Monty Hall, Erdos, and Our Limited Minds(wired.com)
wired.com
Monty Hall, Erdos, and Our Limited Minds
http://www.wired.com/2014/11/monty-hall-erdos-limited-minds/
82 comments
This is exactly right. It drives me nuts that most explanations gloss over this.
If you're a mathematician you don't tend to introduce any additional assumptions unless they are stated. Assuming the host has knowledge of what is behind all of the doors (or at least where the car is) changes everything. Some wordings of the question include it, others don't.
Of course a non-mathematician might assume things about the game show host based on their knowledge of game show hosts but a mathematician is entirely correct to simply take the question at face value and not inject additional unstated assumptions. Otherwise what is preventing them from putting in other completely wacky assumptions like "the host wants you to win the car and knows you know about the monty hall problem and is using reverse psychology"?
If you're a mathematician you don't tend to introduce any additional assumptions unless they are stated. Assuming the host has knowledge of what is behind all of the doors (or at least where the car is) changes everything. Some wordings of the question include it, others don't.
Of course a non-mathematician might assume things about the game show host based on their knowledge of game show hosts but a mathematician is entirely correct to simply take the question at face value and not inject additional unstated assumptions. Otherwise what is preventing them from putting in other completely wacky assumptions like "the host wants you to win the car and knows you know about the monty hall problem and is using reverse psychology"?
And then there's this version
> There are two doors with goats and one with a car. You choose one door from the three. The host may select one of the doors with a goat from the remaining two doors, and open it. Should you switch doors if given the chance?
Now the probability can be anything from 0 to 1, depending on what makes the host tick...
Considering the usual ambiguity in the problem formulation, the only safe reply is "It depends".
> There are two doors with goats and one with a car. You choose one door from the three. The host may select one of the doors with a goat from the remaining two doors, and open it. Should you switch doors if given the chance?
Now the probability can be anything from 0 to 1, depending on what makes the host tick...
Considering the usual ambiguity in the problem formulation, the only safe reply is "It depends".
It's definitely fascinating! Though I feel like attributing this to the intentions of the host is misleading, because it implies something mystical or quantum might be going on. It's more straightforward to just say it depends on the actions of the host.
Specifically, the host in the first formulation never reveals a car, even if she wants to. The host in the second formulation reveals a car roughly 1/3 of the time, even if she'd rather not. Thus, intuitively, it makes sense that the former leaks information about the place of the car, and the latter doesn't. [0]
Of course that's just a rewording of your point, based on my sense that "knowledge and intentions" are concepts that apply readily to game show hosts, but awkwardly to rule-following systems in general.
[0] To the extent that revealing something can not leak information about it...
Edit: To elucidate my quibble further, as a third formulation: "The host chooses one of the two remaining doors at random, which, by coincidence, is always a goat." I assert we should always switch: The intention of the host is the same as in the second case, but the behavior of the system, magically known to us, is that of the first. We make decisions based on that.
Specifically, the host in the first formulation never reveals a car, even if she wants to. The host in the second formulation reveals a car roughly 1/3 of the time, even if she'd rather not. Thus, intuitively, it makes sense that the former leaks information about the place of the car, and the latter doesn't. [0]
Of course that's just a rewording of your point, based on my sense that "knowledge and intentions" are concepts that apply readily to game show hosts, but awkwardly to rule-following systems in general.
[0] To the extent that revealing something can not leak information about it...
Edit: To elucidate my quibble further, as a third formulation: "The host chooses one of the two remaining doors at random, which, by coincidence, is always a goat." I assert we should always switch: The intention of the host is the same as in the second case, but the behavior of the system, magically known to us, is that of the first. We make decisions based on that.
Lots of quantum things are surely going on...
But really, it has to do with the information content of the actions of the host.
But really, it has to do with the information content of the actions of the host.
Just speaking for me, I'm not confident there's a "really" about this, so I'm not too concerned with finding it. I'm just trying to find the easiest way for our co-commenters to understand why the correct answer is, in fact, correct :)
Right. My point was just that the actions of the host, as far as we observe, can be the same in both cases.
Sure, I see what you're saying. In a technical sense the information in the host's head does "matter". My comment is more along the lines that "information content in a system" is a sufficiently abstracted idea from "knowledge of a human being" that, especially if you equivocate them, it's not a useful metaphor for someone who doesn't already get why the right answer is right.
Yeah, it's not even strictly "does Monty know?" It's whether his action embodied that knowledge or not. Obviously (I hope) Monty knowing but still picking based on a coin flip is the same as Monty not knowing.
Which would explain why Erdos might be confounded by the answer. He was an expert at numbers, not psychology.
