether you want to stay in your first group or move to the other group, knowing that the other group claimed ⅔ of the chances. I see these alternatives as showing how the states entangle, meaning you have matches and, as you note, the outcome of the first match may or may not convey useful information.

+1 on this.

The most intuitive way for me to get and explain to others the MH problem has always been to make people think of the allocation of probability across groups, and then point them to the fact that one of the two groups has shrunk in size.

The second most successful explanation in my sample has been the dynamics one — was your first guess right or wrong? and your second guess?

Hi Jonathan, Hi Paolo

The issue is that there are two versions of the Monty Hall problem out there.

In one version it is framed before the door is opened. In this case contestant can form a complete contingent plan, and the solution is to always switch. I think your solutions are the best solutions to this problem, and makes it clear that always switch leads you to win 2/3 of the time.

In the other version, including the most famous version, the contestant sees the door opened first, and then decides. This problem is different. It is asking for a conditional probability. The previous approach, while giving the correct numerical answer, and perhaps shaking people out of their intuition, is not a valid argument in this case. It is easy to test this with minimal modifications. We have a discussion of this on pp.151–153 of this paper.

ps. Jonathan, the goal here was to get idea in the other version. If you prefer the boxers stand in the same physical place, you can put them behind door A, B, C, and then open the door revealing the loser.

> This problem is different. It is asking for a conditional probability.

The problem looks different (that’s why people get confused) but that “conditional probability” is conditioning on an event with 100% probability so nothing changes. There is no new information if it was known ex-ante that Monty would unveil a goat.

By the way, I find strange the reference to “two versions” of the problem. It seems to me they are ways of looking at the same problem to make clear that there is no information and nothing to really condition on.

In formal terms, using Bayes’ theorem, you’re looking for P( car picked | goat shown ) = P( goat shown | car picked ) P( car picked) / P (goat shown).

Given that P(goat shown) and P( goat shown | car picked) are both 1 it’s easy to see that P( car picked | goat shown ) = P(car picked) = 1/3.

HI Carlos:

This is the crux: “There is no new information if it was known ex-ante that Monty would unveil a goat.”

The explanation for the ex ante framing does not provide the reasoning that yields a conditional probability. You seem to agree on that much.

If you are on the game show, and you are told door C has been opened, and now you need to decided between door A and door B, you don’t have enough information until you know how Monty chooses if there were two goats. This is not a controversial point. The original version asks for this posterior probability. Because the explanation does not account for how Monty chooses, it cannot provide the conditional probability of interest.

A way to see this is imagining that Monty always opens the door further in the alphabet if he can (provided it is a goat). In this case always switching yields an ex-ante probability of winning of 2/3, but conditional of revealing door C, the condition probability is 1/2. The argument doesn’t work here, which means even in the original problem under natural assumptions, the argument was not providing the requested Pr(Goat behind door B| door C opened).

> The explanation for the ex ante framing does not provide the reasoning that yields a conditional probability. You seem to agree on that much.

I don’t. I think that the problem implicitely assumes that Monty would never open the door hiding the car and that we are completely ignorant about how does he choose when a choice has to be made. That’s enough to calculate the conditional probabily as I did. Where would you say I got it wrong?

Of course we can always modify the problem to include new information (like “Monty smiles if we pick a goat and frowns if we pick the car”) that would render any previous analysis invalid.

> If you are on the game show, and you are told door C has been opened, and now you need to decided between door A and door B, you don’t have enough information until you know how Monty chooses if there were two goats. This is not a controversial point. The original version asks for this posterior probability. Because the explanation does not account for how Monty chooses, it cannot provide the conditional probability of interest.

That’s not correct. If I don’t know anything about how the choice is made – and the problem doesn’t say anything at all – there is no need to account for it. How Monty chooses is irrelevant for the problem as can be easily understood from invariance considerations: assuming I pick door A there can be no difference between “Monty showed a goat behind door B” and “Monty showed a goat behind door C”. Both are equivalent to “Monty showed a goat behind one of the other doors” and the analysis in my previous comment applies.

The analysis changes (but the switching decision still leads to better, or at least equivalent, outcomes) when we do know how Monty chooses. We still assume the car can be behind any door with equal probability and without loss of generality let’s say we pick door A and he shows a goat behind door B. We now assume that we know the probability X of opening door B when there are goats both behind B and C.

The conditional probabily can by calculated as before P( A = car | B = visible goat ) = P( B = visible goat | A = car ) P(A = car) / P ( B = visible goat )

P( A = car ) is 1/3 as before, but now the other terms are slightly more complicated. P( B = visible goat | A = car ) = X by hypothesis.

P ( B = visible goat ) = P ( B = visible goat | A = car ) P(A = car) + P ( B = visible goat | B = car ) P(B = car) + P ( B = visible goat | C = car ) P(C = car) = X * 1/3+ 0 * 1/3 + 1 * 1/3 = (X+1)/3

The conditional probability is therefore P( A = car | B = visible goat ) = X / (X + 1) where we assume X is known (between 0 and 1) so this conditional probabilty is a number between 0 and 1/2.

Switching is always better (or indifferent, when the probability of having picked the right door conditional on the available information is maximal at 1/2).

Carlos

we are operating under different assumptions.

My points were assuming that the reader has taken the intended assumptions of the canonical Monty Hall problem. (1) Monty always opens one of the doors that the contestant didn’t choose, and reveals a goat, (2) if there are two goats, then Monty chooses one at random (50=50), (3) the car has a 1/3 chance of being behind any of the doors.

I believe (1)-(3) are natural assumptions, but they were never stated explicitly. Even when you state them explicitly, people have issues.

Perhaps now we can converge under these assumptions.

I’m not sure if you disagree with anything that I said.

I agree strongly with two of your “natural” assumptions: the unconditional probability for the car being behind any of the dors is 1/3 (this may represent just our ignorance, it doesn’t have to be necessarily the procedure actually used to assign the car to one of the doors) and one of the two doors you didn’t picked will be opened and there will be a goat behind.

The other assumption “if there are two goats, then Monty chooses one at random (50=50)” is not really needed to have a well-posed problem. We get exactly the same solution with the weaker assumption I used in my previous dicussion (total ignorance about how the choice is made).

And even if we assume that a different choice policy (which may depend on which door was chosen, breaking the symmetry) is used the answer to the question “Is it to your advantage to switch your choice of doors?” is essentially the same (in the edge case there is indifference about switching, so at least it’s never to your disadvantage to switch regardless of the assumptions made about Monty’s choice).

