Monty Hall, Erdos, and Our Limited Minds

http://www.wired.com/2014/11/monty-hall-erdos-limited-minds/

by laacz • 12 years ago

30 78 12 years ago

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.

simonh2• 12 years ago

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.

oneeyedpigeon2• 12 years ago

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

jonsen3• 12 years ago

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.

hamburglar2• 12 years ago

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.

hamburglar2• 12 years ago

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.

kybernetikos2• 12 years ago

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?

simonh2• 12 years ago

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.

throwaway2837193• 12 years ago

The Monty Hall problem is actually a bit more subtle than this article makes out. The exact wording of the problem can change the answer. For example, in this version

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.

tromp• 12 years ago

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".

randomstring• 12 years ago

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!

____a2• 12 years ago

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.

throwaway2837193• 12 years ago

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.

Sandman• 12 years ago

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.

jayvanguard10• 12 years ago

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.

Cushman7• 12 years ago

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.

dllthomas18• 12 years ago

Lots of quantum things are surely going on...

But really, it has to do with the information content of the actions of the host.

jayvanguard10• 12 years ago

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"?

ghusbands• 12 years ago

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.

jayvanguard10• 12 years ago

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).

Philadelphia• 12 years ago

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...

epidemian• 12 years ago

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:

  #!/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)

Sample output:

  Monty knows & keep choice: 0.33
  Monty knows & change choice: 0.67
  Monty ignores & keep choice: 0.50
  Monty ignores & change choice: 0.50

So, in the case of Monty picking at random, even if we ignore the cases where he picks the car, your chances of winning do not improve by changing your first choice or not.

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

The probability of winning is 1/3 (2/3 * 1/2).

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

Same tree, but with win and lose cases inverted. The probability of winning is also 1/3.

In conclusion, my intuition sucks. Thanks for showing me so!

jayvanguard10• 12 years ago

Nice write-up.

throwaway2837193• 12 years ago

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 ;)

GregBuchholz• 12 years ago

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.

nnq• 12 years ago

...am I the only person for whom the intuition about this problem has always yelled: "neither! choose whether to keep the choice or switch it with 50% probability - in other words, flip a coin." ?

dllthomas18• 12 years ago

That's a good way to make sure your odds are 50/50...

barbudorojo• 12 years ago

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.

mindvirus2• 12 years ago

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.

dllthomas18• 12 years ago

Could you elaborate? This seems less clear, to me.

mindvirus2• 12 years ago

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?

madsushi• 12 years ago

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!

ajvdarlington• 12 years ago

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.

jonsen3• 12 years ago

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.

Avernar• 12 years ago

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.

Monty Hall, Erdos, and Our Limited Minds

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