Proofs are built with an adversarial mind set. Always assuming the worst possible case and proving that a theorem holds in all conditions. The "hand of god" (or in this case the game show host) doesn't move to make theorems easier to prove. That isn't the universe works!
Proofs are built with an adversarial mind set. Always assuming the worst possible case and proving that a theorem holds in all conditions. The "hand of god" (or in this case the game show host) doesn't move to make theorems easier to prove. That isn't the universe works!
Hmm.. I don't think there's a difference. In two out of three cases, one of the remaining two doors contains a car. If the host opens the one that does, ok, bummer, you lost. But if not, there's a good chance that the other door contains a car. Remember that you always had only one in three chances of choosing the right door, this doesn't change.
There is a difference. Write a simulation in your favorite language. You'll be forced to decide whether the door with the car ever gets revealed to the contestant. In those cases does the contestant get to switch and win the car? Or did you just throw those cases out in your code? If so, you made an assumption that isn't aways stated clearly in the problem.
But your question was if the host chooses a door at random and reveals a goat. Which means it doesn't matter if he had knowledge or not. You told us the outcome. If you were writing a computer simulation there would be no rgn for the host picking his door. The host would always open the door with the goat, same as if he knew since you already told us he opens the door with the goat.
See my cousin comment[0] about this, I think you're getting tripped up with the idea of the host's "knowledge" mattering. That's a distraction here, what's important is the probabilities.
Two game shows, two hosts, both reveal a goat. The random host revealed a goat with 1/3 probability; the canonical host revealed a goat with 1 probability.
We know that both hosts revealed a goat. We also know (for the purposes of this problem) that they had different likelihoods of doing so before the fact.
Does it make intuitive sense to you that the prior probability of an event could affect how you react to it? That's all that's going on here.
[0] https://news.ycombinator.com/edit?id=8664771
Two game shows, two hosts, both reveal a goat. The random host revealed a goat with 1/3 probability; the canonical host revealed a goat with 1 probability.
We know that both hosts revealed a goat. We also know (for the purposes of this problem) that they had different likelihoods of doing so before the fact.
Does it make intuitive sense to you that the prior probability of an event could affect how you react to it? That's all that's going on here.
[0] https://news.ycombinator.com/edit?id=8664771
The premises of the riddle in both cases is that the host reveals a goat. There is no randomness. The fact that there was a chance the host might show a car doesn't matter. He showed a goat. The probability of the host showing a goat in both cases is 1.
I understand the math. I know what the op was trying to say. But they worded it wrong. The 2nd riddle is the same as the first.
I understand the math. I know what the op was trying to say. But they worded it wrong. The 2nd riddle is the same as the first.
There are two separate things here. First, the premise of the riddle. Second, the procedure of the game show (as fictionalized within the riddle).
If "... and shows you a goat" is a part of the procedure of the game show, there is additional information imparted by the host's selection compared to the situation where "... and shows you a goat" is an artifact of the particular play through the game that is being described in the riddle and the procedure by which that situation was arrived at did not incorporate knowledge of where the car was.
If "... and shows you a goat" is a part of the procedure of the game show, there is additional information imparted by the host's selection compared to the situation where "... and shows you a goat" is an artifact of the particular play through the game that is being described in the riddle and the procedure by which that situation was arrived at did not incorporate knowledge of where the car was.
The question is phrased ad-hoc. We know the host picked a goat in both scenarios. Decision time. Do we switch or no? It didn't ask: "On an infinite timeline where the host might not always pick the goat, should we switch and if so when?" Which I think is what op was trying to say.
It matters why the host picked the goat. (Or, more precisely, what the odds were of him doing otherwise when he picked). That we're looking at a goat doesn't tell us everything.
Again, write the simulation. Either the switch strategy will win 2/3 of the time or 1/2 of the time, depending on how you write Monty. Only one of those approaches matches the results we saw IRL - Monty never revealed the car (so far as I'm aware) - but either is consistent with many phrasings of the question, and OP correctly specified the other case above.
Again, write the simulation. Either the switch strategy will win 2/3 of the time or 1/2 of the time, depending on how you write Monty. Only one of those approaches matches the results we saw IRL - Monty never revealed the car (so far as I'm aware) - but either is consistent with many phrasings of the question, and OP correctly specified the other case above.
No. Even in the case of a single trial, it matters.
If the host picks randomly, and the host picks a goat, and you switch, your odds of a car are 1/3, just as if you'd stuck. There's no point.
If the host always chooses a goat, and the host picks a goat, and you switch, your odds are 2/3 in favor of now having a car. You should always switch.
Intuition suggests this. Statistics proves it. If you implement this in software, you will see it happening to you in black and white, right before your eyes.
If the host picks randomly, and the host picks a goat, and you switch, your odds of a car are 1/3, just as if you'd stuck. There's no point.
