About the revised Monty Hall problem. Prior to choosing or any doors being opened, you know there will be a 2/3 chance Monty has the prize you really want (the special goat). He opens a door and shows you a prize you don’t want (the non-special goat). He still has a 2/3 chance of having the prize you really want. Switch. Maybe I am not understanding what this variant is supposed to add to the original problem.
Even with the original one, if he intentionally reveals a door you don't want and you didn't select, you get some information about the door you didn't select, but none about the door you selected, while if he just randomly opens a door and it happens to be one you don't want and didn't select, then you learned something about the door you selected.
This one changes things yet again - he's trying to avoid one particular door, but it turns out that's not the door you care about, so you learn something different about each of the doors than in either of the other variants.
> He still has a 2/3 chance of having the prize you really want.
I think this is where intuition fails. Before he opens the door, you're right, there's a 2/3 chance you don't have the special goat.
But after he opens the door, you've learned something - that it's not possible you have the normal goat! And that 2/3 chance of you not having the special goat was predicated on there being a 1/3 chance that you **could** have the normal goat! So it's no longer true.
And more, since there's a 100% chance he opens the normal goat door if you have a special goat, and a 50% chance he opens the normal goat door if you have the car, you've learned even more! You've learned it's more likely that you have the special goat, since it's less likely he would have opened the normal goat door if you have the car, and more likely if you have the special goat.
Here's my solution to the revised monty hall problem, from tabulating all the options. Just to explain the notation, assume without loss of generality that you always select door 1 and the host then opens either door 2 or 3.
Let's say the true situation is:
c g g
That is, door 1 has a car and doors 2 & 3 have goats. After the host reveals a door, the situation becomes:
c g g -> c g
That is, the result of staying is car and the result of switching is goat. Let's warm up with the original monty hall problem. In this case, the 6 possibilities are:
c g g - > c g
c g g - > c g
g c g -> g c
g g c -> g c
g c g -> g c
g g c -> g c
Perusing the right hand side, if you stay 2 out of the 6 options give you a car, but if you switch 4 out of the 6 options give you a car.
Now switching to the revised monty hall problem, where g* is the desired goat. The 6 options are:
c g g* - > c g*
c g* g - > c g*
g* c g -> g* c
g* g c -> g* c
g c g* -> NA
g g* c -> NA
But notice that options 5 and 6 are not possible by the problem setup. The host will not reveal the car (because they are playing original monty hall), but you know that they also did not reveal g*. Out of the remaining 4 options, 2 of them have the desired goat behind the original door and 2 of them have it behind the alternate door. It's 50-50.
Your mistake is in not considering the possibilities where the host reveals g*. The problem states "_host is unaware of this_". In the c g g* case, both c g* and c g are equally likely - these two worlds should get half the probability as g* c g -> g* c world (host has no choice).
Does that mean that if I take the normal Monty Hall problem and simply paint the goats different colors so you know which one Monty opened -- then I would also be indifferent between switching and not switching?
Yes! Because described situation is different - the goat X was shown instead of one of the goats was shown. This removes possibility of you selecting goat X, so 1/3 of possible cases are removed.
EDIT: more than 1/3 cases are removed:
- you selecting bad goat (1/3)
- Monty selecting good goat after you selected the prize (1/6)
It feels wrong and unintuitive that the optimal strategy changes because of an arbitrary difference that doesn't seem to affect anything physically. Can you describe it in a way that makes it intuitive?
The change is that you and the host no longer care about the same thing. The host is still making decisions based on the car, but now you're making decisions based on the good goat.
It's like playing chess, but secretly you win if you take the opposing queen. Your opponent plays the same, because they don't know anything is different. But your strategy should change because your victory condition changed. (Maybe you aim for an early queen exchange.)
Back to the Monty Hall problem, let's say there's 1,000,002 doors, and 1 car, 1 good goat, and 1 million bad goats. You pick a door at random. The host opens 1 million other doors, all of which have bad goats behind them. The only doors left are the one you picked, and this one door way over there that you didn't even notice. That other door almost certainly has the car, which means that (by a amazing coincidence) the door you picked almost certainly has the good goat. So if you want the car, you switch doors, but if you want the good goat, you keep the same door.
(It'd be much more likely for one of the million doors that the host opened to have the good goat, but sometimes we get lucky.)
No, you did it wrong. In your scenario, the host is FORCED to open a million doors (and only show bad goats). This can ONLY happen if you first luckily picked the good goat or the car. The entire scenario is very, very unlikely; if you randomly pick your door, almost certainly you would have picked a bad goat and then the host would have revealed the good goat elsewhere. But that got ruled out by the scenario! All those likely worlds didn't happen. You are forced into a world where your initial lucky pick MUST have been either the good goat or the car. And it's 50/50 which it was ... and so it would also be 50/50 if you switched. Whether you're interested in the good goat OR the car. Switching doesn't matter.
This is very different from the original Monty Hall problem. In the original problem, the goats are identical, so the host can ALWAYS find an extra goat to reveal, no matter what you pick first. That doesn't happen in this new scenario. If you "accidentally" pick the (or "a") bad goat ... the host CANNOT complete the new scenario. So you have to begin the analysis by RULING OUT any possible world where your first choice involves a bad goat. Those worlds didn't happen.
No, that person's math is wrong. If you picked goat A, the host will always reveal goat B. But if you picked the car, there's only a 50% chance the host reveals goat B. So the chance you picked goat A is 2 times the chance you picked the car - thus a 2/3 chance switching will get you the car and a 1/3 chance it doesn't.
Your solution contains an error - the 4 cases don't actually have equal probability.
Consider the original Monty Hall problem, but with numbered goats. It actually has 8 cases, not 6. In case the player chooses the car, Monty has 1/2 chance to reveal either goat. So the actual cases are:
c g1 g2 -> c g1 (1/12 probability)
c g1 g2 -> c g2 (1/12 probability)
c g2 g1 -> c g2 (1/12 probability)
c g2 g1 -> c g1 (1/12 probability)
g1 c g2 -> g1 c (1/6 probability)
g1 g2 c -> g1 c (1/6 probability)
g2 c g1 -> g2 c (1/6 probability)
g2 g1 c -> g2 c (1/6 probability)
When you add up probabilities, you still end up with 2/3 chance of a goat when not switching and 1/3 when switching.
In the revised problem, the initial probabilities are the same, but the reveal excludes every possibility where the good goat would have been revealed. So the cases are as follows:
c g* g -> c g* (1/12 initial probability, 1/6 normalized)
c g* g -> c g (excluded)
c g g* -> c g (excluded)
c g g* -> c g* (1/12 initial probability, 1/6 normalized)
g* c g -> g* c (1/6 initial probability, 1/3 normalized)
g* g c -> g* c (1/6 initial probability, 1/3 normalized)
g c g* -> g c (excluded)
g g* c -> g c (excluded)
So in the end, you still get 2/3 chance to get the goat when not switching and 1/3 when switching, just like in the original.
Assuming you're convinced that in the original Monty Hall problem you can increase your chance of winning (getting the car/million bucks/whatever), you don't actually need to do any probability calculations; they're already done, you just want the inverse.
It is obvious that with the new alternate win condition (getting the billionaire's goat) you maximise your chance of success by doing the opposite of what would increase your probability of getting the car.
The way I rationalized it is to say that because the host doesn't distinguish between the goats, then without loss of generality he picks the first goat door available.
c g g* -> c g* switch
c g* g -> NA (host would have shown g* - this is the tricky one)
My understanding is this problem condition says that:
1) Monty opens another door
2) Behind it is an ordinary goat
There is only one ordinary goat in this variant of the problem, so we can conclude from the condition that we have not chosen the door with the ordinary goat in any of the worlds in which the condition places us.
Hence, behind our door there is a car and a valuable goat with equal probability. Hence, it's 50/50.
Edit: no it's not. My thinking was: You can't have chosen the bad goat, since we assume Monty reveals it. Therefore if you chose the car, you need to switch, and if you chose the goodgoat, you need to stay. 50/50
It's wrong; following possibilities don't have the same probability:
1. You choose the car, Monty reveals the badgoat
2. You choose the goodgoat, Monty reveals the badgoat
If you choose car, Monty either reveals badgoat or goodgoat, one of which is excluded by the assumption. Therefore #1 has half the probability of #2
It seems to me that this reasoning is wrong. We should consider probability as a measure of our ignorance. That is, if a die fell and we didn't see it, the probability that it rolled a 3 is 1/6. If the die fell and we saw it, it is either 0 or 1.
In this case, our ignorance is reduced by the condition itself. We know a priori that Monty will not open the car, the prized goat, or the door we chose. This completely rules out situations in which we chose a door with an ordinary goat. In effect (due to some magical intervention by the author of the condition) we are choosing between two doors: the one with the car and the valuable goat. And the probability of choosing one of them is the same.
The point where we find out that monty won't reveal the good goat is causally downstream of the randomization of the doors. So it carries information about the doors.
No, I withdraw my objection. I still can't grasp it by intuition, but my own calculations show that player should not switch. So there's the Monty Hall version of the problem I broke down on!
I realized where my intuition was wrong! I had assumed that since the condition excluded the choice of a door with a regular goat, the total probability of the worlds available in the problem was 2/3. Since the choices of the door with the car and the door with the nice goat both have probability 1/3, it turns out that they are equally likely: (1/3)/(2/3)=1/2
But the condition rules out more than this! It also rules out “I chose the door with the car and Monty opened the door with the good goat”, which has probability 1/6. Then everything converges: the total probability is 1/2, the door with the good goat has probability (1/3)/(1/2)=2/3, and the option “the door with the car + opening the door with the bad goat” has probability (1/6)/(1/2)=1/3.
doesn't the total probability being 1/2 constitute the same rule out as choosing the door with the car and then the good goat being revealed?
that is, if we take Monty showing us a bad goat as a given, we can't use that given twice to estimate probabilities of other things, i thought.
alternately phrased, since seeing the good goat is ruled out, we can't use a scenario where we see the good goat to get a probability, is my intuition.