In any case, using your three assumptions everything I wrote remains valid. Due to the symmetry in the problem P( car behind choosen door A | door C opened) = P( car behind choosen door A | door B opened) = P( car behind choosen door A | any door opened) = P( car behind choosen door A) because the conditional probability you want cannot depend on which door that was opened. It depends on any door being opened, which is a sure event, so the conditional probability is equal to the unconditional probability.

Hi Carlos

That’s fine, but then the ex ante argument needs an additional argument that the posterior probabilities are equal in each case so that you can say they are equal to the prior. To do that you need to account for Monty’s choice rule. There are two ways to do this: (1) assume 50-50 for goat-goat, (2) allow for bias, but apply principle of indifference what this bias is (i.e. Uniform(0,1).

Hi Joshua,

I think it all comes down to the different definitions of probability. Your remarks may be relevant for the frequentist notion of probability, but using the epistemic interpretation we can find the solution to the problem in a rather straightforward manner.

“If you are on the game show, and you are told door C has been opened, and now you need to decided between door A and door B, you don’t have enough information until you know how Monty chooses if there were two goats.”

Using Bayesian inference you can decide to switch based on the information you have. You know that there is a car behind one of the doors (but you don’t know more about how the door was chosen), you know that after you picked door A he had to open one of the other doors to show you a goat, you know that he opened door B. You don’t need to know anything more, you don’t need to know how Monty chose if a choice was made. You can decide to switch based on that information. You can calculate the conditional probability P( car behind picked door A | there was a goat behind door B ) = P( car behind picked door A | there was a goat behind one of the other doors ) = P( car behind picked door A ) as I showed before. You knew already that there was a goat behind one of the other doors, Bayes’ formula trivially gives you again the unconditional probability because the update is not informative at all.

Using a frequentist approach what is the probability supposed to mean anyway? The car is already somewhere, it’s either behind the door you picked A or the remaining door B. The probability of the car being behind door A doesn’t mean anything until you define a universe of “identical” scenarios, and you have to define carefully what variables you keep constant, which ones will vary, and how all of them are related. Do you want to calculate the probability conditional on your choice of door A and his choice of door B? Or on your choice of door A and his choice of any other door? Or on your choice of any door and his choice of the following door in the sequence [A B C A]?

Once you have defined the problem, say you want P(the car is behind door A | you have chosen door A, Monty has chosen door C) you still need to define lot of details about the sampling distribution. You need the joint probability P(car position, your choice, Monty’s choice). And you need to assume that any other variable is irrelevant, or include it explicitly in the analysis. Maybe Monty’s choice doesn’t depend just on your choice and the position of the car, but also on the weather…

In summary, the frequentist solution requires extra assumptions while the Bayesian solution is based only on the information that we are given: we pick one door (A, but as far as we know the three are identical), Monty opens another door (B, but the only thing we know about his choice is that he always opens one of the remaining doors and he newer shows the car). You don’t need to account for Monty’s choice rule because you don’t know what it is. Maybe it does actually depend on the weather, but that’s irrelevant.

Hi Carlos

It seems we have lost the thread. This thread began with me pointing out that this argument—“to make people think of the allocation of probability across groups, and then point them to the fact that one of the two groups has shrunk in size.”—is not a valid way to get the conditional probability. My only goal was to establish this, and I think we have. There are different ways to fix this argument, as I have mentioned. All the fixes involve accounting for Monty’s choice rule.

I am not a fan of this approach; it is not as easy to generalize to non-symmetric versions of the problem (see the Monty fall paper, someone just introduced me to it on Twitter).

Perhaps you haven’t noticed, but my proposed solution is a Bayesian argument.

Also, you write “Bayesian solution is based only on the information that we are given… You don’t need to account for Monty’s choice rule because you don’t know what it is. Maybe it does actually depend on the weather, but that’s irrelevant.”

If you are referring to versions on the problem in which you pick door A, and then Monty reveals C, and now you need to decide whether it is strictly advantageous to switch, or if you want to know the precise probability of winning if you switch, then I have to disagree here. Knowledge of Monty’s choice rule matters a great deal, and it is relevant if you have information about it. Even if you don’t have information about it, any argument that claims to find the posterior probability asked for by arguing that it must match the obvious prior of 2/3 needs to claim symmetry to get identical posteriors, which implicitly assumes the principle of indifference on Monty’s choice rule, so it accounts for Monty’s choice rule.

Hi Carlos-

I have a post that is being moderated (due to hyperlinks I presume). It should appear above this comment once they release it from the review queue.

> Knowledge of Monty’s choice rule matters a great deal, and it is relevant if you have information about it.

But it’s irrelevant when you don’t. And in this problem, in the standard formulation, you don’t.

> Even if you don’t have information about it, any argument that claims to find the posterior probability asked for by arguing that it must match the obvious prior of 2/3 needs to claim symmetry to get identical posteriors, which implicitly assumes the principle of indifference on Monty’s choice rule, so it accounts for Monty’s choice rule.

Maybe we don’t disagree. But I wouldn’t say you claim symmetry to get indentical posteriors, I’d say that you use symmetry to justify that the conditional probability you need to calculate is conditional on any door being opened because you can ignore the irrelevant information provided in the problem statement (like the labels of the doors or the name of the host). See my other message at the bottom, we’re too deep in the thread and it’s easier to continue at the top level.

Best explanation I’ve ever heard:

If you pick the wrong door initially, then switching is certainly right because Monty avoids revealing the prize when opening a door. If you pick the right door initially, switching is certainly wrong. You live in the former world with a 2/3 chance so you get: 2/3*1+1/3*0=2/3.

Justsomeguy:

Yeah, I think that is the best answer for the ex ante version of the problem, where you commit to what you will do. On the other hand it doesn’t solve the most famous version of the problem, which is a conditional probability problem. At the bottom of the comments in this response, I have a discussion.

I like to explain this by telling people to consider playing the game 100 times.

0. Note that there are 100 cars up for grabs.

1. If you don’t switch, you get 33 cars.

2. Where did the other 66 cars go?

Most people realize quite quickly that the other 66 are all behind the doors Monty didn’t open, so switching gets you twice as many cars.