If the host always chooses a goat, and the host picks a goat, and you switch, your odds are 2/3 in favor of now having a car. You should always switch.
Intuition suggests this. Statistics proves it. If you implement this in software, you will see it happening to you in black and white, right before your eyes.
"If the host picks randomly, and the host picks a goat, and you switch, your odds of a car are 1/3, just as if you'd stuck. There's no point."
1/2, once you see the goat. One of 1/3, 2/3, and 1/2 beforehand, depending on what happens when Monty shows a car: whether you automatically win, automatically lose, or start over from the top, respectively.
1/2, once you see the goat. One of 1/3, 2/3, and 1/2 beforehand, depending on what happens when Monty shows a car: whether you automatically win, automatically lose, or start over from the top, respectively.
> The probability of the host showing a goat in both cases is 1.
Are you familiar with the concept of prior probability? The premise of the meta-riddle is that both hosts showed a goat, but one host could have not shown a goat. So there must be a difference in the information conveyed by that goat.
Again, there isn't anything special about this problem, it's just how probabilities work. Respectfully-- there are a number of different people trying to explain this to you in different ways, and you can verify it on Wikipedia. Are you positive you do understand the math?
Are you familiar with the concept of prior probability? The premise of the meta-riddle is that both hosts showed a goat, but one host could have not shown a goat. So there must be a difference in the information conveyed by that goat.
Again, there isn't anything special about this problem, it's just how probabilities work. Respectfully-- there are a number of different people trying to explain this to you in different ways, and you can verify it on Wikipedia. Are you positive you do understand the math?
[deleted]
He does in this single instance, but in order to calculate the probably you need to consider all the possible cases. Does the case where opens the door with the car calculate in? Why or why not? Seriously, write the simulation to calculate the probabilities. It makes the assumptions very obvious.
You don't need to because you're asking what is the probability of winning by switching once the goat has been revealed. We want to know:
(# of possible worlds in which we win after switching) / (all possible worlds in which we are shown a goat after the initial pick)
vs
(# of possible worlds in which we win without switching) / (all possible worlds in which we are shown a goat after the initial pick)
in this case 2/3 vs 1/3. We are only considering the cases where a goat has been shown so what happens when a goat isn't shown or what the host's intentions are when a goat is show is irrelevant.
(# of possible worlds in which we win after switching) / (all possible worlds in which we are shown a goat after the initial pick)
vs
(# of possible worlds in which we win without switching) / (all possible worlds in which we are shown a goat after the initial pick)
in this case 2/3 vs 1/3. We are only considering the cases where a goat has been shown so what happens when a goat isn't shown or what the host's intentions are when a goat is show is irrelevant.
But 1 in 3 times the host opens a door with a car. That changes the probabilities. Or are you assuming this never happens? Why?
Let's suppose we are doing a computer simulation.
1. We simulate a million trials where the contestant chooses one of three door, one of which has a car and two a goat
2. We simulate the host randomly choosing one of the remaining doors to open.
3. We discard all trials that resulted in a car. We are left with the number of trials that resulted in a goat and store that number in a variable 'total'
4. In the remaining trails we switch and reveal what was behind the door. We store the number of times we saw a car in a variable called 'wins'.
5. The probability of winning after switching is 'wins' / 'total'
It doesn't matter how many times the host shows a car because those trials are discarded.
1. We simulate a million trials where the contestant chooses one of three door, one of which has a car and two a goat
2. We simulate the host randomly choosing one of the remaining doors to open.
3. We discard all trials that resulted in a car. We are left with the number of trials that resulted in a goat and store that number in a variable 'total'
4. In the remaining trails we switch and reveal what was behind the door. We store the number of times we saw a car in a variable called 'wins'.
5. The probability of winning after switching is 'wins' / 'total'
It doesn't matter how many times the host shows a car because those trials are discarded.
Did you actually write that simulation? Because once up on a time (many years ago) I followed the same line of reasoning, wrote a simulation to show that I was right, and that simulation showed me I was wrong.
If the host picks randomly and you discard rounds with a car, your odds switching are 50/50. If the host uses knowledge of where the car is to definitely reveal a goat, your odds switching are better.
If the host picks randomly and you discard rounds with a car, your odds switching are 50/50. If the host uses knowledge of where the car is to definitely reveal a goat, your odds switching are better.
@dllthomas
I've written the simulation, and you're right. It is 50% when the host chooses randomly!
I've written the simulation, and you're right. It is 50% when the host chooses randomly!
You told us he opens the door with the goat. All cases consist of the host opening the door with a goat after you have picked.
But it matters whether you have the guarantee or not.
If the host picked at random and happened to get a goat, it is 1/2 vs 1/2. If the host picked a door he knew had a goat, it's 2/3 vs 1/3.