Edit: I think I get what you're saying, the phrasing just confused me you mean that given that we can rule out the situation in which you picked the car and Monty shows the good goat because that's not included in the problem, there has to be more opportunities to pick the good goat because you can't see it when Monty shows you a goat. I don't know why this is so confusing to read or explain, given that you're obvious intuition is to stay.
Yes, You haven’t chosen the ordinary goat. And if you have chosen the car Monty hasn’t shown you the valuable goat which would be unplayable anyway and also isn’t part of the initial constraints.
So there are two options - you are on the door with the valuable goat, or you have chosen the car. 50-50
My thought with the revised Monty Hall problem: we know from the normal one that you should switch if you want the car, which implies you should stay if you want the (other) goat. The only difference is that you can tell the goats apart, but I can't imagine adding that to the original would change anything.
I'm never sure why there is ever even a debate about this kind of thing. Not because the math and logic isn't tricky (it is), but because it's very straightforward to simulate this situation one hundred times with simple python code, so we could just... check.
In science, one shouldn't believe an experimental result for certain until it is confirmed by theory. One should also not believe a scientific theory until it is backed up by experiment.
In any case, testing it by experiment shows you the answer in this case, as Naremus demonstrated. But what do you learn, other than the answer?
I found it paradoxical that if you get at least 23 random people together that at least two will likely (>50%) share a birthday. When I delved into this to understand why, and understood its truth, I concluded that it is more likely, if one buys three easy-pick lottery tickets, it is more likely that one ticket will match another than for any of them to have the winning combination.
I don't disagree. But if you look at the other discussions, people do not just disagree about which is the right theory to reach the correct result, they also disagree about which is the correct result. And that just seems like a waste of brainpower, when it's so easy to check.
Also sometimes you just need the result for some practical purpose, for example to win the game described if you actually participate. This is less applicable here, since the entire point of a thought experiment is obviously to get you to think.
That's also the impression I got. Seems pretty straightforward to me. So I can't tell if I'm under-thinking it or if everyone else is over-thinking it.
There are two 'playable' options but they aren't equally likely to occur: having chosen the good goat initially is twice as likely as having picked the car initially, because half the worlds in which you picked the car and the host shows the good goat didn't happen per the problem definition, but could have happened because the host doesn't know not to pick the good goat.
Assuming the doors are random, and you pick door 1 (the door you pick doesn't change any probabilities), there are 6 possible configurations
G C B #+1/6
G B C #+1/6
B C G #X
B G C #X
C G B
C B G
The important bit, I hope you agree the last two worlds have a 1/3 chance of occurring (1/6 for each ordering) prior to the host revealing a door. Now, the last two break down (using [] to indicate the hosts random choice) into two further possible worlds based on the hosts random choice after your initial choice:
C [G] B #X
C G [B] #+1/12
and
C [B] G #+1/12
C B [G] #X
So adding up possible worlds that you could still be in, we get 1/3 you've picked the Good goat initially, and 1/6 that you picked the car and the host showed you the bad goat. Since having picked the good goat initially is more likely, we should stay.
You can do the same cheat than with the original problem, to make it more intuitive:
There are 1000 doors, with one car, one very special goat, and 998 not so special ones.
You choose a door and Monty opens 998, revealing 998 regular goats. Since Monty is working under the impression that you want the car, the probability that the car is in the door you didn't choose is 999/1000, so you should definitely not switch.
Monty doesn’t have enough information to open only doors with the rubbish goats, so you can’t generalise this problem in the same way as you could the original Monty test. In the original, Monty has to open only goat doors, in this case he doesn’t care about the valuable goat.
So in the original Monty is not going to show the car in any of the N -2 (998 here) doors he opens, in this case the valuable goat will appear most of the time in his opened doors.
Yes, Monty wouldn't care about the type of goat, and in most cases he would show it with the rest. But as in the 3 door problem, there just isn't a problem to solve if he shows you the good goat: you can't get the goat and just pick the (door most likely to have the) car. However, if he hasn't shown you the GOAT goat, it's more likely to be in the door you already picked.
This version of the problem seems to place us in a universe with six possible combinations, but the twist is that we actually find ourselves in one with only four possibilities: it is impossible that we initially chose the bad goat. In the original version, we don’t learn anything that changes the odds of switching; in this version, we don’t learn anything at all, although the game is phrased in a manner that leads us to believe that we have.
Play it this way: I’ll be Monty and start by opening the door with the bad goat and then let you pick another door. Now I’ll offer you the chance to switch. Rather obviously, you’ve learned nothing since your original choice and so the odds cannot have not changed.
The original version can be rephrased to make the correct choice obvious: I’ll let you choose one door, and then immediately give you the option of holding or switching for both other doors, one of which is certainly worthless. What fools us is thinking that being told precisely which door is worthless changes anything important.
In the classic formulation of the Monty Hall problem, there are three outcomes.
1. You initially pick car. Monty reveals a goat at random. If you stay you win, if you swap you lose.
2. You initially pick goat A. Monty reveals goat B. If you stay you lose, if you swap you win.
3. You initially pick goat B. Monty reveals goat A. If you stay you lose, if you swap you win.
So in 2 out of 3 equally likely scenarios you win if you swap.
In the revised version there are three different outcomes.
1. You initially pick good goat. Monty reveals bad goat. If you stay you win. If you swap you lose.
2. You initially pick bad goat. Monty reveals good goat. You cannot win.
3. You initially pick car. Monty reveals either good goat or bad goat with 50% probability. If he reveals good goat, you cannot win. If he reveals bad goat, you win if you swap and lose if you stay.
By revealing bad goat, Monty has inadvertently eliminated all of option 2's probability weight and half of option 3's weight. Those are worlds you could have been in but now know you are not. So you now know you are in either scenario 1 or scenario 3. But because scenario 3 has only a 50% chance of revealing the bad goat while scenario 1 has a 100% chance, you are twice as likely to be in scenario 1. If you stay you are therefore twice as likely to win compared to swapping.
What makes the Monty Hall problem so counterintuitive is that people tend to think about a probability as a property of the specific scenario. Eg. 3 doors 1 prize must always have the same probability. In actuality, a probability is a statement about *how much we know* about a scenario. In the original, Monty knows where the car is and will always reveal a goat. In the revised version, Monty does not know which goat is good and which is bad, and his actions reveal different information.
Of course the real best play is to try to win the car then offer to buy the good goat after the game is over.
A) 1/3 chance you picked the correct goat and the host showed you the wrong goat. The situation where you picked the correct goat and the host showed you the car cannot occur based on the rules. In 100% of cases here you will see the wrong goat, so seeing the wrong goat gives you no information on whether you picked the correct goat or the car.
B) 1/3 chance you picked the car, but when the host shows you the wrong goat, you can eliminate the scenario where you picked the car and the host showed you the correct goat. This is 50/50.
C) 1/3 chance you picked the wrong goat but when you see the wrong goat you can eliminate these scenarios.
Of the scenarios not eliminated by assumption, in 2 you picked the correct goat, in one you picked the wrong goat.
He's more likely to have shown you the normal goat in the "you picked special goat" situation (100%) than in the "you picked car" situation (50%) or the "you picked normal goat" situation (0%), so you do get to update your probabilities: You're more likely in the scenario where you're more likely to have seen what you actually saw. (Unlike in the original problem, where you see "goat" 100% of the time.)
It's actually equivalent to the original problem: 1/3 odds your original pick was car, 2/3 odds your original pick was _the other goat_.
I think the easiest way to express what's going on is with the odds form of bayes theorem.
In the original monty hall problem, say we hypothesize that we picked the right door from the start. We believe this to be true with 1:2 odds (since there's 1 car and 2 goats).
Then, monty shows us a goat. If we picked the car, that'll happen 100% of the time, so we don't adjust the 1. If we picked a goat, that'll also happen 100% of the time, so we don't adjust the 2. Thus, after showing us a goat, we still believe that our initial pick is correct with 1:2 odds. This implies that the *alternative* is correct with 2:1 odds, so we should switch.
Then, consider the revised problem.
There's 3 cases: We picked the special goat, we picked the normal goat, or we picked the car, each with equal odds, so 1:1:1.
Then Monty reveals the normal goat.
Under the hypothesis that we picked the special goat, this is unsurprising. That's what *must* happen, so we don't adjust.
Under the hypothesis that we picked the normal goat, this is impossible, so the odds go to 0.
Under the hypothesis that we picked the car, this is *surprising*. We'd expect to be shown the normal goat only half the time (we'd see the special goat the other half), so our 1 gets multiplied by 1/2.
After the information, our odds are at 1:0:0.5. We can re-normalize to 2:0:1, meaning that we expect there's a 2/3rds chance that our door has the special goat and thus we should stay.
I think the easiest way to solve the revised Monty Hall problem is to use the solution to the traditional Monty Hall problem.
No matter what you desire in your heart of hearts, since Monty doesn't know about it, it will always be the case that switching has a 2/3 chance of winning you the car. Therefore not switching has a 2/3 chance of winning you whichever goat has not been seen yet.
In the situation described, the goat that hasn't been seen yet is the valuable one, and so you have a 2/3 chance of getting it by not switching.