(Of course, there’s the problem that any sensible person would prefer a goat to a car. Goats make great cheese or soup. Cars are dangerous, stinky, expensive to maintain, and slower than public transportation since you have to park the damn thing.)

Hi David

That’s a really nice one, never seen it.

How would you adapt it to the case in which door A is chosen, door C is opened, and you believe that when the car is behind A, Montly will always show the goat behind C?

In this case your argument correctly says you get 66 out of 100 if you switch, but it doesn’t give you the conditional probability if C or B is opened. That isuggest that your original argument is not complete, it is not giving the conditional probability, even if it happens to coincide numerically.

“Monty will always show the goat behind C”

Truth in advertising: my math/stat background is weak by local standards. While the 100 cars trick is something that I came up with trying to explain this to some non-math friends, I’d be surprised if no one else has ever come up with it.

I don’t understand the game in which Monty has a choice. I’ve only seen the version in which Monty always shows you a goat and offers you the option to switch. I could blather on here (and started to do so and deleted it) but my foot would be in my mouth if I did.

As to “conditional probability if C or B is opened”, I don’t think that’s relevant in the original form of the game. C and B are symmetrical, so handling those cases separately just makes the intermediate math messier. But, again, writing this up formally is beyond me.

Hi David

The version I was describing is explicit about what Monty does if he has two goats to choose from. For an argument to yield the posterior (conditional) probability, it has to account for how Monty chooses at some point, or it is at best incomplete. For example, if Monty has a bias to reveal a goal behind a higher numbered door (or lower in the alphabet), then the posterior probability will change, even if the prior probability remains the same. Since the argument above pertained to the prior probability, we can see that it does not yield the posterior probability without extra justification.

I agree with you on the appeal to symmetry as a justification. Symmetry implies that the posterior probability of as win after a switching must be the same regardless of which door is opened. The law of total probability, in turn, implies that each posterior must be equal to the prior probability of winning under a commit-to-switch strategy, which you showed was ~67/100.

The appeal to symmetry implicitly assumes that Monty is not biased when opening a door (reasonable assumption), or that even if he were, you have no information about the direction of the bias (e.g. principle of indifference on choice of door if goat-goat, or uniform beliefs over Monty’s door bias). This means accounting for how Monty chooses is going on somewhere, just not mentioned. Not mentioning this, and the chain of reasoning means that the explanation is incomplete with respect to justifying that it yields the posterior.

Why is a complete explanation (not necessarily this one!) important? Well people may walk away with the illusion of understanding that they found the posterior, when really they just understood the prior. They would be at a loss at what to do if there were minor modifications of to the problem, e.g. (i) making the prizes behind each door goat/car determined by independent coin flips, (ii) allowing Monty hall to have a specific bias towards a door.

That is why I am less keen on this explanation, I prefer the ones that illustrate the information you received when a door was opened.

Ha, actually I preferred this one (without the random):

I doubt people can get the probability right though.

I came to this problem after I knew game theory. The tree for this one’s so shallow that it’s trivial. So I’ve always been a bit baffled why people find it so hard.

Maybe the right route to explaining this to people is to teach them probabilistic game trees. It’s sort of the route Richard McElreath takes to introduce conditional probability in

Statistical Rethinking. Or that a lot of intros to Bayesian inference use to discuss diagnostic accuracy.THE GAME: Monty Hall randomly places a prize behind one of three doors and nothing behind the other two doors. The player picks a door with no knowledge of what’s behind them, but does not open it yet. Monty Hall then uniformly at random opens one of the remaining doors that does not contain the prize. The player gets a final choice to switch doors or stick with their original choice.

Without loss of generality, we can assume the player chooses door 1 initially.

So there’s a 2/3 chance of winning by switching and thus a 1/3 chance of winning by staying. You could add graphs for choosing door 2 or 3 first, but they’d look exactly the same.

Why is this so hard?

IMO it is counter-intuitive (‘hard’) because the problem is so simple that it seems like our intuition should grasp that it is advantages to switch, without writing out the entire tree, and then grinding out the probability to find Pr(Switch)>Pr(Stay).

By the way, you are (correctly) solving a different version of the problem, the doesn’t apply to the most famous version of the problem discussion here.

Suppose there are three black boxes. A dice is thrown in each of the boxes. Two of them are fair 1-6 dices and the last one has a 7 on each face. Your goal is to find the box containing the dice with 7 on each face. Your first pick a box. From the other two boxes, the one with the lowest number is then removed (in case of match, dices are thrown again but you do not know it). Your are then asked if you want to switch box.

There are now two variants of this problem. If the value of the losing dice is not shown to you, then we are back to the Monty Hall problem. But if you are shown the value of this dice, you get an additional information. My guess is, the higher the value of the losing dice, the more likely you are to win by changing boxes.

You’re correct; a quick Monte Carlo suggests that the odds of stick-to-switch go 11:12, 9:12, 7:12, 5:12, 3:12, 1:12.

(“A dice is thrown” does my head in — took me a couple of readings to understand that there are not two dice in each box but rather one die. Whine whine whine, ignore me.)

Sorry, that’s odds of failure-to-success if switching.

Apologies, there is too much too read here to easily check, but I always thought the thinking that Monte’s choice between the two goat doors was random (1/2) was a mistake. That’s not humanly possible without a good random device to ensure it. Now if he favors somewhat the higher numbered door, the conditional probability of winning by switching changes quantitatively but not qualitatively (still advantageous to switch.)

To me this then becomes – try not intuit if the door opened is informative but rather see if the conditional probability changes (or likelihood is not flat or posterior changes).

I think the basic paradox of the Monty Hall problem involves the sequencing, and the tennis/boxing metaphor just obscures what’s actually going on.

The confusing aspect of the Monty Hall problem is the fact that it’s different from a closely related problem. You’re on a show with three doors: two goats and a car. You write down your guess for the car on a sheet of paper. Monty randomly selects and reveals one of the two goat doors (say Door B). What’s the probability that A and C respectively have a car/should you change your guess? Here, our intuition serves us well and the answer is 50-50 between A and C. There’s no point changing your guess if you chose one of those initially (obviously, you should change if you initially chose B). At first glance, the Monty Hall problem appears to be exactly the same as this. What the problem fails to make obvious is that what Monty does *depends* on what you do, and that’s why it matters. I think the tennis/boxing framing makes this even less obvious because that version is *only* the same as the Monty Hall problem if Boxers 2 and 3 fight *because* you blindly chose Boxer 1 at first.