Again, write the simulation. It shouldn't take you 5 minutes.
If the host picked at random and happened to get a goat, it is 1/2 vs 1/2. If the host picked a door he knew had a goat, it's 2/3 vs 1/3.
Again, write the simulation. It shouldn't take you 5 minutes.
You do have a guarantee. The premises of #2 above is that the host shows a goat.
Yes, _this time_. But how on earth are you evaluating probabilities without considering all the priors and/or doing multiple trials? Is this some new kind of mathematics you've invented?
The hidden assumption is that the odds that he opens a door containing a car is zero. That is what changes the odds in the second step. Without that they don't change.
Seriously. Write the simulation.
The hidden assumption is that the odds that he opens a door containing a car is zero. That is what changes the odds in the second step. Without that they don't change.
Seriously. Write the simulation.
Read your second question again. It says that the host shows a goat. Then asks if you would switch. Yes, you would switch.
I understand what you were trying to ask but you worded the second question wrong. You can't tell the listener that the host picks a goat. That defeats the purpose of the "randomness" which is meaningless since we know the host picks a door with a goat.
I understand what you were trying to ask but you worded the second question wrong. You can't tell the listener that the host picks a goat. That defeats the purpose of the "randomness" which is meaningless since we know the host picks a door with a goat.
I found this on the internet and modified it so you can just change the commented out options to run it under the different scenarios.
car = wins = 0
many = 100000
actual = 0
many.times do
choice1 = rand(3)
car = rand(3)
#host_opts = [0, 1, 2] - [choice1, car] # host knows what is behind
host_opts = [0, 1, 2] - [choice1] # host doesn't know what is behind
#choice2 = [choice1] # don't switch
choice2 = [0, 1, 2] - [choice1, host_opts.first] # switch
if host_opts.first == car then
# Discard? Automatic win? It doesn't actually matter!
# It only changes the denominator.
else
# According to the puzzle, it is decision time!
wins += 1 if choice2.first == car
actual += 1
end
end
I believe it doesn't change the result, but this is technically Monty always picking "the lower (leftmost?) door you didn't pick" not "a random door you didn't pick", right?
Also, "it only changes the denominator" may be misleading, as it does not change the only denominator actually visible in the code. It changes the total numbers of wins and losses, but not the total numbers of wins and losses given that you saw a goat.
Also, "it only changes the denominator" may be misleading, as it does not change the only denominator actually visible in the code. It changes the total numbers of wins and losses, but not the total numbers of wins and losses given that you saw a goat.
> if host_opts.first == car then
That is false 100% of the time given the parameters that the outcome of the host's choice is a goat (as the question states)
That is false 100% of the time given the parameters that the outcome of the host's choice is a goat (as the question states)
WRITE THE SIMULATION
No doubt. See seriuslyguys post above. He did it and now agrees.
You didn't write the simulation.
On top of this, if you're at all unsure about the wording, you can apply the meta-solution that it's best to switch, because the chance of it giving you the car is always at least 1/2 and has a non-zero probability of being 2/3. That gives an expected cross-universe probability of more than 1/2.
This is true but it isn't really a mathematical question then with a single correct answer.
If I was in a culture where gameshow hosts behaved differently the question would have different priors and a different answer.
What if all gameshow hosts shot contestants immediately after they won? Would you be better of switching? No, you'd be dead, so you didn't "win" the car. Completely silly example but it shows why this can't be taken seriously as a mathematical question with probabilities without injecting additional assumptions (unless of course it is stated clearly but then I bet few mathematicians would get tripped up).
If I was in a culture where gameshow hosts behaved differently the question would have different priors and a different answer.
What if all gameshow hosts shot contestants immediately after they won? Would you be better of switching? No, you'd be dead, so you didn't "win" the car. Completely silly example but it shows why this can't be taken seriously as a mathematical question with probabilities without injecting additional assumptions (unless of course it is stated clearly but then I bet few mathematicians would get tripped up).
My favorite overview of this is a New York Times article that asked Monty Hall himself about the problem [1]. I think I first saw it linked from from his biographical page on Wikipedia.
[1] http://www.nytimes.com/1991/07/21/us/behind-monty-hall-s-doo...
[1] http://www.nytimes.com/1991/07/21/us/behind-monty-hall-s-doo...
My understanding is that in both situations as you described above you should switch. You are still in a new conditional state. Telling you that the host opened another door by chance and it is a goat removes the possibility that he opened the door with the car behind it by chance. You have eliminated one of the options and you have the 2/3 chance of winning by switching.
No. In the first case, the fact that the host shows you a goat has not put you in any new state of the world (the host was going to show you a goat regardless of what happened).