No, wrong. Because in the original, the host can ALWAYS open another goat door. But in the new scenario, if you originally picked the bad goat, it is no longer possible for the host to open a bad goat door. In the new scenario (unlike the original), that possibility cannot happen.
Kindly was right. The new problem is trivial to solve in terms of the old problem. The host never seeks to open a bad goat door. All goats are alike to him.
But does the host EVER open a door with a good goat? The description says no, it didn't happen. But it ... "could have" happened? You have to be careful in probabilities, with assigning weights to possible worlds that you know did not occur. In what precise sense "might" Monty have opened a good goat door -- given that we know he didn't?
It matters a lot what space of games you're drawing from. We KNOW that, in the games we're considering, you CANNOT have originally chosen the bad goat door. This is different from the original Monty Hall problem, where originally choosing the "bad goat" WAS a possibility. So the possibilities (potentially) have changed. You have to be careful, solving it "in terms of the old problem". It isn't the old problem any more.
Here's a possible repeated game: FIRST, Monty opens a door with a bad goat. THEN you choose from the remaining two doors. (One has a good goat, one has a car.) THEN Monty offers you the chance to switch. Do you agree, in this game, that the odds are 50/50, whether you switch or not? You could play it 1000 times, and you'll get the car 50%, and the good goat 50%, whether you switch or not.
You want to say that the game that we are talking about is different than the game in my previous paragraph. Different how, exactly? Here's one way it might be different: if you played the original Monty Hall problem, and your first choice was the bad goat, then the scenario described CANNOT happen. (Monty doesn't have a bad goat to show you.) So it must be clear that you cannot be playing the original game. The remaining question is: if you choose the car first, might Monty have shown you the good goat then? According to the puzzle, he didn't. But ... "could" he have? What does that actually mean? "Could" you have chosen the bad goat first? It seems, somehow, that you "couldn't" have chosen the bad goat ... yet Monty "could" have chosen the good goat (if you chose the car).
What exactly is the difference in analysis, between the "couldn't" (you first choosing the bad goat), and the "could" (Monty choosing the good goat when you choose the car first)? According to the scenario, neither one happened. Why is one still "possible", somehow, but the other isn't?
This only matters if you believe that the optimal strategy in the *original* Monty Hall problem depends on the quality of the goat behind the opened door.
To put it differently, suppose that we are in the scenario with the eccentric billionaire's goat, but you are philosophically opposed to helping billionaires retrieve their lost pets, and so both goats are equally worthless to you: you want the car. When Monty opens the door and you see the bad goat behind it, that's vaguely interesting but doesn't affect you in any way: by the standard Monty Hall strategy, you have a 2/3 chance of getting the car if you switch, and a 1/3 strategy of getting the (billionaire's pet) goat.
Similarly, there is a different world, in which Monty opens the door with the billionaire's pet goat behind it. Again, that's vaguely interesting but doesn't affect you in any way: by the standard Monty Hall strategy, you have a 2/3 chance of getting the car if you switch, and a 1/3 strategy of getting the (bat) goat.
The problem is that your last paragraph didn't happen, by assumption in the new scenario. So it matters a lot about why it didn't happen. It depends a bit on when you eliminate those possible worlds. If you do it before you analyze the scenario at all (originally choosing the bad goat can never be part of this new game), then it's only 50/50 for the remaining choices.
You're basically making an analysis based on a counterfactual that didn't happen (you choose the bad goat, and Monty shows the good goat). You're assigning it some weight, some probability that it "might have" happened -- even though we know it didn't.
That's where the disagreement is. With a single iteration, there aren't any probabilities at all. You either get the goat, or the car, 100%. Probabilities are about your state of knowledge, and they can only be verified by a repeated experiment where you set up some scenario over and over again, and then count the various outcomes.
What is the scenario that is being set up over and over again, in this case? If I run the example 1000 times, what are the 1000 examples? Does Monty show the good goat in some of those 1000 runs? Or not?
You need an "example generation machine", that can generate scenarios with outcomes you can count. This new problem is a little bit ambiguous about what the space of possible games is.
We're assuming that Monty doesn't know anything about the goats, so he is equally likely to choose either goat. Therefore the two scenarios are equally likely, and more importantly P(your door has car | Monty picks good goat) = P(your door has car | Monty picks bad goat).
I agree that if we don't assume this, then the answer to the entire problem is different. But I think it's unambiguous that we should assume it.
The difference is that the host opens a GOAT door, chosen uniformly at random from the two goats. The uniform part is the active ingredient here. In the original show, the two goats were indistinguishable, but if the host opens a goat at random then the two are still indistinguishable TO THE HOST, although not to you.
The thing about the car is that the host never opens a car door (they know where the car is after all) which gives you the 2/3 probability of the prize you really want which is guaranteed not to be a goat in the original case.
The revised problem as stated (you've already seen a goat but not the one you want) has cut off the branch in the probability tree where you get the good goat, so you're conditioning on seeing the other one.
If you switch, you still get probability 2/3 of a car. Therefore if you don't switch, you get probability 2/3 of "a goat" - but in this case you know it's the goat you want.
You pick a door. Monty randomly opens a door you didn't pick (not worrying about whether it has a goat or a car). He happens to reveal a door with a goat. Now what should you do?
Before you saw what you revealed, there's a 1/3 chance your first pick was the car (and he definitely reveals a goat) and a 1/3 chance your first pick was a goat and he reveals a goat, and a 1/3 chance your first pick was a goat and he reveals the car.
We can eliminate the last one, now that we've seen the goat, so it's 50/50.
It's interesting that it's different again when he definitely chooses not to reveal the car, not knowing that one of the goats is the prize you want!
How do you feel about the original Monty Hall problem? If we apply the same chart but instead want the car instead of either G or B, then stay/switch is also 50/50, right?
The lesson I take is from both the original and this one is that probability is not about it's about information. Think of these as a Bayesian. (In fact, a straightforward application of Bayes Theorem really shows it here.)
In this case, when you get lucky enough to see the bad goat, that tells you that you probably picked the good goat. Worlds in which Monty reveals the bad goat are simply less common when the bad goat is not the one you chose.
This is what I was thinking (50/50 at the end), but I wrote a simulation that says you should only switch 1/3 of the time at the end: https://pastebin.com/YtvBWpc5
Another way of working the problem that works out the same is to reduce it to the already-known solution to the standard Monty Hall problem, that there is a 2/3 chance that the car is behind the last door and a 1/3 chance the goat is there. Your door has a 2/3 chance of goat and a 1/3 chance of car. Since the still-hidden goat is Rosebud, you should stay with your door to maximize your chance of finding him.
Your door is the bad goat, monty shows you the good goat(he wont show you the car) and you lose immediately 1/3
Your door is the car 50/50 monty shows you the good goat 1/6
Your door is the car monty shows the bad goat. 1/6
The switch strategy has 1/6 winning. The keep strategy has 1/3 winning. 1/2 you lose immediately. Conditional on seeing the bad goat, you have 2/3 chance winning by not switching.
This feels bizarre to me. Why would the hosts intentions matter? But I think your reasoning is correct.
When you see new information, it's not enough just to eliminate impossible options.
You also have to take into account, that the options that made the new evidence more likely, are themselves more likely than options that had a lesser chance of showing you that information.
Having the host open a goat door is more likely if you originally chose a car door, than if you chose a goat door. Which bumps the probability that you originally chose a car door from 1/3 to 1/2.
It's not precisely that the host's *intentions* matter, but rather that their *algorithm* matters. In order to understand the significance of the information the host reveals, you need to understand what *could* have been revealed. Since the host's actions depend on what door you originally picked, you can't calculate the counterfactual observations without knowing the host's decision algorithm.
For example, suppose the hosts acts like this:
- If your original pick was the car, the host reveals a goat and asks if you want to switch
- If your original pick was a goat, the host treats that as your final choice and does not give you the opportunity to switch
In this case, if the host reveals a goat, then there is 100% chance that you already picked the car, and a 0% chance that you will get the car if you switch.
You can make it even more extreme. Say that the host says, “I will open the lowest-number door that is not the one you pick, and not the one with the car”. If you pick 1 and the host opens 3, it’s obvious that 2 has the car. But if you pick 1 and the host opens 2, then you’re now 50/50.
> You also have to take into account, that the options that made the new evidence more likely, are themselves more likely than options that had a lesser chance of showing you that information.
That's true, but you're not doing it.
> Having the host open a goat door is more likely if you originally chose a car door, than if you chose a goat door.
That's false. By definition, the probability that the host opens a goat door is 100% in all cases.
I just want to commend the OP for rephrasing to at least implicitly specify the host's decision algorithm, The host opens a door showing a goat "as is traditional", meaning always or nearly always. This rules out algorithms like "Host shows you a goat and offers a chance to switch if you picked the car, but doesn't open any door or give you a chance to switch, because cars are expensive and goats are cheap". Or, "Host decides what to do based on what he feels the audience would most enjoy, and since you're a telegenic and enthusiastic young woman he thinks they'd be happiest if he nudged you towards switching your choice to the car (which isn't being paid out of *his* salary)".
The original Monty Hall problem, what divided the intertubes back in the day, only specified that the host revealed a goat and offered a switch on this one occasion, and much of the dispute was based on differing assumptions as to the algorithm.
And, of course, this was when Monty Hall was still alive and could tell you what the algorithm was if you asked. It was not, "host always opens a door to reveal a goat".
Agreed, I've always felt the problem tells us more about how easy it is make, rely on, and be confused by unspoken assumptions, than it does about probability.
As Jeffrey's says in the Theory Of Probability:
"The most beneficial result that I can hope for as a consequence of this work is that more attention will be paid to the precise statement of the alternatives involved in the questions asked. It is sometimes considered a paradox that the answer depends not only on the observations but on the question: it should be a platitude."