Here’s another variant on Monty Hall:

1) You’re watching along at home. You make your own prediction that the car is behind Door C.

2) You get up to grab a snack. You miss the contestant’s selection, but you return in time to see Monty revealing a goat behind Door B.

Should you change your choice? No. This is the same as the first scenario. Even though the contestant, if they chose Door C at the same time as you, should change. They have information that you don’t (which doors were compared leading up to the reveal).

The problem is confusing because we think of the player’s initial action as guessing where the car is. That’s not the work being done at the initial step. The initial action is a request to be shown a goat (I think there’s also some kind of anchoring bias when we ask people about “switching” their guess rather than making one de novo).

So, taking a cut at the tennis analogy. There are the three players as designated. Initially, you form an arbitrary judgement that A is the pro. You show up to watch the tennis matches but your tickets are invalid, so you’re stuck in the parking lot. The matches are part of an elimination tournament. A little bit later, you see B walk out into the parking lot to drive home because he has been eliminated. Should you change your guess? No for the same reason as above. You have to also know or control the schedule.

Here’s my intuitive Monty Hall. Monty is going to let you designate two doors that he will search behind and reveal a goat. For whatever reason, going into this, you have a good feeling about Door B. It’s your lucky letter. Should you include Door B as one of the two to search? I think everyone immediately recognizes that the answer is “yes” which isn’t the whole intuition but gets you most of the way and that after the search is Door B did not reveal a goat, then you ought to be more confident.

Hi Joe

I think your search version of Monty hall is really nice, though I’d worry about people’s intuition if Door B did reveal goat. I think what you may be exploiting here is confirmation bias. On the other hand perhaps it could be adapted to a literal game of hide-and-seek where you can send your friend (or family member) to inspect the boxes/doors but they are only allowed to tell you where it isn’t (and just one box). Maybe people would get that information is revealed. Not sure.

Your comment: “that version is *only* the same as the Monty Hall problem if Boxers 2 and 3 fight *because* you blindly chose Boxer 1 at first… This is the same as the first scenario.”

Your variant of Monty hall not equivalent. It doesn’t matter if the choice if two boxers is independent of your initial choice, if by chance boxers 2 & 3 fight after you choose boxer 1, then you have learned something. Agreed that those would be different rules than the original Monty, they were not intended that way. Could make the implicit (natural) assumption explicit.

If you really want to go down the MH rabbit hole, assume that Monty’s objective is to make the contestant LOSE.

So, if the contestant picks a goat, Monty does nothing and the contestant loses. But, if the contestant initially picks the car door, then it gets more complicated. Monty would try to get the contestant to switch by opening a goat door (so everyone on this blog would switch), in which case the contestant LOSES AGAIN, so Monty would make the contestant ALWAYS LOSE. But then contestants would realize that when Monty reveals a goat, if they switch, they will lose, so the contestant does not change their choice. Therefore, the optimal strategy for the contestant is to never switch and they will win the car 1/3 of the time.

Moral? To solve this problem you HAVE TO completely specify Monty’s strategy in every state of the world. If you don’t, you’re just reasoning based on whims.

(Hat tip to Bob Carpenter who went right to the heart of the problem by recognizing this as a basic game theory problem. A game theory problem is not complete without a complete description of the player’s objectives.)

> HAVE TO completely specify Monty’s strategy in every state of the world. If you don’t, you’re just reasoning based on whims.

Agree, including how Monte makes the choice between two doors with probability exactly equal 50%.

Good point.

There is no reason to assume Monty’s strategy is a coin flip. Perhaps Monty wants to tilt the outcome one way of the other a bit by deviating from the 50/50 assumption. Perhaps Monty’s strategy is time-varying. Perhaps management wants contestants to win x% of the time and a lot of contestants have won recently, so Monty will favor an action that will tend to make the contestant lose.

Monty’s actions could be dependent on all sorts of other considerations. Maybe Monty just doesn’t like the cut of this contestant’s jib.

Fun fact:

If Monty randomly chooses which door to show the contestant (revealing the car half the time) then the expected payoff to the contestant is 1/3 regardless of whether the contestant chooses to switch or not switch.

Proof:

Case 1: Contestant always switches.

One-third of the time, the contestant initially picks the car, Monty reveals a goat, the contestant switches and gets a goat.

One-third of the time, the contestant initially picks a goat, Monty reveals the car, game ends and contestant gets a goat.

One-third of the time, the contestant initially picks a goat, Monty reveals a goat, the contestant switches and gets a car.

Expected payoff is 1/3*0 + 1/3*0 + 1/3*1 = 1/3.

Case 2: Contestant never switches.

One-third of the time, the contestant initially picks the car, Monty reveals a goat, the contestant doesn’t switch and gets a car.

One-third of the time, the contestant initially picks a goat, Monty reveals the car, game ends and contestant gets a goat.

One-third of the time, the contestant initially picks a goat, Monty reveals a goat, the contestant doesn’t switch and gets a goat.

Expected payoff to contestant is 1/3*1 + 1/3*0 + 1/3*0 = 1/3.

Joshua,

I’m basing this response on the formulation of the problem you give in your paper:

Monty Hall problem: Suppose you’re on a game show, and you’re given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #3, and the host, who knows what’s behind the doors, opens another door, say #1, which has a goat. He says to you, “Do you want to pick door #2?” Is it to your advantage to switch your choice of doors?

The natural assumption is that the mechanic of the game is that after one door is chosen one of the other doors, hiding a worthless prize, is opened and at that point we are offered to switch. That’s the only assumption required to solve the problem which is essentially identical to the following problems:

Monty Hall problem: Suppose you’re on a game show, and you’re given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat. He says to you, “Do you want to pick door #2?” Is it to your advantage to switch your choice of doors?

Monty Hall problem: Suppose you’re on a game show, and you’re given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door, say #2, which has a goat. He says to you, “Do you want to pick door #3?” Is it to your advantage to switch your choice of doors?

Alex Trebek problem: Suppose you’re on a game show, and you’re given the choice of three boxes. Inside one box is a diamond, inside the others, rocks. You pick a box, say the yellow one, and the host, who knows what’s inside the boxes, opens another box, say the blue one, which has a rock. He says to you, “Do you want to pick the red box?” Is it to your advantage to switch your choice of boxes?