In the second case, you now know that you are not in one of the states of the world where the host accidentally opened a door that had a car behind it. They could have opened a door with a car, but they didn't. That gives you some information - it stands to reason that since they didn't open a door with a car, it was probably a bit more likely that they would pick a door with a goat, which makes it more likely that there are two goats than that there is one car and one goat, which is equivalent to saying that it's a bit more likely that you have already chosen the car.
In the second case, you now know that you are not in one of the states of the world where the host accidentally opened a door that had a car behind it. They could have opened a door with a car, but they didn't. That gives you some information - it stands to reason that since they didn't open a door with a car, it was probably a bit more likely that they would pick a door with a goat, which makes it more likely that there are two goats than that there is one car and one goat, which is equivalent to saying that it's a bit more likely that you have already chosen the car.
Okay, yes. Had to work it out on paper to convince myself that is works out to 1/2 in the second case. Thank you.
[deleted]
Wow. As many others here, i was sure there wouldn't be a difference if Monty picked at random AND we ignored the cases where he picks the door with the car. But after reading "write the damn simulation!" too many times, i wrote one, being sure it would prove me right.
Of course, i was wrong :D
The simulation testing the four possible scenarios:
After verifying that the simulation makes sense, i tried to figure out why it makes sense.
In an information theory kinda way, it makes sense that if Monty picks at random then he is not providing any new information: you are just being shown what was behind a random door you didn't choose and you can't choose now. But that explanation is not intuitive to my brain.
I found it more intuitive to actually draw a decision tree with the possibilities and their probabilities. The tree is actually very simple:
With the Always Change Door strategy:
With the Always Keep Door strategy:
In conclusion, my intuition sucks. Thanks for showing me so!
Of course, i was wrong :D
The simulation testing the four possible scenarios:
#!/usr/bin/env ruby
def run_games(monty_knows:, change_choice:)
options = [:goat1, :goat2, :car]
total_games = 0
wins = 0
100000.times do
choice = options.sample
monty_choice = (options - [choice, (:car if monty_knows)]).sample
# Don't consider cases where Monty shows a car.
next if monty_choice == :car
if change_choice
choice = (options - [monty_choice, choice]).first
end
total_games += 1
wins += 1 if choice == :car
end
wins / total_games.to_f
end
puts 'Monty knows & keep choice: %.2f' %
run_games(monty_knows: true, change_choice: false)
puts 'Monty knows & change choice: %.2f' %
run_games(monty_knows: true, change_choice: true)
puts 'Monty ignores & keep choice: %.2f' %
run_games(monty_knows: false, change_choice: false)
puts 'Monty ignores & change choice: %.2f' %
run_games(monty_knows: false, change_choice: true)
Monty knows & keep choice: 0.33
Monty knows & change choice: 0.67
Monty ignores & keep choice: 0.50
Monty ignores & change choice: 0.50
After verifying that the simulation makes sense, i tried to figure out why it makes sense.
In an information theory kinda way, it makes sense that if Monty picks at random then he is not providing any new information: you are just being shown what was behind a random door you didn't choose and you can't choose now. But that explanation is not intuitive to my brain.
I found it more intuitive to actually draw a decision tree with the possibilities and their probabilities. The tree is actually very simple:
With the Always Change Door strategy:
car (1/3) -> LOSE (Monty always reveals goat,
/ we change and we always lose)
first choice
\ goat (1/2) -> WIN (we change to the
\ / car door)
goat (2/3) -> Monty choice
\
car (1/2) -> IGNORE
With the Always Keep Door strategy:
car (1/3) -> WIN (doesn't matter what Monty chooses)
/
first choice
\ goat (1/2) -> LOSE (we keep our
\ / goat-hiding door)
goat (2/3) -> Monty choice
\
car (1/2) -> IGNORE
In conclusion, my intuition sucks. Thanks for showing me so!
Kudos for having enough doubt that your intuition was correct to actually write the simulation!
Self-doubt is a quantity often found wanting among smart people, computer programmers, and especially among the intersection ;)
Self-doubt is a quantity often found wanting among smart people, computer programmers, and especially among the intersection ;)
Nice write-up.
[deleted](1)
My favourite bit of this article is right at the end:
"Pigeons repeatedly exposed to the problem show that they rapidly learn always to switch, unlike humans (Herbranson and Schroeder, 2010)."
This just blows my mind and, as you might tell from my username, I'm a big fan of the bird that some denigrate as 'rats with wings'; I prefer to think of them as rat angels.
"Pigeons repeatedly exposed to the problem show that they rapidly learn always to switch, unlike humans (Herbranson and Schroeder, 2010)."
This just blows my mind and, as you might tell from my username, I'm a big fan of the bird that some denigrate as 'rats with wings'; I prefer to think of them as rat angels.