Cool. I didnt realize "the monty hall problem" was not in fact a regularly occuring game on the show. Never watched it. History has been rewritten by this brain teaser. Everyone will think, as I did, that was the whole show
> The original Monty Hall problem, what divided the intertubes back in the day, only specified that the host revealed a goat and offered a switch on this one occasion, and much of the dispute was based on differing assumptions as to the algorithm.
This is just you not knowing what you're talking about. The problem is older than popular awareness of the internet. It divided the letter-writing public.
> Savant was asked the following question in her September 9, 1990, column:
> 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?
> This question is called the Monty Hall problem due to its resembling scenarios on the game show Let's Make a Deal, hosted by Monty Hall. It was a known logic problem before it was used in "Ask Marilyn".
> If the host merely selects a door at random, the question is likewise very different from the standard version. Savant addressed these issues by writing the following in Parade magazine, "the original answer defines certain conditions, the most significant of which is that the host always opens a losing door on purpose. Anything else is a different question."
And her second followup pointed out:
>> We've received thousands of letters, and of the people who performed the experiment by hand as described, the results are close to unanimous: you win twice as often when you change doors. Nearly 100% of those readers now believe it pays to switch. (One is an eighth-grade math teacher who, despite data clearly supporting the position, simply refuses to believe it!)
>> But many people tried performing similar experiments on computers, fearlessly programming them in hundreds of different ways. Not surprisingly, they fared a little less well. Even so, about 97% of them now believe it pays to switch.
>> And a very small percentage of readers feel convinced that the furor is resulting from people not realizing that the host is opening a losing door on purpose. (But they haven't read my mail! The great majority of people understand the conditions perfectly.)
Isn't this version of the problem simpler than the original one? You've distinguished the goats, so now there are just 3 doors with 3 prizes. You've already been told that Monty opens a door to reveal a prize you're not interested in so you should stick to win 2/3 of the time. (Because, as in the classic Monty Hall problem, the car is behind the third door 2/3 of the time.)
Your point is absolutely right though - a lot of the confusion on these problems hinges on Monty's behaviour not being completely specified. I think in your scenario the key information is that Monty doesn't know/care where the car is, so in opening a door he's not giving you any information about where the car is or isn't. You can swap the order of events: Monty reveals a goat, you choose from the remaining doors, and have a 50/50 chance of winning. Conversely if Monty avoids the car, he reveals information and improves your chances.
> You've already been told that Monty opens a door to reveal a prize you're not interested in so you should stick to win 2/3 of the time.
No, the opposite. This is the same as the original problem, with no differences except in which prize you want. Monty opens a door to reveal a goat. What he reveals is a prize you don't want, but that's not why he opened the door.
The results are identical to the original problem, because it's the same problem. After Monty opens the one door, the final door has a car 2/3 of the time, and your door has a car 1/3 of the time. Since you don't want the car, you don't switch.
I don’t know if non-Anglo countries also believe this but the idea that being cold gives you a cold is very widely held here despite people also knowing that it’s transmitted by a virus. I know there’s some research about seasonality etc but a lot of people seem to go way beyond that
…the cold water thing isn’t entirely wrong either though?- drinking the local tap water without boiling it in places where that belief is prevalent carries a risk of making you ill? It’s lumping the fan thing in there that seems unfair - that’s the belief that isn’t like the others.
There are two stable equilibria. If everyone believes that drinking non-boiled tap water is safe, then you want regulations in place regarding the amount of bacteria in the water, so it is indeed safe.
If nobody thinks it is safe, then nobody cares and it will probably be somewhat unsafe by default.
(Source: am Chinese, but don't have a scholarly understanding of this or anything) I do think it's more than just believing that the water needs to be boiled in order to be safe. Like if you boiled water, let it cool, and added ice to it, many Chinese people would probably think that it would hurt them to drink it. Like it will affect your digestion or fertility.
I suspect that the second thing you mentioned is the real origin of the belief that cold causes colds, and the first was something that people came up with afterward.
I think it's got a lot more to do with the fact that people get colds at the stat of winter. (Because the folk belief is true, at least in certain ways.)
What makes your nose run is the counter-current-heat-exchange-system that helps you avoid wasting lung water. The air you exhale is warm and full of water; in the nose it meets cold blood through very thin nose walls, cooling down and making water to gather into droplets. Some of those run out of your nose, as a side effect. The system works more strongly when the air (and the nose) are colder. The nose running when it's cold is not a disease, the nose is not producing more mucus. Heat exchange systems like this are common among mammals.
This in interesting but are you sure? Would the nose stop running (would the nose stop, haha) if you started breathing through your mouth? Or if you either inhaled or exhaled through your mouth consistently? If this is true, it would have to stop (maybe after drying the nostrils). I may test that if it gets cold again.
It's standard animal physiology. You can test if your nose only starts running outdoors, but not indoors (it must be actually warm indoors; some people keep 16 centigrade at home in winter, this can make the nose run). If it also runs at a warm temperature, you probably have an infection, allergy, or other irritation in the nose, producing extra mucus.
Good to know. I think you missed part of my point: the body doesn't have a reserve of cold blood, right? If the nose is cold, it's because the nose is cold. So within a couple minutes of coming indoors, the dehumidifier effect should stop. And give it a few minutes more for the nasal passages to drip a bit. Though to know how long this takes, you'd have to get a reference measurement: use some nasal spray and time how long you nose takes you dry.
Yeah, I don't know exactly how long it will keep dripping after coming indoors. Probably it's also slightly different person-by-person. Some have naturally colder noses even indoors, etc.
I don't know about the US, but in Europe people believe that it's ok to blow your nose when you have a cold. Instead of swallowing it or spitting it on the street. In medical terms, that's gross.
We find it gross because our parents have told us that it's gross (unless we do it on the soccer place, if you live in Europe).
Mucus on the street will not infect anyone. It does not find its way to the faces of people, and the street is a terrible place for germs to survive.
Asian people don't mind anyone spitting on the street, but they find the thought abhorrent to blow mucus into a handkerchief, put this into your pocket where it has perfect breeding conditions, touch it with your hand and then perhaps you even want to shake hands with them. From hygienic perspective, they are right that each of these steps is gross.
In the U.S., standard practice is to blow your nose into tissue paper made for that purpose and then throw it in the trash. And, ideally, washing your hands afterward.
About the revised Monty Hall problem. Prior to choosing or any doors being opened, you know there will be a 2/3 chance Monty has the prize you really want (the special goat). He opens a door and shows you a prize you don’t want (the non-special goat). He still has a 2/3 chance of having the prize you really want. Switch. Maybe I am not understanding what this variant is supposed to add to the original problem.
Even with the original one, if he intentionally reveals a door you don't want and you didn't select, you get some information about the door you didn't select, but none about the door you selected, while if he just randomly opens a door and it happens to be one you don't want and didn't select, then you learned something about the door you selected.
This one changes things yet again - he's trying to avoid one particular door, but it turns out that's not the door you care about, so you learn something different about each of the doors than in either of the other variants.
The difference is that Monty thinks you want the car, and might have opened the door to the special goat. That he didn't is information.
> He still has a 2/3 chance of having the prize you really want.
I think this is where intuition fails. Before he opens the door, you're right, there's a 2/3 chance you don't have the special goat.
But after he opens the door, you've learned something - that it's not possible you have the normal goat! And that 2/3 chance of you not having the special goat was predicated on there being a 1/3 chance that you **could** have the normal goat! So it's no longer true.
And more, since there's a 100% chance he opens the normal goat door if you have a special goat, and a 50% chance he opens the normal goat door if you have the car, you've learned even more! You've learned it's more likely that you have the special goat, since it's less likely he would have opened the normal goat door if you have the car, and more likely if you have the special goat.
Here's my solution to the revised monty hall problem, from tabulating all the options. Just to explain the notation, assume without loss of generality that you always select door 1 and the host then opens either door 2 or 3.
Let's say the true situation is:
c g g
That is, door 1 has a car and doors 2 & 3 have goats. After the host reveals a door, the situation becomes:
c g g -> c g
That is, the result of staying is car and the result of switching is goat. Let's warm up with the original monty hall problem. In this case, the 6 possibilities are:
c g g - > c g
c g g - > c g
g c g -> g c
g g c -> g c
g c g -> g c
g g c -> g c
Perusing the right hand side, if you stay 2 out of the 6 options give you a car, but if you switch 4 out of the 6 options give you a car.
Now switching to the revised monty hall problem, where g* is the desired goat. The 6 options are:
c g g* - > c g*
c g* g - > c g*
g* c g -> g* c
g* g c -> g* c
g c g* -> NA
g g* c -> NA
But notice that options 5 and 6 are not possible by the problem setup. The host will not reveal the car (because they are playing original monty hall), but you know that they also did not reveal g*. Out of the remaining 4 options, 2 of them have the desired goat behind the original door and 2 of them have it behind the alternate door. It's 50-50.
Your mistake is in not considering the possibilities where the host reveals g*. The problem states "_host is unaware of this_". In the c g g* case, both c g* and c g are equally likely - these two worlds should get half the probability as g* c g -> g* c world (host has no choice).
if the host reveals g* then I definitely know what door to choose.
true, host's equally likely to show g or g* - 1/2 each. each of these scenarios are half as likely as the one in which you pick g* and he shows you g.
Does that mean that if I take the normal Monty Hall problem and simply paint the goats different colors so you know which one Monty opened -- then I would also be indifferent between switching and not switching?
Yes! Because described situation is different - the goat X was shown instead of one of the goats was shown. This removes possibility of you selecting goat X, so 1/3 of possible cases are removed.