If the original problem, without introducing extra assumptions, has a solution at all then it has the same solution for all the variants above. That fact that one particular door was chosen or opened cannot change the answer and the minimal problem can also be formulated as

You’re given the choice of three boxes. Inside one of the boxes there is a prize. You will pick a box and the host will open one box but not the one you picked nor the one containing the prize. You will have the choice between the box you picked first and the remaining box. You pick a box and the host opens a prizeless box. Is the prize more likely to be in the box you picked or in the remaining box?

If you think that solving this problem is different from solving the original problem is because you’re including additional assumptions that are not included in the original problem. That’s fine, but then I would say it’s you who is giving the answer to a different question.

If you think that to calculate a conditional probability you need to take into account that the door opened was B rather than C then maybe you need to take into account also that the host was called Monty rather than Alex, the prize was a car rather than a diamond, it was hidden behind a door rather than inside a box, etc.

Hi Carlos (if anyone is following along, ha!, we are continuing from here)

To circle back on where I think we are. My original claim was that the following prior probability arguments are incomplete:

(1) Staying yields the contestant’s original door. The prior probability that hides a car is 1/3, that cannot change when Monty opens one of the other two doors.

(2) Switching yields the best door of the other two. The prior probability that door hides a car is 2/3, that cannot change when Monty opens one of the other two doors.

There is no explanation for why the prior probability doesn’t change. My stated view was that as the problems are formulated, a posterior probability is requested. In order to get that, you need an argument for why the probability doesn’t change. I think we agree on that much. I argued that accounting for how Monty chooses is necessary. You argue that it is not.

One argument is that the two posterior probabilities must be equal to each other because we don’t have any information about the label of the door being opened; the label was arbitrary. We can treat the labels as if they were permuted randomly (or sampled without replacement, or whatever). This implies that the contingencies are symmetric. We can apply to the law of total probability to say the respective posteriors are equal to the prior and we are done. [obviously over-explained here]

You would stop here. We have ignored how Monty chooses, and that is just fine.

That seems pretty convincing at first glance. I mean it feels right. But then I have some hesitation. I have an image of three doors next to each other, 1,2,3. I know that Monty, the game show host knows these door numbers. The number labels may not arbitrary and irrelevant in this case. I can’t just assume symmetry, I need to check. To check I model this environment. There are two things I don’t control: (1) where the prize is placed, (2) what door the host opens. I need to specify a data generating process, which means I need a model for Monty. Does he have a bias for lower numbered doors? The problem didn’t specify, so it is most natural to say he doesn’t have a bias because I have infinitely many choices if I say yes. Now it is each to see that we get that desired symmetry. I think this needs to be there.

By the way, I think the need to make the prior/posterior argument limits these explanations. Before the argument is made, the the reasoning is misleading by suggesting that Monty cannot influence the probability that the originally picked door is right (or the remaining door of the other two is right). That is clearly not true unless you make an assumption about how Monty chooses. Which again argues for the need to address it.

ps. I don’t find your Monty/Alex car/diamond, door/box comments on point. I don’t see how they highlight an inconsistency in my reasoning.

Do you agree that the following problem is equivalent to the original one or not?

“You’re given the choice of three doors. Behind one of the doors there is a prize. You will pick a door and the host will open one door (not the one you picked nor the one hiding the prize). You will have the choice between the staying with door you picked first and switching to the remaining door. You pick a door and the host opens a prizeless door, as expected. Is the prize more likely to be behind the door you picked or behind the other one?”

Wasn’t the theme of this blog post to find a alternative, more intuitive, formulation of Monty Hall problem? Well, there you have it.

I think that this restatement of the problem is implicit in the solution that you don’t like and makes the calculation of the conditional probability trivial and the intuitive answer correct. The following argument do not seem incomplete to me (number (2) is of course similar):

(1) Staying yields the contestant’s original door. The prior probability that hides a car is 1/3, that cannot change when Monty opens one of the other two doors because we knew already that Monty would open one of the other two doors and we are not gaining any new information at all.

P(car behind original door | Monty opens one of the two other doors) = P(Monty opens one of the two other doors | car behind original door) * P(car behind original door) / P(Monty opens one of the two other doors) = 1*P(car behind original door)/1 = 1/3

I don’t think it is equivalent for the reasons I mentioned in my previous comment.

I do agree that with this version of the problem you present we can appeal to symmetry as I did above and ignore Monty’s choice rule.

That said, I don’t think an intuitive answer will immediately pop into peoples minds.

Also, I believe the solution you wrote will not be obvious to people for two reasons:

(i) they may think “if I am not gaining any information” as you say, then why is the other door suddenly more likely? The paradox still sits there. Incidentally, in the original Monty Hall problem, the Three prisoners problem, this argument was put in the problem statement. That was the paradox. On the one hand, if our initial chances where 1/3, why should that change? On the other hand, it doesn’t feel like we gained information about the door he left closed, so it should be 1/2. What explains the difference?

(ii) The need to explicitly argue that the prior they accept must be equal to the posterior they want.

IMO: I think the best explanation highlight the information gained from Monty’s choice, i.e. Bayesian updating rather than the connection to the conditional via the law of probability.

> Bayesian updating rather than the connection to the conditional via the law of probability.

Bayesian updating is all about conditional probabilities so I don’t know what to make of that statement.

If you don’t think the formulation I proposed is equivalent to the original problem, I guess you also find that the original problem

“Monty Hall problem: Suppose you’re on a game show, and you’re given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #3, and the host, who knows what’s behind the doors, opens another door, say #1, which has a goat. He says to you, “Do you want to pick door #2?” Is it to your advantage to switch your choice of doors?”

is not equivalent to the problem where he “opens another door, say #2” and says to you “do you want to pick door #3?” which seems a remarkable position.

(Of course I meant “do you want to pick door #1?”, door #3 is the one you picked…)

Carlos:

You wrote: “Bayesian updating is all about conditional probabilities so I don’t know what to make of that statement.”

My point an argument that proves that Pr(switching yields prize)=Pr(switching to B yields prize| C is revealed) by appealing to law of total probability to connect the probability to the conditional is different from a Bayesian updating argument. There is no updating factor.

Further, conditional probability is a definition, Bayes rule is a Theorem. What makes the two clearly different is thinking about the different procedures/algorithms they entail. You can use the definition to calculate the conditional probability, or you can use Bayesian update and calculate the posterior (conditional) probability by applying the updating factor to the prior. Hope that is clear.