The Monty Hall problem is a real classic, I remember puzzling over it for hours and discussing it with family and friends for days after hearing of it.
It amazes me how many people believe that human beings are primarily logical beings.
The best way I've come up to explain the Monty Hall problem is that in the first place you have a 2/3 chance of choosing a Goat and in that case switching guarantees you will win the car. Alternatively you have a 1/3 chance you chose the car to start with and switching guarantees you will win a goat. Therefore a strategy of always switching gives you a 2/3 chance of winning the car.
It amazes me how many people believe that human beings are primarily logical beings.
The best way I've come up to explain the Monty Hall problem is that in the first place you have a 2/3 chance of choosing a Goat and in that case switching guarantees you will win the car. Alternatively you have a 1/3 chance you chose the car to start with and switching guarantees you will win a goat. Therefore a strategy of always switching gives you a 2/3 chance of winning the car.
An interesting thing to realize about the problem is that there actually is a way to make your odds the "obvious" 50/50, and that's to flip a coin to decide whether to switch doors at the end. The probability of a win on the two doors isn't equal, but the probability on the first door is 1/3 and the probability on the door you can switch to is 2/3, so using a coin toss to decide whether to switch would result in a 50% win rate.
This is because the way the problem is structured gives you six choices, not three: [A, A], [A, switch], [B, B], [B, switch], [C, C], [C, switch]. 3 of these are wins and three of these are losses. If you choose from these six options randomly you will win 50% of the time, but if you only choose from the "switch" options you will win 2/3 of the time.
This is because the way the problem is structured gives you six choices, not three: [A, A], [A, switch], [B, B], [B, switch], [C, C], [C, switch]. 3 of these are wins and three of these are losses. If you choose from these six options randomly you will win 50% of the time, but if you only choose from the "switch" options you will win 2/3 of the time.
Wow - I think that's the most straightforward explanation I've ever heard. I think, to rephrase it, you're saying:
a) because a goat is revealed, swapping will always give you the opposite of what you had
b) you're more likely to pick a goat in the first place, so you're more likely to switch to a car
a) because a goat is revealed, swapping will always give you the opposite of what you had
b) you're more likely to pick a goat in the first place, so you're more likely to switch to a car
The notion of revealing the goat before you make your second choice is a bit of a distraction. It becomes a lot more clear if you realize that in your first choice, you're dividing the doors into two groups: the group you choose has a 1/N chance of having a car in it, and the group you didn't choose has a (N-1)/N chance of having a car in it. Now, imagine that without opening any doors, Monty said: if you're willing to switch to the N-1 group, you win if there's a car in any of them. Do you take it? Of course.
Yeah this is it I think. If I can restate more simply for the 3-door case:
You pick. The doors are now in two groups: your door (1/3 chance) and the other two doors (2/3 chance).
The host picks. The second group is now reduced to a single door, still with 2/3 chance.
Now you decide - stick with the door with 1/3 chance, or switch to the door with 2/3 chance?
You pick. The doors are now in two groups: your door (1/3 chance) and the other two doors (2/3 chance).
The host picks. The second group is now reduced to a single door, still with 2/3 chance.
Now you decide - stick with the door with 1/3 chance, or switch to the door with 2/3 chance?
The most straightforward explanation I've heard is this:
The chance is 1/3 the car is behind the door chosen at first. The chance is 0 for the opened door. As the three chances must sum to 1, the last door has a chance of 2/3.
The chance is 1/3 the car is behind the door chosen at first. The chance is 0 for the opened door. As the three chances must sum to 1, the last door has a chance of 2/3.
That's great for defusing the "there are two options and it's random so it must be 50/50" initial of some, but it doesn't quite work for the "Monty chooses randomly, and happens to have revealed a goat" situation. I think the next point is that, when the host picks randomly, it is then correspondingly less likely that you've found yourself in that situation.
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The best way of explaining it that I've found is to describe it as a gameshow where there are 100 doors, and you choose one door, then the host (who knows what's behind each of the doors) opens 98 doors with goats behind them. Do you stay with the door you started with, or switch to the remaining one?
That sort of hints at the answer, but it's not watertight. Wouldn't a more close analogy be if the host opened one of the other 99 doors? In which case it's still not obvious. After all he only opened one in the orriginal version of the puzzle. Not everyone will necessarily agree that your new version is similar enough to be directly applicable to the original.
The analogy is not chosen merely for being similar, but to fulfill a purpose. It's based on the idea that in the original puzzle what you are faced with is a choice between your original door, or all the other doors. That's why you have a 1/3 or 2/3 chance of being right if you stick or switch respectively - because you are either choosing one door, or effectively two doors. The host showing you whats behind a door is just an implementation detail that causes you to have a stick or switch decision that has that effect. That's why a nice analogy is one that gives you a choice between 1 door or 99 doors, i.e. a chance of 1/100 or 99/100 of winning.