EDIT: more than 1/3 cases are removed:
- you selecting bad goat (1/3)
- Monty selecting good goat after you selected the prize (1/6)
It feels wrong and unintuitive that the optimal strategy changes because of an arbitrary difference that doesn't seem to affect anything physically. Can you describe it in a way that makes it intuitive?
It doesn't change anything, supposing a red and blue goat and you pick the first door (door choice doesn't matter):
R B C -> blue revealed, should switch
R C B -> blue revealed, should switch
B R C -> red revealed, should switch
B C R -> red revealed, should switch
C R B -> red or blue revealed, should stay
C B R -> red or blue revealed, should stay
Switch is still 2/3 compared to Stay at 1/3
You just colored goats. You did not specify which goat is presented.
For example if goat B is presented, then both cases 1 and 2 are removed and one of the cases 5 6 is removed too.
The change is that you and the host no longer care about the same thing. The host is still making decisions based on the car, but now you're making decisions based on the good goat.
It's like playing chess, but secretly you win if you take the opposing queen. Your opponent plays the same, because they don't know anything is different. But your strategy should change because your victory condition changed. (Maybe you aim for an early queen exchange.)
Back to the Monty Hall problem, let's say there's 1,000,002 doors, and 1 car, 1 good goat, and 1 million bad goats. You pick a door at random. The host opens 1 million other doors, all of which have bad goats behind them. The only doors left are the one you picked, and this one door way over there that you didn't even notice. That other door almost certainly has the car, which means that (by a amazing coincidence) the door you picked almost certainly has the good goat. So if you want the car, you switch doors, but if you want the good goat, you keep the same door.
(It'd be much more likely for one of the million doors that the host opened to have the good goat, but sometimes we get lucky.)
No, you did it wrong. In your scenario, the host is FORCED to open a million doors (and only show bad goats). This can ONLY happen if you first luckily picked the good goat or the car. The entire scenario is very, very unlikely; if you randomly pick your door, almost certainly you would have picked a bad goat and then the host would have revealed the good goat elsewhere. But that got ruled out by the scenario! All those likely worlds didn't happen. You are forced into a world where your initial lucky pick MUST have been either the good goat or the car. And it's 50/50 which it was ... and so it would also be 50/50 if you switched. Whether you're interested in the good goat OR the car. Switching doesn't matter.
This is very different from the original Monty Hall problem. In the original problem, the goats are identical, so the host can ALWAYS find an extra goat to reveal, no matter what you pick first. That doesn't happen in this new scenario. If you "accidentally" pick the (or "a") bad goat ... the host CANNOT complete the new scenario. So you have to begin the analysis by RULING OUT any possible world where your first choice involves a bad goat. Those worlds didn't happen.
It feels wrong and unintuitive that the optimal strategy changes because of an arbitrary difference that doesn't seem to affect anything physically.
It is wrong
No, that person's math is wrong. If you picked goat A, the host will always reveal goat B. But if you picked the car, there's only a 50% chance the host reveals goat B. So the chance you picked goat A is 2 times the chance you picked the car - thus a 2/3 chance switching will get you the car and a 1/3 chance it doesn't.
Your solution contains an error - the 4 cases don't actually have equal probability.
Consider the original Monty Hall problem, but with numbered goats. It actually has 8 cases, not 6. In case the player chooses the car, Monty has 1/2 chance to reveal either goat. So the actual cases are:
c g1 g2 -> c g1 (1/12 probability)
c g1 g2 -> c g2 (1/12 probability)
c g2 g1 -> c g2 (1/12 probability)
c g2 g1 -> c g1 (1/12 probability)
g1 c g2 -> g1 c (1/6 probability)
g1 g2 c -> g1 c (1/6 probability)
g2 c g1 -> g2 c (1/6 probability)
g2 g1 c -> g2 c (1/6 probability)
When you add up probabilities, you still end up with 2/3 chance of a goat when not switching and 1/3 when switching.
In the revised problem, the initial probabilities are the same, but the reveal excludes every possibility where the good goat would have been revealed. So the cases are as follows:
c g* g -> c g* (1/12 initial probability, 1/6 normalized)
c g* g -> c g (excluded)
c g g* -> c g (excluded)
c g g* -> c g* (1/12 initial probability, 1/6 normalized)
g* c g -> g* c (1/6 initial probability, 1/3 normalized)
g* g c -> g* c (1/6 initial probability, 1/3 normalized)
g c g* -> g c (excluded)
g g* c -> g c (excluded)
So in the end, you still get 2/3 chance to get the goat when not switching and 1/3 when switching, just like in the original.
ohhh .... nice
Yeah the optimal strategy is to stay as there is a 2/3 chance you picked the good goat given the bad one is shown. Here are several arrays showing this. Initially I screwed up myself but someone quickly pointed out my fallacy: https://x.com/pwalshmusic/status/1880355906569003337?s=46&t=wbf9WfI-2y6NxCrWwiqnNQ
I just realised this myself.
Assuming you're convinced that in the original Monty Hall problem you can increase your chance of winning (getting the car/million bucks/whatever), you don't actually need to do any probability calculations; they're already done, you just want the inverse.
It is obvious that with the new alternate win condition (getting the billionaire's goat) you maximise your chance of success by doing the opposite of what would increase your probability of getting the car.
The way I rationalized it is to say that because the host doesn't distinguish between the goats, then without loss of generality he picks the first goat door available.
c g g* -> c g* switch
c g* g -> NA (host would have shown g* - this is the tricky one)
g* c g -> g* c stay
g* g c -> g* c stay
g c g* -> NA (host would have shown g*)
g g* c -> NA (host would have shown g*)
My understanding is this problem condition says that:
1) Monty opens another door
2) Behind it is an ordinary goat
There is only one ordinary goat in this variant of the problem, so we can conclude from the condition that we have not chosen the door with the ordinary goat in any of the worlds in which the condition places us.
Hence, behind our door there is a car and a valuable goat with equal probability. Hence, it's 50/50.
Where am I wrong?
agreed, it's 50/50
Edit: no it's not. My thinking was: You can't have chosen the bad goat, since we assume Monty reveals it. Therefore if you chose the car, you need to switch, and if you chose the goodgoat, you need to stay. 50/50
It's wrong; following possibilities don't have the same probability:
1. You choose the car, Monty reveals the badgoat
2. You choose the goodgoat, Monty reveals the badgoat
If you choose car, Monty either reveals badgoat or goodgoat, one of which is excluded by the assumption. Therefore #1 has half the probability of #2
It seems to me that this reasoning is wrong. We should consider probability as a measure of our ignorance. That is, if a die fell and we didn't see it, the probability that it rolled a 3 is 1/6. If the die fell and we saw it, it is either 0 or 1.
In this case, our ignorance is reduced by the condition itself. We know a priori that Monty will not open the car, the prized goat, or the door we chose. This completely rules out situations in which we chose a door with an ordinary goat. In effect (due to some magical intervention by the author of the condition) we are choosing between two doors: the one with the car and the valuable goat. And the probability of choosing one of them is the same.
The point where we find out that monty won't reveal the good goat is causally downstream of the randomization of the doors. So it carries information about the doors.
No, I withdraw my objection. I still can't grasp it by intuition, but my own calculations show that player should not switch. So there's the Monty Hall version of the problem I broke down on!
I realized where my intuition was wrong! I had assumed that since the condition excluded the choice of a door with a regular goat, the total probability of the worlds available in the problem was 2/3. Since the choices of the door with the car and the door with the nice goat both have probability 1/3, it turns out that they are equally likely: (1/3)/(2/3)=1/2
But the condition rules out more than this! It also rules out “I chose the door with the car and Monty opened the door with the good goat”, which has probability 1/6. Then everything converges: the total probability is 1/2, the door with the good goat has probability (1/3)/(1/2)=2/3, and the option “the door with the car + opening the door with the bad goat” has probability (1/6)/(1/2)=1/3.
doesn't the total probability being 1/2 constitute the same rule out as choosing the door with the car and then the good goat being revealed?
that is, if we take Monty showing us a bad goat as a given, we can't use that given twice to estimate probabilities of other things, i thought.
alternately phrased, since seeing the good goat is ruled out, we can't use a scenario where we see the good goat to get a probability, is my intuition.
Edit: I think I get what you're saying, the phrasing just confused me you mean that given that we can rule out the situation in which you picked the car and Monty shows the good goat because that's not included in the problem, there has to be more opportunities to pick the good goat because you can't see it when Monty shows you a goat. I don't know why this is so confusing to read or explain, given that you're obvious intuition is to stay.
Yes, You haven’t chosen the ordinary goat. And if you have chosen the car Monty hasn’t shown you the valuable goat which would be unplayable anyway and also isn’t part of the initial constraints.
So there are two options - you are on the door with the valuable goat, or you have chosen the car. 50-50
Except it's half as likely that you picked the car:
If you picked the valuable goat, 100% of the time Monty shows you the non-valuable goat.
If you picked the car, 50% of the time Monty shows you the non-valuable goat. By assumption we eliminate the other possibility.
So in two out of three cases where you see a non-valuable goat, you've picked the valuable goat.
Yes indeed this is correct. Optimal choice is to stay. Sometimes it’s easier to grasp these conditional probabilities with a tree diagram with hard numbers (1,000 simulations for instance) rather than pure probabilities. Here is one of helpful: https://x.com/pwalshmusic/status/1880355906569003337?s=46&t=wbf9WfI-2y6NxCrWwiqnNQ
My thought with the revised Monty Hall problem: we know from the normal one that you should switch if you want the car, which implies you should stay if you want the (other) goat. The only difference is that you can tell the goats apart, but I can't imagine adding that to the original would change anything.
