As a mathematician, I've long followed the MH problem. I basically used the exhaustive approach when explaining it to others, but your "grouping" approach is far better--a perfect conceptual shortcut. Thanks!
I think it's relevant that you're not trying to convince a naive person who's a total blank slate with no idea what the answer might be — good luck finding someone like that! — but rather, you want to convince someone who has good reason to think the answer is other than it is.
So while it's definitely helpful to have multiple explanations for why the right answer is right, I think you'll have more success if you also explain why the intuition goes awry, so that the person can successfully set their intuition aside and be open to a different answer.
To that end, I'd suggest something like:
> The obvious answer is "It doesn't matter": each door originally had equal probability of hiding the car, and Monty Hall hasn't moved the car, so they still have equal probability.
> But by that argument, you could also say that it's just as good to switch to the door where Monty Hall just revealed a goat! After all, he didn't change whether it had hidden a car, so surely it still has a one-third chance of hiding a car, even though you can plainly see that it doesn't?
> The resolution is to see that the car never literally had a one-third chance of being behind each door — it was behind a specific door, so that door had a 100% chance and the others had a 0% chance — it's just that you had no information about which door it was behind, so from your perspective it was *as if* each door had the same chance. If you played this game many times, your first guess would be right about one-third of the time.
> Monty Hall has now changed the situation: he's given you some information (that such-and-such door hid a goat), and crucially, he decided what information to give you *based on what door you had chosen*. By choosing one door, you guaranteed that Monty Hall wouldn't open it, so you affected what information he could give you. If you chose a door with a goat, then you forced him to show you the other door with a goat, whereas if you chose a door with a car, then you let him freely choose which goat to show you. So he's more likely to show you a given goat if you chose the door with the other goat than if you chose the door with the car; and now he's showing you a goat, so working backwards, this means it's more likely that you picked the other goat than that you picked the car.
> (Of course, the car is still behind one specific door, so it still technically has a 100% probability of being behind that door and a 0% probability of being behind the other. But given the information you now have, it's *as if* it has a one-third chance of being behind one door and a two-thirds chance of being behind the other. If you play this game many times using this strategy, you'll win about two-thirds of the cars.)
Once you understand that ” If you played this game many times, your first guess would be right about one-third of the time ”, which means it would be false two-third of the time, it’s easy to understand you had better always change your guess so that it would be right two-third of the time!
I tend to use the proof by exhaustion method to convince people, since they often struggle with calculations of probabilities and maths (as do I).
I use the analogy of a deck of cards. I ask them to pick one [and hold on to it], and ask what is the chance that it was the ace of spades. They generally understand the probability is 1 in 52 that it is the ace of spades, and you can get them to accept those odds don't change when you flip 50 of the remaining 51 and none of those is the ace either.
To start, contestant has a 1 in 3 chance of having picked the right door. Three doors, one prize. If that doesn't resonate with the student, it's unlikely that anything that follows will.
Often a useful heuristic is to restate the beginning condition. To start, the contestant has a 2 in 3 change of having picked the wrong door, which means that the two doors he didn't pick have a 2 in 3 chance of being the right door.
The host reveals that one of the doors does not contain the prize. This is not a surprise, because we knew from the start that only one of the two could contain the prize. What is new is that we now know that the remaining door is the sole possessor of the 2 in 3 chance,
Nothing else has changed except for our information about the two doors we didn't pick. With that information we can double our odds from 1 in 3 to 2 in 3 by switching.
There's a psychological barrier to do this, which is loss aversion bias. Would I be happier to have swapped the losing door for the winning door than I would be sadder than swapping the winning door for the losing door? Fear of regret is a powerful barrier to rationality.
I worked out what I believe is a succinct and convincing explanation for my entry on the Monty Hall Problem for the Oxford Dictionary of Psychology. Here it is: There is one chance in three that the car is behind your originally chosen door, and accepting the invitation to switch wins the car if your original choice was wrong. The probability that your original choice was wrong is 2/3, and if it was wrong, then you are certain to win the car if you switch.
The initial player choice imposes a limit on Monty's choice of door in 2 out 3 plays. The information conferred by Monty's constraint will inform a players second choice in 2 out of 3 plays.
Thanks. Reframing as groups really works for me. Only one door has a car behind it. Choose to open just one door, or choose to open BOTH of the other doors?
It mysteriously becomes a no-brainer when previously it seemed baffling.
Great post. I'm happy enough with the proof by exhaustion, but have always yearned for a more intuitive explanation (I gather I'm in exalted company with Paul Erdös here), and your proofs 3&4 help here. Do you have any similar wisdom to offer on the Sleeping Beauty problem? Thirder or halfer?
While I'm here, I very much enjoyed "Proof", but this sentence on spherical triangles in Chapter 2 has been really bugging me: “Eventually we’ll end up with a triangle that has one right angle (at the North Pole), and two 45° angles”. Surely we'll never end up there? Isn't this a geometrical equivalent of Cauchy's "this last is called the limit of all the others” which Weierstrauss took exception to, and which you'd just been discussing?
Well, I have for many years thought I needed to learn a much more rigorous approach to probability and statistics, but this argument, and everyone who supports it, have convinced me that it would be a complete waste of time. In addition, where our climate is heading, a goat might be a greater asset than a car.
How do I know that Monty always reveals a goat after my door choice? Maybe Monty's strategy is a) if I choose a goat door go shucks you chose a goat b) if I choose the car door to show me a goat and hope I switch?