Your variant is a valid analogy too, but the probabilities in your scenario between sticking and switching are so close to each other, it's not going to trigger any new intuition in someone who is already confused.
Don't think of the 100 door puzzle as an analogy with the 3 door puzzle, think of it as a class of puzzle where there are N doors, you can pick one, and then the host (who knows what's behind the doors) reveals N-2 fails. It's clear that the 3 door puzzle is a specific instance of that, and if you have understood something intuitively about that puzzle class through considering N being a large number, that will still hold for N being a small number.
The class of puzzles that you describe is M doors, where you pick one and then the host (who knows what's behind the doors) reveals one fail. The three door puzzle is also a specific instance of that puzzle, but I doubt that you've learned anything from that, since examining M at other numbers probably hasn't given you any insight (still, if it did, those insights are indeed applicable to 3 door Monty Hall).
I've tried a number of ways of explaining this problem to people, and empirically the 100 doors one is the way that seems to work best. Interested to hear any versions you tell that seem to help people understand whats going on.
Your variant is a valid analogy too, but the probabilities in your scenario between sticking and switching are so close to each other, it's not going to trigger any new intuition in someone who is already confused.
Don't think of the 100 door puzzle as an analogy with the 3 door puzzle, think of it as a class of puzzle where there are N doors, you can pick one, and then the host (who knows what's behind the doors) reveals N-2 fails. It's clear that the 3 door puzzle is a specific instance of that, and if you have understood something intuitively about that puzzle class through considering N being a large number, that will still hold for N being a small number.
The class of puzzles that you describe is M doors, where you pick one and then the host (who knows what's behind the doors) reveals one fail. The three door puzzle is also a specific instance of that puzzle, but I doubt that you've learned anything from that, since examining M at other numbers probably hasn't given you any insight (still, if it did, those insights are indeed applicable to 3 door Monty Hall).
I've tried a number of ways of explaining this problem to people, and empirically the 100 doors one is the way that seems to work best. Interested to hear any versions you tell that seem to help people understand whats going on.
I've been wondering about another somewhat intuitive way to see the problem: suppose a third person enters the room and he is allowed to finish the game, knowing nothing of what went on before. In particular, he doesn't know your initial choice. Now it is obvious that his chances of winning are 1/2. It is also obvious that if you stick to your first choice, your chance of winning is still 1/3. Now it is odd that somebody who has LESS information about the problem has a better chance of winning than you. Meaning that you must have some way of getting at least the odds of the other guy. But there's only one thing you can do other than sticking to your first choice: taking the other door. So that seems the correct thing to do.
More intuitive explanation.
Let's play a similar game with three doors, two goats and a car. But you get to pick one of two rules to play by:
A) You get to pick one door you want and get what's behind that door.
B) You get to pick one door you don't want and get the better prize behind the other two doors.
Pretty much everyone is going to pick rule B. So after picking the door you don't want I will open the other doors. If the car is behind one of those doors I will always open that door last. Does the last sentence make you wish you picked rule A to play by? Heck no. We already know that one of those two will be a goat. The order I open the doors won't change anything.
Even if I let you retroactively switch to rule A after opening the first door? This is what trips people up. They think they're getting new information and that their original door choice may be incorrect.
But in this version of the game it's easy to see that we're not changing our choice of door, we're changing from the better game rule B to the worse rule A.
So the correct choice is to always stick with rule B.
Now we map the rules onto the real Monty Hall game. Rule A is pick the door you want and stick with it. Rule B is pick the door you don't want and switch.
So the correct choice is to always switch.
Let's play a similar game with three doors, two goats and a car. But you get to pick one of two rules to play by:
A) You get to pick one door you want and get what's behind that door.
B) You get to pick one door you don't want and get the better prize behind the other two doors.
Pretty much everyone is going to pick rule B. So after picking the door you don't want I will open the other doors. If the car is behind one of those doors I will always open that door last. Does the last sentence make you wish you picked rule A to play by? Heck no. We already know that one of those two will be a goat. The order I open the doors won't change anything.
Even if I let you retroactively switch to rule A after opening the first door? This is what trips people up. They think they're getting new information and that their original door choice may be incorrect.
But in this version of the game it's easy to see that we're not changing our choice of door, we're changing from the better game rule B to the worse rule A.
So the correct choice is to always stick with rule B.
Now we map the rules onto the real Monty Hall game. Rule A is pick the door you want and stick with it. Rule B is pick the door you don't want and switch.
So the correct choice is to always switch.