That is my impression as well.
I'm never sure why there is ever even a debate about this kind of thing. Not because the math and logic isn't tricky (it is), but because it's very straightforward to simulate this situation one hundred times with simple python code, so we could just... check.
Indeed:
import numpy as np
staywin = 0
switchwin = 0
total = 0
for i in range(10000):
player = np.random.choice(np.array([10,20,30]),1,replace=False)[0]
if player == 10: #per problem definition, we ignore all worlds where this happened
continue
if player == 30: #you picked the special goat, host will always reveal normal goat and not car
staywin += 1
if player == 20: #you picked car, host will reveal one of the two goats randomly
host = np.random.choice(np.array([10,30]),1,replace=False)[0]
if host == 30: #host picked special goat, we ignore worlds where this happened per problem definition
continue
else: #host picked normal goat
switchwin += 1
total += 1 #increment on possible worlds
print("Stay Win: {:.2f}%".format(staywin/total*100))
print("Switch win: {:.2f}%".format(switchwin/total*100))
>>
Stay Win: 66.87%
Switch win: 33.13%
Thanks. Of course then the trouble is to check if the code makes sense, but that seems easier than to check the math.
Here's a simpler implementation: https://pastecode.io/s/o5scz95i
In science, one shouldn't believe an experimental result for certain until it is confirmed by theory. One should also not believe a scientific theory until it is backed up by experiment.
In any case, testing it by experiment shows you the answer in this case, as Naremus demonstrated. But what do you learn, other than the answer?
I found it paradoxical that if you get at least 23 random people together that at least two will likely (>50%) share a birthday. When I delved into this to understand why, and understood its truth, I concluded that it is more likely, if one buys three easy-pick lottery tickets, it is more likely that one ticket will match another than for any of them to have the winning combination.
I don't disagree. But if you look at the other discussions, people do not just disagree about which is the right theory to reach the correct result, they also disagree about which is the correct result. And that just seems like a waste of brainpower, when it's so easy to check.
Also sometimes you just need the result for some practical purpose, for example to win the game described if you actually participate. This is less applicable here, since the entire point of a thought experiment is obviously to get you to think.
That's also the impression I got. Seems pretty straightforward to me. So I can't tell if I'm under-thinking it or if everyone else is over-thinking it.
If you see the good goat when the host opens the door there’s nothing playable. That rules out two options.
1) you’ve chosen the bad goat. Bad luck you can’t play. You can only switch to the car. You can’t chose an opened door.
2) you’ve chosen the car and the host shown the good goat. Bad luck. You can’t play.
You have two remaining workable chances. So there are only two playable options.
A) You have chosen the car and the host has shown the bad goat. Switching here will get you the good goat.
B) You have chosen the good goat. Switching here will get you the car.
So it’s 50-50 of all playable options.
There are two 'playable' options but they aren't equally likely to occur: having chosen the good goat initially is twice as likely as having picked the car initially, because half the worlds in which you picked the car and the host shows the good goat didn't happen per the problem definition, but could have happened because the host doesn't know not to pick the good goat.
Assuming the doors are random, and you pick door 1 (the door you pick doesn't change any probabilities), there are 6 possible configurations
G C B #+1/6
G B C #+1/6
B C G #X
B G C #X
C G B
C B G
The important bit, I hope you agree the last two worlds have a 1/3 chance of occurring (1/6 for each ordering) prior to the host revealing a door. Now, the last two break down (using [] to indicate the hosts random choice) into two further possible worlds based on the hosts random choice after your initial choice:
C [G] B #X
C G [B] #+1/12
and
C [B] G #+1/12
C B [G] #X
So adding up possible worlds that you could still be in, we get 1/3 you've picked the Good goat initially, and 1/6 that you picked the car and the host showed you the bad goat. Since having picked the good goat initially is more likely, we should stay.
You can do the same cheat than with the original problem, to make it more intuitive:
There are 1000 doors, with one car, one very special goat, and 998 not so special ones.
You choose a door and Monty opens 998, revealing 998 regular goats. Since Monty is working under the impression that you want the car, the probability that the car is in the door you didn't choose is 999/1000, so you should definitely not switch.
Monty doesn’t have enough information to open only doors with the rubbish goats, so you can’t generalise this problem in the same way as you could the original Monty test. In the original, Monty has to open only goat doors, in this case he doesn’t care about the valuable goat.
So in the original Monty is not going to show the car in any of the N -2 (998 here) doors he opens, in this case the valuable goat will appear most of the time in his opened doors.
Yes, Monty wouldn't care about the type of goat, and in most cases he would show it with the rest. But as in the 3 door problem, there just isn't a problem to solve if he shows you the good goat: you can't get the goat and just pick the (door most likely to have the) car. However, if he hasn't shown you the GOAT goat, it's more likely to be in the door you already picked.
This version of the problem seems to place us in a universe with six possible combinations, but the twist is that we actually find ourselves in one with only four possibilities: it is impossible that we initially chose the bad goat. In the original version, we don’t learn anything that changes the odds of switching; in this version, we don’t learn anything at all, although the game is phrased in a manner that leads us to believe that we have.
Play it this way: I’ll be Monty and start by opening the door with the bad goat and then let you pick another door. Now I’ll offer you the chance to switch. Rather obviously, you’ve learned nothing since your original choice and so the odds cannot have not changed.
The original version can be rephrased to make the correct choice obvious: I’ll let you choose one door, and then immediately give you the option of holding or switching for both other doors, one of which is certainly worthless. What fools us is thinking that being told precisely which door is worthless changes anything important.
In the classic formulation of the Monty Hall problem, there are three outcomes.
1. You initially pick car. Monty reveals a goat at random. If you stay you win, if you swap you lose.
2. You initially pick goat A. Monty reveals goat B. If you stay you lose, if you swap you win.
3. You initially pick goat B. Monty reveals goat A. If you stay you lose, if you swap you win.
So in 2 out of 3 equally likely scenarios you win if you swap.
In the revised version there are three different outcomes.
1. You initially pick good goat. Monty reveals bad goat. If you stay you win. If you swap you lose.
2. You initially pick bad goat. Monty reveals good goat. You cannot win.
3. You initially pick car. Monty reveals either good goat or bad goat with 50% probability. If he reveals good goat, you cannot win. If he reveals bad goat, you win if you swap and lose if you stay.
By revealing bad goat, Monty has inadvertently eliminated all of option 2's probability weight and half of option 3's weight. Those are worlds you could have been in but now know you are not. So you now know you are in either scenario 1 or scenario 3. But because scenario 3 has only a 50% chance of revealing the bad goat while scenario 1 has a 100% chance, you are twice as likely to be in scenario 1. If you stay you are therefore twice as likely to win compared to swapping.
What makes the Monty Hall problem so counterintuitive is that people tend to think about a probability as a property of the specific scenario. Eg. 3 doors 1 prize must always have the same probability. In actuality, a probability is a statement about *how much we know* about a scenario. In the original, Monty knows where the car is and will always reveal a goat. In the revised version, Monty does not know which goat is good and which is bad, and his actions reveal different information.
Of course the real best play is to try to win the car then offer to buy the good goat after the game is over.
Most concise analysis, so far. (in my book, anyway)
A) 1/3 chance you picked the correct goat and the host showed you the wrong goat. The situation where you picked the correct goat and the host showed you the car cannot occur based on the rules. In 100% of cases here you will see the wrong goat, so seeing the wrong goat gives you no information on whether you picked the correct goat or the car.
B) 1/3 chance you picked the car, but when the host shows you the wrong goat, you can eliminate the scenario where you picked the car and the host showed you the correct goat. This is 50/50.
C) 1/3 chance you picked the wrong goat but when you see the wrong goat you can eliminate these scenarios.
Of the scenarios not eliminated by assumption, in 2 you picked the correct goat, in one you picked the wrong goat.
He's more likely to have shown you the normal goat in the "you picked special goat" situation (100%) than in the "you picked car" situation (50%) or the "you picked normal goat" situation (0%), so you do get to update your probabilities: You're more likely in the scenario where you're more likely to have seen what you actually saw. (Unlike in the original problem, where you see "goat" 100% of the time.)
It's actually equivalent to the original problem: 1/3 odds your original pick was car, 2/3 odds your original pick was _the other goat_.
I think the easiest way to express what's going on is with the odds form of bayes theorem.
In the original monty hall problem, say we hypothesize that we picked the right door from the start. We believe this to be true with 1:2 odds (since there's 1 car and 2 goats).
Then, monty shows us a goat. If we picked the car, that'll happen 100% of the time, so we don't adjust the 1. If we picked a goat, that'll also happen 100% of the time, so we don't adjust the 2. Thus, after showing us a goat, we still believe that our initial pick is correct with 1:2 odds. This implies that the *alternative* is correct with 2:1 odds, so we should switch.
Then, consider the revised problem.
There's 3 cases: We picked the special goat, we picked the normal goat, or we picked the car, each with equal odds, so 1:1:1.
Then Monty reveals the normal goat.
Under the hypothesis that we picked the special goat, this is unsurprising. That's what *must* happen, so we don't adjust.
Under the hypothesis that we picked the normal goat, this is impossible, so the odds go to 0.
Under the hypothesis that we picked the car, this is *surprising*. We'd expect to be shown the normal goat only half the time (we'd see the special goat the other half), so our 1 gets multiplied by 1/2.
After the information, our odds are at 1:0:0.5. We can re-normalize to 2:0:1, meaning that we expect there's a 2/3rds chance that our door has the special goat and thus we should stay.
I think the easiest way to solve the revised Monty Hall problem is to use the solution to the traditional Monty Hall problem.