Hi, Adam,
As a mathematician, I've long followed the MH problem. I basically used the exhaustive approach when explaining it to others, but your "grouping" approach is far better--a perfect conceptual shortcut. Thanks!
I think it's relevant that you're not trying to convince a naive person who's a total blank slate with no idea what the answer might be — good luck finding someone like that! — but rather, you want to convince someone who has good reason to think the answer is other than it is.
So while it's definitely helpful to have multiple explanations for why the right answer is right, I think you'll have more success if you also explain why the intuition goes awry, so that the person can successfully set their intuition aside and be open to a different answer.
To that end, I'd suggest something like:
> The obvious answer is "It doesn't matter": each door originally had equal probability of hiding the car, and Monty Hall hasn't moved the car, so they still have equal probability.
> But by that argument, you could also say that it's just as good to switch to the door where Monty Hall just revealed a goat! After all, he didn't change whether it had hidden a car, so surely it still has a one-third chance of hiding a car, even though you can plainly see that it doesn't?
> The resolution is to see that the car never literally had a one-third chance of being behind each door — it was behind a specific door, so that door had a 100% chance and the others had a 0% chance — it's just that you had no information about which door it was behind, so from your perspective it was *as if* each door had the same chance. If you played this game many times, your first guess would be right about one-third of the time.
> Monty Hall has now changed the situation: he's given you some information (that such-and-such door hid a goat), and crucially, he decided what information to give you *based on what door you had chosen*. By choosing one door, you guaranteed that Monty Hall wouldn't open it, so you affected what information he could give you. If you chose a door with a goat, then you forced him to show you the other door with a goat, whereas if you chose a door with a car, then you let him freely choose which goat to show you. So he's more likely to show you a given goat if you chose the door with the other goat than if you chose the door with the car; and now he's showing you a goat, so working backwards, this means it's more likely that you picked the other goat than that you picked the car.
> (Of course, the car is still behind one specific door, so it still technically has a 100% probability of being behind that door and a 0% probability of being behind the other. But given the information you now have, it's *as if* it has a one-third chance of being behind one door and a two-thirds chance of being behind the other. If you play this game many times using this strategy, you'll win about two-thirds of the cars.)
Once you understand that ” If you played this game many times, your first guess would be right about one-third of the time ”, which means it would be false two-third of the time, it’s easy to understand you had better always change your guess so that it would be right two-third of the time!
What happens if, like me, you’d prefer to win a goat?
I tend to use the proof by exhaustion method to convince people, since they often struggle with calculations of probabilities and maths (as do I).
I use the analogy of a deck of cards. I ask them to pick one [and hold on to it], and ask what is the chance that it was the ace of spades. They generally understand the probability is 1 in 52 that it is the ace of spades, and you can get them to accept those odds don't change when you flip 50 of the remaining 51 and none of those is the ace either.
To start, contestant has a 1 in 3 chance of having picked the right door. Three doors, one prize. If that doesn't resonate with the student, it's unlikely that anything that follows will.
Often a useful heuristic is to restate the beginning condition. To start, the contestant has a 2 in 3 change of having picked the wrong door, which means that the two doors he didn't pick have a 2 in 3 chance of being the right door.
The host reveals that one of the doors does not contain the prize. This is not a surprise, because we knew from the start that only one of the two could contain the prize. What is new is that we now know that the remaining door is the sole possessor of the 2 in 3 chance,
Nothing else has changed except for our information about the two doors we didn't pick. With that information we can double our odds from 1 in 3 to 2 in 3 by switching.
There's a psychological barrier to do this, which is loss aversion bias. Would I be happier to have swapped the losing door for the winning door than I would be sadder than swapping the winning door for the losing door? Fear of regret is a powerful barrier to rationality.
I worked out what I believe is a succinct and convincing explanation for my entry on the Monty Hall Problem for the Oxford Dictionary of Psychology. Here it is: There is one chance in three that the car is behind your originally chosen door, and accepting the invitation to switch wins the car if your original choice was wrong. The probability that your original choice was wrong is 2/3, and if it was wrong, then you are certain to win the car if you switch.
The initial player choice imposes a limit on Monty's choice of door in 2 out 3 plays. The information conferred by Monty's constraint will inform a players second choice in 2 out of 3 plays.
Thanks. Reframing as groups really works for me. Only one door has a car behind it. Choose to open just one door, or choose to open BOTH of the other doors?
It mysteriously becomes a no-brainer when previously it seemed baffling.
Great post. I'm happy enough with the proof by exhaustion, but have always yearned for a more intuitive explanation (I gather I'm in exalted company with Paul Erdös here), and your proofs 3&4 help here. Do you have any similar wisdom to offer on the Sleeping Beauty problem? Thirder or halfer?
While I'm here, I very much enjoyed "Proof", but this sentence on spherical triangles in Chapter 2 has been really bugging me: “Eventually we’ll end up with a triangle that has one right angle (at the North Pole), and two 45° angles”. Surely we'll never end up there? Isn't this a geometrical equivalent of Cauchy's "this last is called the limit of all the others” which Weierstrauss took exception to, and which you'd just been discussing?
Well, I have for many years thought I needed to learn a much more rigorous approach to probability and statistics, but this argument, and everyone who supports it, have convinced me that it would be a complete waste of time. In addition, where our climate is heading, a goat might be a greater asset than a car.
How do I know that Monty always reveals a goat after my door choice? Maybe Monty's strategy is a) if I choose a goat door go shucks you chose a goat b) if I choose the car door to show me a goat and hope I switch?