My criticism of the problem is that the presentation is usually poor, and it isn't always noted that the host knows what door the prize is behind, and always opens a goat door. I wonder what the responses would be like if it was presented as examples, in the same manner as was presented to the pigeons.
The way that I've tried to explain/understand the Monty Hall problem is to expand it to 100 doors.
You're allowed to pick 1 door out of 100. Your chance of being correct is 1%. Then, the host gives you the option of swapping to the entire other set of 99 doors. Of course, your chance of hitting the car would be 99%, so you swap.
Opening 98 doors with goats behind them is expected, and so just because the host chooses to open 98 goats in a row doesn't mean that the chance of the last 2 doors (your final door out of the 99, and your original 1 door) both have an equal chance. Your set of 99 doors was opened in a specific order and that's what actually matters. The 99th door still has a 99% chance to contain the car, because your set had a 99% chance to contain the car. The order is everything!
You're allowed to pick 1 door out of 100. Your chance of being correct is 1%. Then, the host gives you the option of swapping to the entire other set of 99 doors. Of course, your chance of hitting the car would be 99%, so you swap.
Opening 98 doors with goats behind them is expected, and so just because the host chooses to open 98 goats in a row doesn't mean that the chance of the last 2 doors (your final door out of the 99, and your original 1 door) both have an equal chance. Your set of 99 doors was opened in a specific order and that's what actually matters. The 99th door still has a 99% chance to contain the car, because your set had a 99% chance to contain the car. The order is everything!
Man the number of people suggesting it matters whether the host knows or not makes my head hurt. It doesn't. The problem is equivalent to "given that the host reveals a goat and you swap doors, what is the probability you get the prize". This will be 2/3. Whether the host knows or not is irrelevant.
There are only 2 possible paths in the problem: Path 1: you originally picked a goat (p = 2/3), now you swap and get the prize
Path 2: you orinally picked the prize (p = 1/3), now you swap and get nothing
This depends in no way on the host having knowledge.
There are only 2 possible paths in the problem: Path 1: you originally picked a goat (p = 2/3), now you swap and get the prize
Path 2: you orinally picked the prize (p = 1/3), now you swap and get nothing
This depends in no way on the host having knowledge.
If you imagine the same game but with a million doors and in the end there are only two doors, then you realize that there is a strong asymmetry and that the initial door is special. Going to the limit, with infinite doors to choose, the best thing you can do is choose the initial door, all the other doors are just to confuse you.
Another way to think about it: Imagine instead of opening the door, Monty points to a door and says that the prize is either behind this door, or the one you chose, and the question is, who is right? This is equivalent, but seems easier to think about. Also imagine this with 5 doors - then it seems much more clear.
Could you elaborate? This seems less clear, to me.
The variant works like this: A player picks a door, and Monty picks a door, and whoever gets the right door gets the prize. Monty knows what is behind every door, so the prize is either behind the player's door or Monty's.
So now, suppose you get to bet on who will win.
Its clear that the chance of the player winning is 1/3. Thus the chance of Monty winning is 2/3. So bet on Monty.
To change it back to the original problem, you are also the player in the game. So now, do you bet on yourself or on Monty?
So now, suppose you get to bet on who will win.
Its clear that the chance of the player winning is 1/3. Thus the chance of Monty winning is 2/3. So bet on Monty.
To change it back to the original problem, you are also the player in the game. So now, do you bet on yourself or on Monty?
This thread needs an argument for the case where the host picks a door at random.
The chance for the host getting the car is of course also 1/3. Then the complement 2/3 chance of the host getting a goat splits the car chance evenly between the two remaining doors as 1/3 for each door.
The chance for the host getting the car is of course also 1/3. Then the complement 2/3 chance of the host getting a goat splits the car chance evenly between the two remaining doors as 1/3 for each door.
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> There are two doors with goats and one with a car. You choose one door from the three. The host selects one of the doors with a goat from the remaining two doors, and opens it. Should you switch doors if given the chance?
has a different answer than this one
> There are two doors with goats and one with a car. You choose one door from the three. The host chooses one of the remaining two doors at random and opens it, showing a goat. Should you switch doors if given the chance?
In the first example the hosts choice is either forced (if you chose a goat initially) or doesn't make a difference (if you chose a car).
In the second example, the host could have opened a door with a car because he was choosing the doors at random. Given that he didn't, your estimate of the probability that you have chosen the car should increase - in fact, it should increase to 1/2, and switching doesn't make any difference.
To me, this is the really interesting thing about the Monty Hall problem - the knowledge and intentions of the host are important in the correct solution to the problem! So there are two ways to come up with the answer "it doesn't matter if you switch" -
1. You understand the problem in the first sense above, and your intuition leads you to an incorrect answer.
2. You understand the problem in the second sense above, and you are correct that it doesn't matter whether you switch or not.