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Monty Hall, Erdos, and Our Limited Minds
- oneeyedpigeon 12y agoMy 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.
- simonh 12y agoThe 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.
- oneeyedpigeon 12y agoWow - 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
- deleted 12y ago[deleted]
- jonsen 12y agoThe 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.
- dllthomas 12y agoThat'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.
- hamburglar 12y agoThe 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.
- lukasb 12y agoYeah 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?
- hamburglar 12y agoAn 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.
- kybernetikos 12y agoThe 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?
- simonh 12y agoThat 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.
- kybernetikos 12y agoThe 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.
- throwaway283719 12y agoThe 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 12y agoAnd 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 12y agoWhich 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!
- GregBuchholz 12y agoMy 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.
- deleted 12y ago[deleted]
- nnq 12y 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." ?
- dllthomas 12y agoThat's a good way to make sure your odds are 50/50...
- barbudorojo 12y agoIf 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.
- mindvirus 12y agoAnother 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.
- dllthomas 12y agoCould you elaborate? This seems less clear, to me.
- mindvirus 12y agoThe 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 12y agoThe 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 12y agoMan 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.
- jonsen 12y agoThis 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 12y agoMore 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.