You wrote: “I guess you also find that the original problem… is not equivalent to the problem”

No, I find that to be equivalent. As I explained, once you provide specific labels that Monty can discriminate between, then the labels can be part of his choice procedure. It is not like a case in which the contestant blindly pulls an item from the urn, which happen to be a car, and Monty looks into the urn and pulls out one of two indistinguishable goats. Instead, the case we are dealing with, if I ask you to justify symmetry, you will have to appeal to some assumption about Monty’s choice procedure, or your beliefs about it.

> No, I find that to be equivalent.

We have a good starting point here! We agree that the problems “You pick #3, Monty shows a goat behind #1, do you stay or switch?” and “You pick #3, Monty shows a goat behind #2, do you stay or switch?” are equivalent. The only assumptions are that the rules of the game are that A) there are two goats and a car, and B) after we pick we are shown a goat behind a different door. To answer we just need to calculate the conditional probability that the car is behind the door we picked (if it’s below 0.5 we should switch, if it’s equal to 0.5 we’re indifferent).

Hopefully we agree that the answer to the two problems is the same P(car behind #3 | #3 picked, #1 opened) = P(car behind #3 | #3 picked, #2 opened) = X.

Now, the answer to the problem “You pick #3, Monty shows a goat behind either #1 or #2, do you stay or switch?” is also the same:

P(car behind #3 | #3 picked, #1 or #2 opened) = P(car behind #3 | #3 picked, #1 opened) * P(#1 opened | #3 picked) + P(car behind #3 | #3 picked, #2 opened) * P(#2 opened | #3 picked) = X * [ P(#1 opened | #3 picked) + P(#2 opened | #3 picked) ] = X

> As I explained, once you provide specific labels that Monty can discriminate between, then the labels can be part of his choice procedure.

But that’s irrelevant because as I just showed above if we solve the problem “You pick #3, Monty shows a goat behind either #1 or #2, do you stay or switch?” we have found the solution to the original problem.

What about the following formulation?

“Monty Hall problem: Suppose you’re on a game show, and you’re given the choice of three doors. Behind one door is a car, behind the others, goats accompanied by their caretakers Mandy and Sandy. You pick a door, say #3, and the host, who knows what’s behind the doors, opens another door, say #1, where Mandy is holding a white goat. He says to you, “Do you want to pick door #2?” Is it to your advantage to switch your choice of doors?”

Now you have to take into account that he opened door #1 and also that Mandy was there. Maybe Monty prefers to give Mandy screen time for some reason, when he has the choice. And you should also take into account that it is a white goat, maybe he likes white goats better… Including the doors and the assistants in the model requires some assumptions, including the color of the goat even more.

And what’s the point, if you can solve the easier problem where you are free to ignore all the details that won’t change the answer?

Carlos

The problem is we didn’t settle on what we meant by equivalent. By equivalent I meant in both cases you will have to solve them the same way, and both involve thinking about Monty’s choice procedure.

You wrote: “Hopefully we agree that the answer to the two problems is the same P(car behind #3 | #3 picked, #1 opened) = P(car behind #3 | #3 picked, #2 opened) = X.”

Sure, but you need to argue this. If you assume that you don’t need Monty’s choice procedure, which you are doing here, then you certainly don’t need to account for it later!

In addition: Notice that if we are in the #1 picked frame, and you bring up the #3 picked frame next to it, then we need to justify anew the symmetry to argue that the posteriors are the same if #1 is opened or #2 is opened, as we have only said what Monty does when facing #2 and #3.

Joshua,

> If you assume that you don’t need Monty’s choice procedure, which you are doing here

I’m not assuming that, I’m proving it. I assume that whenever I pick one door another door is opened and there is a goat. I assume that P(car behind #3 | #3 picked, #1 opened) = P(car behind #3 | #3 picked, #2 opened).

Under those assumptions, which I think you accept, P(car behind #3 | #3 picked, #1 or #2 opened) is also equal to the conditional probabilities in the previous paragraph.

If you don’t agree, it would help your position if you could point to a flaw in the proof.

Rereading you’re answer I understand now that you don’t think that the problem has a solution at all unless additional assumptions are made. But then it’s not even equivalent to itself!

My point of view is the opposite, we should try to solve the problem with as little assumptions as possible. Assuming there is a solution with te assumptions I gave it has to be the same for P(car behind #3 | #3 picked, #1 ) and P(car behind #3 | #3 picked, #2 opened). As we can find a unique and consistent solution, I find satisfactory say that the problem can be solved without additional assumptions.

Carlos

you wrote: “Rereading you’re answer I understand now that you don’t think that the problem has a solution at all unless additional assumptions are made”

Yes, of course, you need tons of assumptions to solve the Month hall problem.

On you previous comment: I don’t accept those assumptions, that is why I wrote that you are assuming what you are trying to prove. By assuming symmetry, as you have done, you can side-step talking about the choice rule. My point is that you have to justify symmetry.

This and your argument that followed sent us in a circle: I have already made this point, and the point about the connection between posteriors and priors after establishing symmetry further above

in this comment.

You wrote: “‘If you don’t agree, it would help your position if you could point to a flaw in the proof”

I pointed out where you assumed symmetry, which implies that the choice rule doesn’t matter. Thing is, symmetry was thing that needs to be justified.

Anyhow, I am fine leaving here. Thanks for you feedback!

Dear Carlos:

you wrote: “Rereading you’re answer I understand now that you don’t think that the problem has a solution at all unless additional assumptions are made. But then it’s not even equivalent to itself!”

Yes, of course, we need tons of assumptions to solve the Month hall problem. Please note what I said I meant by equivalent.

On your previous comment: I don’t accept those assumptions, that is why I wrote that you are assuming what you are trying to prove. By implicitly assuming symmetry (via equal posteriors), as you have done, you can side-step talking about the choice rule. My point is that you have to justify symmetry.

You wrote: “‘If you don’t agree, it would help your position if you could point to a flaw in the proof”

I did point it out. That is where you assumed symmetry, which implies that the choice rule doesn’t matter. Thing is, symmetry was thing that we are trying to justify.

In my comment further above I already made the point about assuming symmetry, and the point about the easy argument/proof that connects priors to posteriors after establishing symmetry.

In any event, I am fine leaving here. I particular enjoyed and benefited from our discussion of distinguishable labels, and exploring the connection between the contestant’s & host’s knowledge of those labels. Thanks for you feedback!