No matter what you desire in your heart of hearts, since Monty doesn't know about it, it will always be the case that switching has a 2/3 chance of winning you the car. Therefore not switching has a 2/3 chance of winning you whichever goat has not been seen yet.
In the situation described, the goat that hasn't been seen yet is the valuable one, and so you have a 2/3 chance of getting it by not switching.
No, wrong. Because in the original, the host can ALWAYS open another goat door. But in the new scenario, if you originally picked the bad goat, it is no longer possible for the host to open a bad goat door. In the new scenario (unlike the original), that possibility cannot happen.
> No, wrong.
Kindly was right. The new problem is trivial to solve in terms of the old problem. The host never seeks to open a bad goat door. All goats are alike to him.
But does the host EVER open a door with a good goat? The description says no, it didn't happen. But it ... "could have" happened? You have to be careful in probabilities, with assigning weights to possible worlds that you know did not occur. In what precise sense "might" Monty have opened a good goat door -- given that we know he didn't?
It matters a lot what space of games you're drawing from. We KNOW that, in the games we're considering, you CANNOT have originally chosen the bad goat door. This is different from the original Monty Hall problem, where originally choosing the "bad goat" WAS a possibility. So the possibilities (potentially) have changed. You have to be careful, solving it "in terms of the old problem". It isn't the old problem any more.
Are you trying to say something? If so, what is it?
Here's a possible repeated game: FIRST, Monty opens a door with a bad goat. THEN you choose from the remaining two doors. (One has a good goat, one has a car.) THEN Monty offers you the chance to switch. Do you agree, in this game, that the odds are 50/50, whether you switch or not? You could play it 1000 times, and you'll get the car 50%, and the good goat 50%, whether you switch or not.
You want to say that the game that we are talking about is different than the game in my previous paragraph. Different how, exactly? Here's one way it might be different: if you played the original Monty Hall problem, and your first choice was the bad goat, then the scenario described CANNOT happen. (Monty doesn't have a bad goat to show you.) So it must be clear that you cannot be playing the original game. The remaining question is: if you choose the car first, might Monty have shown you the good goat then? According to the puzzle, he didn't. But ... "could" he have? What does that actually mean? "Could" you have chosen the bad goat first? It seems, somehow, that you "couldn't" have chosen the bad goat ... yet Monty "could" have chosen the good goat (if you chose the car).
What exactly is the difference in analysis, between the "couldn't" (you first choosing the bad goat), and the "could" (Monty choosing the good goat when you choose the car first)? According to the scenario, neither one happened. Why is one still "possible", somehow, but the other isn't?
This only matters if you believe that the optimal strategy in the *original* Monty Hall problem depends on the quality of the goat behind the opened door.
To put it differently, suppose that we are in the scenario with the eccentric billionaire's goat, but you are philosophically opposed to helping billionaires retrieve their lost pets, and so both goats are equally worthless to you: you want the car. When Monty opens the door and you see the bad goat behind it, that's vaguely interesting but doesn't affect you in any way: by the standard Monty Hall strategy, you have a 2/3 chance of getting the car if you switch, and a 1/3 strategy of getting the (billionaire's pet) goat.
Similarly, there is a different world, in which Monty opens the door with the billionaire's pet goat behind it. Again, that's vaguely interesting but doesn't affect you in any way: by the standard Monty Hall strategy, you have a 2/3 chance of getting the car if you switch, and a 1/3 strategy of getting the (bat) goat.
The problem is that your last paragraph didn't happen, by assumption in the new scenario. So it matters a lot about why it didn't happen. It depends a bit on when you eliminate those possible worlds. If you do it before you analyze the scenario at all (originally choosing the bad goat can never be part of this new game), then it's only 50/50 for the remaining choices.
You're basically making an analysis based on a counterfactual that didn't happen (you choose the bad goat, and Monty shows the good goat). You're assigning it some weight, some probability that it "might have" happened -- even though we know it didn't.
That's where the disagreement is. With a single iteration, there aren't any probabilities at all. You either get the goat, or the car, 100%. Probabilities are about your state of knowledge, and they can only be verified by a repeated experiment where you set up some scenario over and over again, and then count the various outcomes.
What is the scenario that is being set up over and over again, in this case? If I run the example 1000 times, what are the 1000 examples? Does Monty show the good goat in some of those 1000 runs? Or not?
You need an "example generation machine", that can generate scenarios with outcomes you can count. This new problem is a little bit ambiguous about what the space of possible games is.
We're assuming that Monty doesn't know anything about the goats, so he is equally likely to choose either goat. Therefore the two scenarios are equally likely, and more importantly P(your door has car | Monty picks good goat) = P(your door has car | Monty picks bad goat).
I agree that if we don't assume this, then the answer to the entire problem is different. But I think it's unambiguous that we should assume it.
Yes, I think you're probably right.
The difference is that the host opens a GOAT door, chosen uniformly at random from the two goats. The uniform part is the active ingredient here. In the original show, the two goats were indistinguishable, but if the host opens a goat at random then the two are still indistinguishable TO THE HOST, although not to you.
The thing about the car is that the host never opens a car door (they know where the car is after all) which gives you the 2/3 probability of the prize you really want which is guaranteed not to be a goat in the original case.
The revised problem as stated (you've already seen a goat but not the one you want) has cut off the branch in the probability tree where you get the good goat, so you're conditioning on seeing the other one.
If you switch, you still get probability 2/3 of a car. Therefore if you don't switch, you get probability 2/3 of "a goat" - but in this case you know it's the goat you want.
We can just compute the answer with Bayes' theorem.
P(you picked good goat | bad goat revealed)
= P(bad goat revealed | you have good goat) P(you picked good goat) / P(bad goat revealed)
= 1 (1/3) / (1/2) = 2/3
On the Monty Hall one - good twist!
I've often thought of this modification:
You pick a door. Monty randomly opens a door you didn't pick (not worrying about whether it has a goat or a car). He happens to reveal a door with a goat. Now what should you do?
Before you saw what you revealed, there's a 1/3 chance your first pick was the car (and he definitely reveals a goat) and a 1/3 chance your first pick was a goat and he reveals a goat, and a 1/3 chance your first pick was a goat and he reveals the car.
We can eliminate the last one, now that we've seen the goat, so it's 50/50.
It's interesting that it's different again when he definitely chooses not to reveal the car, not knowing that one of the goats is the prize you want!
How do you feel about the original Monty Hall problem? If we apply the same chart but instead want the car instead of either G or B, then stay/switch is also 50/50, right?
The lesson I take is from both the original and this one is that probability is not about it's about information. Think of these as a Bayesian. (In fact, a straightforward application of Bayes Theorem really shows it here.)
In this case, when you get lucky enough to see the bad goat, that tells you that you probably picked the good goat. Worlds in which Monty reveals the bad goat are simply less common when the bad goat is not the one you chose.
Rrr, on that last line... when the *good* goat is not the one you chose.
This is what I was thinking (50/50 at the end), but I wrote a simulation that says you should only switch 1/3 of the time at the end: https://pastebin.com/YtvBWpc5
Not sure yet what the problem is.
Monty is randomly sampling the goat doors remaining, so revealing Smith gives you Bayesian information about the distribution of remaining goats.
There are three prizes:
Rosebud, the billionaire's pet goat
Smith, a regular goat
Herbie, a car
P(Monty shows Smith | you chose Herbie) = 50%, since there are two goat doors for Monty to choose between.
P(Monty shows Smith | you chose Rosebud) = 100%, since Smith is the only goat available for Monty to show.
P(Monty shows Smith | you chose Smith) = 0%, since Rosebud is the only goat available for Monty to show.
So Monty actually showing Smith should have you update your estimate of P(you chose Rosebud) from 33.3% to 66.7%.
Another way of working the problem that works out the same is to reduce it to the already-known solution to the standard Monty Hall problem, that there is a 2/3 chance that the car is behind the last door and a 1/3 chance the goat is there. Your door has a 2/3 chance of goat and a 1/3 chance of car. Since the still-hidden goat is Rosebud, you should stay with your door to maximize your chance of finding him.
Possible scenarios:
Your door is the good goat 1/3
Your door is the bad goat, monty shows you the good goat(he wont show you the car) and you lose immediately 1/3
Your door is the car 50/50 monty shows you the good goat 1/6
Your door is the car monty shows the bad goat. 1/6
The switch strategy has 1/6 winning. The keep strategy has 1/3 winning. 1/2 you lose immediately. Conditional on seeing the bad goat, you have 2/3 chance winning by not switching.
One of those is not playable, Monty showing you the good goat when you pick the car.
2 scenarios are not winnable. Both involve monty showing you the good goat. In those cases you switch to have a maximum chance to win the car.
This feels bizarre to me. Why would the hosts intentions matter? But I think your reasoning is correct.
When you see new information, it's not enough just to eliminate impossible options.
You also have to take into account, that the options that made the new evidence more likely, are themselves more likely than options that had a lesser chance of showing you that information.
Having the host open a goat door is more likely if you originally chose a car door, than if you chose a goat door. Which bumps the probability that you originally chose a car door from 1/3 to 1/2.
It's not precisely that the host's *intentions* matter, but rather that their *algorithm* matters. In order to understand the significance of the information the host reveals, you need to understand what *could* have been revealed. Since the host's actions depend on what door you originally picked, you can't calculate the counterfactual observations without knowing the host's decision algorithm.
For example, suppose the hosts acts like this:
- If your original pick was the car, the host reveals a goat and asks if you want to switch
- If your original pick was a goat, the host treats that as your final choice and does not give you the opportunity to switch
In this case, if the host reveals a goat, then there is 100% chance that you already picked the car, and a 0% chance that you will get the car if you switch.
You can make it even more extreme. Say that the host says, “I will open the lowest-number door that is not the one you pick, and not the one with the car”. If you pick 1 and the host opens 3, it’s obvious that 2 has the car. But if you pick 1 and the host opens 2, then you’re now 50/50.
> You also have to take into account, that the options that made the new evidence more likely, are themselves more likely than options that had a lesser chance of showing you that information.
That's true, but you're not doing it.
> Having the host open a goat door is more likely if you originally chose a car door, than if you chose a goat door.
That's false. By definition, the probability that the host opens a goat door is 100% in all cases.
??? The comment I responded to modifies the usual problem, to allow the host to open a car door if you didn't choose it.
I apologize. It's a long thread.
I just want to commend the OP for rephrasing to at least implicitly specify the host's decision algorithm, The host opens a door showing a goat "as is traditional", meaning always or nearly always. This rules out algorithms like "Host shows you a goat and offers a chance to switch if you picked the car, but doesn't open any door or give you a chance to switch, because cars are expensive and goats are cheap". Or, "Host decides what to do based on what he feels the audience would most enjoy, and since you're a telegenic and enthusiastic young woman he thinks they'd be happiest if he nudged you towards switching your choice to the car (which isn't being paid out of *his* salary)".
The original Monty Hall problem, what divided the intertubes back in the day, only specified that the host revealed a goat and offered a switch on this one occasion, and much of the dispute was based on differing assumptions as to the algorithm.
And, of course, this was when Monty Hall was still alive and could tell you what the algorithm was if you asked. It was not, "host always opens a door to reveal a goat".
Agreed, I've always felt the problem tells us more about how easy it is make, rely on, and be confused by unspoken assumptions, than it does about probability.
As Jeffrey's says in the Theory Of Probability:
"The most beneficial result that I can hope for as a consequence of this work is that more attention will be paid to the precise statement of the alternatives involved in the questions asked. It is sometimes considered a paradox that the answer depends not only on the observations but on the question: it should be a platitude."
Cool. I didnt realize "the monty hall problem" was not in fact a regularly occuring game on the show. Never watched it. History has been rewritten by this brain teaser. Everyone will think, as I did, that was the whole show
> The original Monty Hall problem, what divided the intertubes back in the day, only specified that the host revealed a goat and offered a switch on this one occasion, and much of the dispute was based on differing assumptions as to the algorithm.
This is just you not knowing what you're talking about. The problem is older than popular awareness of the internet. It divided the letter-writing public.
https://en.wikipedia.org/wiki/Marilyn_vos_Savant#Monty_Hall_problem
> Savant was asked the following question in her September 9, 1990, column:
> 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?
> This question is called the Monty Hall problem due to its resembling scenarios on the game show Let's Make a Deal, hosted by Monty Hall. It was a known logic problem before it was used in "Ask Marilyn".
> If the host merely selects a door at random, the question is likewise very different from the standard version. Savant addressed these issues by writing the following in Parade magazine, "the original answer defines certain conditions, the most significant of which is that the host always opens a losing door on purpose. Anything else is a different question."
And her second followup pointed out:
>> We've received thousands of letters, and of the people who performed the experiment by hand as described, the results are close to unanimous: you win twice as often when you change doors. Nearly 100% of those readers now believe it pays to switch. (One is an eighth-grade math teacher who, despite data clearly supporting the position, simply refuses to believe it!)
>> But many people tried performing similar experiments on computers, fearlessly programming them in hundreds of different ways. Not surprisingly, they fared a little less well. Even so, about 97% of them now believe it pays to switch.
>> And a very small percentage of readers feel convinced that the furor is resulting from people not realizing that the host is opening a losing door on purpose. (But they haven't read my mail! The great majority of people understand the conditions perfectly.)
( https://web.archive.org/web/20100310140547/http://www.marilynvossavant.com/articles/gameshow.html )
Isn't this version of the problem simpler than the original one? You've distinguished the goats, so now there are just 3 doors with 3 prizes. You've already been told that Monty opens a door to reveal a prize you're not interested in so you should stick to win 2/3 of the time. (Because, as in the classic Monty Hall problem, the car is behind the third door 2/3 of the time.)
Your point is absolutely right though - a lot of the confusion on these problems hinges on Monty's behaviour not being completely specified. I think in your scenario the key information is that Monty doesn't know/care where the car is, so in opening a door he's not giving you any information about where the car is or isn't. You can swap the order of events: Monty reveals a goat, you choose from the remaining doors, and have a 50/50 chance of winning. Conversely if Monty avoids the car, he reveals information and improves your chances.
> You've already been told that Monty opens a door to reveal a prize you're not interested in so you should stick to win 2/3 of the time.
No, the opposite. This is the same as the original problem, with no differences except in which prize you want. Monty opens a door to reveal a goat. What he reveals is a prize you don't want, but that's not why he opened the door.
The results are identical to the original problem, because it's the same problem. After Monty opens the one door, the final door has a car 2/3 of the time, and your door has a car 1/3 of the time. Since you don't want the car, you don't switch.
Bayes theorem is the easiest way to solve all the variants.
For your variant (door is opened randomly):
P(you pick car | goat revealed) = P(goat revealed | you pick car) P(you pick car) / P(goat revealed)
= 1 (1/3) / (2/3)
= 1/2
I don’t know if non-Anglo countries also believe this but the idea that being cold gives you a cold is very widely held here despite people also knowing that it’s transmitted by a virus. I know there’s some research about seasonality etc but a lot of people seem to go way beyond that
I can confirm that it's a common folk belief in India as well.
In a similar vein, do non-Americans believe that vitamin C helps with colds?
That's a fairly common belief here in the UK.
Also in Spain.
That's not entirely wrong though - you are more susceptible to diseases when you're cold. So I don't think that's equivalent.
…the cold water thing isn’t entirely wrong either though?- drinking the local tap water without boiling it in places where that belief is prevalent carries a risk of making you ill? It’s lumping the fan thing in there that seems unfair - that’s the belief that isn’t like the others.
This.
There are two stable equilibria. If everyone believes that drinking non-boiled tap water is safe, then you want regulations in place regarding the amount of bacteria in the water, so it is indeed safe.
If nobody thinks it is safe, then nobody cares and it will probably be somewhat unsafe by default.
Well, my understanding is that Chinese tap water contains a lot more heavy metal than you might hope, which boiling won't help with.
(Source: am Chinese, but don't have a scholarly understanding of this or anything) I do think it's more than just believing that the water needs to be boiled in order to be safe. Like if you boiled water, let it cool, and added ice to it, many Chinese people would probably think that it would hurt them to drink it. Like it will affect your digestion or fertility.
I think it's more likely that the belief is due to the fact that being cold can make your nose run.
I thought it was that being cold can put your immune system down a level making you more susceptible to the actual cause of the sickness.
But for me being cold definitely makes my nose run which feels like having a mild cold.
I suspect that the second thing you mentioned is the real origin of the belief that cold causes colds, and the first was something that people came up with afterward.
I think it's got a lot more to do with the fact that people get colds at the stat of winter. (Because the folk belief is true, at least in certain ways.)
What makes your nose run is the counter-current-heat-exchange-system that helps you avoid wasting lung water. The air you exhale is warm and full of water; in the nose it meets cold blood through very thin nose walls, cooling down and making water to gather into droplets. Some of those run out of your nose, as a side effect. The system works more strongly when the air (and the nose) are colder. The nose running when it's cold is not a disease, the nose is not producing more mucus. Heat exchange systems like this are common among mammals.
This in interesting but are you sure? Would the nose stop running (would the nose stop, haha) if you started breathing through your mouth? Or if you either inhaled or exhaled through your mouth consistently? If this is true, it would have to stop (maybe after drying the nostrils). I may test that if it gets cold again.
It's standard animal physiology. You can test if your nose only starts running outdoors, but not indoors (it must be actually warm indoors; some people keep 16 centigrade at home in winter, this can make the nose run). If it also runs at a warm temperature, you probably have an infection, allergy, or other irritation in the nose, producing extra mucus.
Good to know. I think you missed part of my point: the body doesn't have a reserve of cold blood, right? If the nose is cold, it's because the nose is cold. So within a couple minutes of coming indoors, the dehumidifier effect should stop. And give it a few minutes more for the nasal passages to drip a bit. Though to know how long this takes, you'd have to get a reference measurement: use some nasal spray and time how long you nose takes you dry.
Yeah, I don't know exactly how long it will keep dripping after coming indoors. Probably it's also slightly different person-by-person. Some have naturally colder noses even indoors, etc.
I don't know about the US, but in Europe people believe that it's ok to blow your nose when you have a cold. Instead of swallowing it or spitting it on the street. In medical terms, that's gross.
In societal terms it’s less gross
Spitting it in the street is *less* gross? What?
We find it gross because our parents have told us that it's gross (unless we do it on the soccer place, if you live in Europe).
Mucus on the street will not infect anyone. It does not find its way to the faces of people, and the street is a terrible place for germs to survive.
Asian people don't mind anyone spitting on the street, but they find the thought abhorrent to blow mucus into a handkerchief, put this into your pocket where it has perfect breeding conditions, touch it with your hand and then perhaps you even want to shake hands with them. From hygienic perspective, they are right that each of these steps is gross.
In the U.S., standard practice is to blow your nose into tissue paper made for that purpose and then throw it in the trash. And, ideally, washing your hands afterward.
I'll defer to LKY on this one:
https://youtu.be/xQbLo-Kt0W4?si=WmmsqhPRXWBowmH9
Not all mucus can be sucked from your nose into your mouth so you can spit it on the street (not gross at all :) )