Hi Joshua,

I think “equivalent problems” typically means that they have the same solution (same values) so either can be solved to find the solution. They are not necessarily solved in the same way, often the idea is to find and solve a simpler problem with the same solution (dual optimization problems, par example).

As far as additonal assumptions go, invariance under permutation of labels is a pretty mild and natural one. I wouldn’t even say it’s an assumption, it’s a basic desideratum. Given the problem statement I cannot imagine how anyone could find acceptable to have different solutions for “You pick #3, Monty shows a goat behind #1, do you stay or switch?” and “You pick #3, Monty shows a goat behind #2, do you stay or switch?”

At least the root of the disagreement is clear now. It was an interesting discussion, thanks.

I agree we found the root of the issue, but you have made some new comments that I do not agree with, and does not accurately characterize what I wrote above:

In particular:

(1) On your definition of equivalent: I did not express well my sense of equivalent further above. By equivalent problem, I meant an equivalent problem “statement,” not an “equivalent” framing of a problem that a problem-solver might use to help solve the original problem in a different way. For example, a problem statement that explicitly asks for a prior probability is not equivalent to a problem statement that explicitly asks for a conditional probability. So in contrast, my view is that having the same numerical value as a solution in each problem is not enough, the value in each problem should correspond to the same mathematical object. If it don’t then an argument needs to be made linking the two.

(2) You wrote:

(a) As I mentioned, I think permutation of labels is fine when the actors of the problem can’t see the labels, and therefore can’t base their choices on them. The Tennis player and Boxer versions are cases of these. In my opinion these version are much better because you don’t have that third party (Monty) choosing, which in turn, misdirects the reader from the fact that the two doors have interacted/collided in the conditional probability version of the problem (this misdirection is the magic trick, so to speak, and is probably a big reason why people find the puzzle so counter-intuitive).

(b) You seem to suggest that my view implies different solutions, and that I find it acceptable. It doesn’t have different solutions on the natural assumptions I outlined. I hope what I wrote above makes that clear. I am sorry if it didn’t.

Anyway, if you want to discuss this further, I would be happy to continue over email (you can click on my name). I don’t think continuing with this thread would be productive for either of us.

Just for the record, I’m not suggesting that you get different solutions. What I imply, for the sake of the argument, is that if you got different solutions that would be acceptable to you: otherwise you wouldn’t need to calculate P(car behind #3 | #3 picked, #1 opened) and P(car behind #3 | #3 picked, #2 opened) to establish that they are equal.

I disagree for reasons stated above.

Another formulation, as intuitive as it gets:

Inside a box there are three balls, one black and two white. You pick one and keep it in your hand, without looking at it. I look into the box and take out a white ball. Is the black ball more likely to be in the box or in your hand?

That is intuitive (for me), though not equivalent. Labels are critical. Here is a verion that makes it even more clear why the labels are important

That version (which is a completely different problem) can also be formulated and solved ignoring labels:

“Three coins are flipped, the outcome is hidden, You pick one, you will win if it is tails up. I will look at the outcome of the other two and unveil a coin which is heads up if possible. I do unveil a heads up coin. Are you now more likely to win if you stay with your choice or if you switch to the remaining coin?”

In this problem the three coin flips are independent. Therefore P(picked coin = tails | heads shown) = P(picked coin = tails | the total number of tails in the other coins is 0 or 1) = P(picked coin = tails) = 1/2.

In this problem the probability of winning when switching is not one minus the probability of winning when keeping the original choice, it has to be calculated. P(remaining coin = tails | head shown) = P(the total number of tails in the two coins not picked is 1 | the total number of tails in the two coins not picked is 0 or 1) . If we call X = “the total number of tails in the two coins that were not picked” then the conditional probability we’re looking for is P(X=1|X=0 or X=1) = P(X=0 or X=1|X=1)*P(X=1)/P(X=0 or X=1). The first term is equal to 1, the others can be calculated using the binomial distribution and the probability of winning if we switch, conditional on a heads flip being shown, is 0.5/0.75=2/3 so switching is advantageous,

It is a version of the Monty Hall with a different probability distribution over prizes, but I agree it is a different problem.

You label-free re-phrasing allows you to appeal to symmetry without any additional justification, but it is not the same as the problem I presented. The reasons are the same as in this comment. Now I agree that this version/new problem doesn’t illustrate the importance of labeling any better than the previous. What it does illustrate is how the procedure used to solve the problem under the no label Monty hall cannot be applied with only a modification of changing the probability distribution over prizes. You need to use an entirely new approach.

Instead, if you use Bayes rule directly on each problem (restricted choice), you can solve them in the same way, and quite easily. In this case: the prior odds in favor of a tails behind door #2 were 1:1, but the probability door 3 is opened revealing a heads in that case is #1, whereas it is 1/2 in the case a heads is behind door #2, so the posterior odds in favor of a tails go to 2:1, i.e. 2 chances out of 3 total chances.

Regarding the equivalence of this formulation with the original one, I think all the following variants are also equivalent:

Inside a box there are three balls, one black and two white. You pick one and keep it in your hand, without looking at it, and then …

… I look into the box and take out one ball in each hand. I tell you one of the balls I’m holding is white. Is the black ball more likely to be in your hand or in one of my hands?

… I look into the box and take out one ball in each hand. I open one of my hands showing a white ball. Is the black ball more likely to be in your hand or in one of my hands?

… I look into the box and take out one ball in each hand. I open my left hand showing a white ball. Is the black ball more likely to be in your hand or in my right hand?

I wonder if all of them are different from each other according to you.

responding quickly: I think this comparison is very interesting. Reading quickly I don’t think #3 is equivalent to #2, both the contestant, and the host know the labels. I think #2, is pretty close to your original urns example, except now Monty knows the labels, but the contestant doesn’t, which means we can ignore the decision rule and argue symmetry from the perspective of the contestant’s information state. #1 feels different still, but it depends on how you formally define “equivalent.”

Joshua:

> There are two ways to do this: (1) assume 50-50 for goat-goat, (2) allow for bias, but apply principle of indifference what this bias is (i.e. Uniform(0,1).

Instead of applying the principle of indifference (Laplace’s folly?) why not a sensitivity analysis with a couple possibilities (e.g. always opens higher/lower numbered door).

Are you “demanding” a single solution?

Hi Keith-

Carlos’ claim was that we don’t need to account for how Monty chooses in order to reach 2/3. I was arguing that we do. If we do, the assumptions we need to make to reach 2/3 are either assumptions on Monty’s choice rule, or on the contestants beliefs about it.

Not sure I followed your other point. Do you mean letting p=P(B remains closed | B & C hide goats) so that posterior odds in favor of B having the car are 1:p, i.e. Pr(B has car| B remains closed) = 1/(1+p). Then after that taking specific values of p, or a higher order distribution?

> Not sure I followed your other point.

No, simply when Monte has a choice between two doors (both with goats behind them) his “rule” can either be always open the higher numbered door or always the lower numbered door. In terms of bias, either the most biased for higher numbered doors or vice versa.

For either, the best strategy is still to switch. So if these are the only biases, they don’t matter for the decision. (This I think is a shame, because if it did matter, it would be a better lessen to not intuit if something is informative but rather discern this by the change in posterior probabilities. Those change with these biases but not enough to change the optimal decision.)

oh, I see, you are asking for a the contestant to make a decision that is robust to different choice rules of Monty.

one thing, you wrote: “For either, the best strategy is still to switch.”

I agree for the prior, but conditional on the door being opened, as in the problem above, it is not necessarily strictly advantageous to switch.

Hi Keith-

Carlos’ claim was that we don’t need to account for how Monty chooses in order to reach 2/3. I was arguing that we do. If we do, the assumptions we need to make to reach 2/3 are either assumptions on Monty’s choice rule, or on the contestants beliefs about it.

Not sure I followed your other point. Do you mean letting p=P(B remains closed | B & C hide goats) so that posterior odds in favor of B having the car are 1:p, i.e. Pr(B has car| B remains closed) = 1/(1+p). Then after that taking specific values of p, or a higher order distribution?

Suppose the contestant caught a glimpse of a car part under the crack in the door and chose that door. Under this state of information the contestant should never switch.

The 1/3 and 2/3 solution apply to a symmetric state of information only.

sure, I agree, but maybe I don’t understand why you wrote this.

It seems that you are interested in the case where you know how Monty chooses the door as being somehow different from the case where you don’t know how Monty chooses the door. I agree with this, except although it changes your probability estimate, it doesn’t change how you should act.

Suppose for example you have no knowledge about where the Car is, but you know that with probability 80% after you guess, Monty will show you the car, and with 20% probability after you guess Monty will show you a goat not behind the door you guess… You guess, and Monty shows you a goat… what should you do?

Suppose for example you have no knowledge about where the Car is, but you know that Monty will 30% show you what’s behind your door, and 70% show you a random other door. You guess, and Monty shows you another door that has a goat… what should you do?

We can all agree that if Monty shows you your own door and it has a Goat, you should switch, and if he shows you another door and it has a Car you should switch… Restricted to the case where you see another door and it has a goat… is there any rule that can make it so that switching is the wrong thing to do?

I should say, if your initial state of information is that you don’t have any idea which door the car is behind.

you write: “I agree with this, except although it changes your probability estimate, it doesn’t change how you should act.”

agreed, though there is the knife-edge case in the original problem in which knowing how Monty chooses allows you to feel comfortable not switching, staying is not the wrong thing to do here.

also: some formulations of the problem (e.g. Three Prisoners), asks for the probability.

On you questions:

1. switch

2. indifferent

you ask: “is there any rule that can make it so that switching is the wrong thing to do?”

yes, if Monty opens a door < 50% of the time in the case that you didn't initially pick the car.

So you are saying there could be a rule where Monty doesn’t always show a door. In this situation the fact that you got shown a door could be indicative of what’s behind your own door.

This is I believe related to another topic that Carlos and I discussed in depth a while back, namely informative stopping rules.

In typical Bayesian experimental analysis the stopping rule “doesn’t matter” which is often reasonable. But some of the time we can infer something about the problem by virtue of the fact that the experimenter stopped the experiment.

For example, suppose you are doing an experiment where you repeatedly measure say pollution in Sea water. You stop after taking 10 measurements. Normally we don’t use this fact of stopping in our own analysis of your data. But suppose you have very precise information about your apparatus measurement precision, and we know you stop when your posterior probability over the mean pollution level is 1 part per billion or less… now we can use the fact that you stopped to infer something about your apparatus, something that can also give us more information about the pollution level than we would normally get from using our own less informed prior over your apparatus precision.

Phone-a-friend lifeline at Monty’s

[Phone rings]

Frequentist Friend : Hello?

Clueless Contestant : Hi, it’s me. I’m calling you from the Monty Hall’s game on TV

FF: The one where the are one car and two goats behind three doors, you pick one door and he shows you a goat behind one of the other doors?

CC: That’s the one! I picked a door at random and he showed me a goat. Now I have to the choice to stay with the door I picked or switch to the remaining door. What should I do?

FF: Well, the car was behind one of the doors and you picked at random so you got it right with probability 1/3. It was sure that he was going to show you a goat, so your probability of having picked the car conditional on that is 1*(1/3)/1 = 1/3.

CC: So I don’t stay with door #3?

FF: You didn’t tell me that you had chosen door #3! That changes everything, because if the car is behind that door you shouldn’t switch!

CC: What? If I knew that the car is behind door #3 I wouldn’t be calling you, what are you talking about?

FF: Don’t panic, I just need to model the probability that the car was behind each of the doors to begin with. Give me a few seconds… Ok, I got it. If I assume that the car is behind any of the doors with equal probability the probability that the car is behind the door you picked is 1/3, so you should switch.

CC: Great! So I switch to door #2.

FF: Wait a moment! Did Monty opened door #1? That changes everything! Monty may have had a choice between opening door #1 and door #2 and we should include in the model that a choice was made and how it was made. He could have a preference for lower numbers…

CC: Or for higher numbers, what don’t know anything!

FF: Ok, I’ll make some additional assumptions: he could open either door for all that we know so if I plug this 50/50 distribution into my model… and… we get that the probability that the car is hidden behind the door you picked is now 1/3.

CC: So the same as before? I’m not surprised, what would it matter that the white goat was behind door #1?

FF: Why didn’t you tell me before that the goat was white? I need to redo my model now…

CC: Is the answer going to change?

FF: Well, no. But to calculate correctly the conditional probability it’s critical that I take into acc…

[CC hangs the phone]

CC: Monty, I’d like to switch!