The thing that is often de-emphasised in the presentation of the problem, in order to make it seem more mysteriously paradoxical, is that the presenter knows where the car is and this knowledge is always used perfectly. If the question always ended with "remember: Monty knows where the car is and will use this information", it would be more obvious.
Imagine a universe with many simultaneous Monty Hall clones playing at once in many studios, where Monty doesn't know and opens another door at random. If that door has the car behind it, Monty and contestant are both shot in the head and the studio burned down and erased from all records. This bloody culling of branches of the probabilities is the same as effected by giving Monty the knowledge and telling him to act on it.
> The thing that is often de-emphasised in the presentation of the problem, in order to make it seem more mysteriously paradoxical, is that the presenter knows where the car is and this knowledge is always used perfectly. If the question always ended with "remember: Monty knows where the car is and will use this information", it would be more obvious.
The associated line of reasoning resolved the paradox for me. If I stick with my original choice, it is as if I ignored the new information. If I switch the choice, I react to the new information.
The extreme of this is to pick among countably infinite doors, having the presenter open countably infinite doors and leaving just yours and another closed. Who could reasonable suggest that the chance is still 50/50, assuming you don't flip a coin and base your choice on that?
> If I switch the choice, I react to the new information
This assumes that him opening a door is new information. It isn't unless you make an assumption about the game masters intentions.
If the host doesn't want you to win then you will lose 100% of the time if you switch, if the host wants you to win then you could win 100% of the time by switching etc. But people think that these assumptions are "obvious" so of course you must make them, but that makes the riddle bad.
If the riddle doesn't say that the host always opens an empty door after you pick a door, then the typical solution isn't correct. People does the typical "X happened once" and assumes it means "X always happens", that is a typical naive assumption but no, just because it happened once doesn't mean it always happens, you can't make that assumption here unless stated in the riddle.
The problem I always struggle with is the premise that presenter not only knows which box is the prize, but also does not want you to win.
In other words, there should never be a scenario where the show host gives you an opportunity to switch boxes unless you have already chosen the prize box, in which your choice should be to not change boxes.
I guess I just naturally assumed that game shows don't want contestants to win and the odds are against you.
I agree that this additional information is necessary to understand the problem. I've seen that show maybe twice in my lifetime, both times when I was a little kid. I did not know that Monty never opens the door with the prize behind it. Not one time in the history of the show did he ever open the prize door. Knowing that is additional information, and can be used to figure out the problem. But you need to watch the show more than twice to know that.
I'm giving you an up vote for creative writing and nothing else. I'm not claiming anything about your data, just your creative writing skills. Thank you.
That presumably quickly gets into a game-theoretic double-triple-etc-bluff affair where you have to assume the host is using his knowledge against you.
If you choose the door with the car, the host will open another door to tempt you, so then you don't switch, unlike in traditional Monty Hall. And if he didn't open a door, well, that means you choose a goat, so you should switch for a 50% chance. Except the host knows you're thinking this. And you know he knows. And he knows you know he knows. And...
Perhaps, iterated infinitely, this eventually resolves into a limiting expected value for switching and not, but I have no idea how to compute such a scenario.
The puzzle is is commonly stated is not how any TV show ever worked. In a real (repeated) TV show the host behavior will be non-deterministic, and will sometimes be benign, sometimes adverse, to increase the suspense and ratings. If the host follows a deterministic set of rules (as is usually implicitly assumed in the puzzle version), the optimal strategy is pretty easy to work out.
I think their explanation is a lot easier to understand
"When we pick the original box, we know that the probability that the keys will be in there is 1/3. The probability that the keys will not be in the box you originally chose is 1 - 1/3 = 2/3. Just from this knowledge alone, you could decide that you will always switch, since the probability that the other boxes have the keys is 2/3."
>= 2/3. Just from this knowledge alone, you could decide that you will always switch, since the probability that the other boxes have the keys is 2/3.
Your sentence the particular way you worded it is not the correct mathematical model.
The player does not get to switch to BOTH OF THE OTHER 2 boxes as an alternative to just the 1st box. Therefore the 2/3rd probability doesn't apply.
Where the non-intuitive 2/3rds probability becomes the answer instead of 50/50 is the host's perfect knowledge of always choosing the door without the car.
This was the one that worked when explaining it to my friends.
It gives a mental image of the host opening 998 boxes, leaving only your selected box and one other. From here it’s easier to see that there must be something special about that one box the host left un-opened!
(Though even then there were people who clung to the “2 boxes means 1-in-2 chance” fallacy, failing to see that the host has revealed information.)
Edit: an other version was to change the hosts proposal: what if he let you choose one box, and then said he would let you switch to having whatever was in the other 999 boxes? Of course you would switch! The crux is understanding that this offer is actually the same as in the first proposal, since the host is not opening the boxes at random.
Consider the Honty Mall problem: it’s like the original problem, except after you pick a box, Honty offers you both of the other boxes. It’s much easier to see swapping is better in this problem, and it’s also easier to see that the chance is 2/3 if you swap. Then you just have to show that the Honty Mall problem is equivalent to the Monty Hall problem, by stipulating that Monty will always open a box that’s empty.
The article gets the 50/50 case subtly wrong. The really naive analysis is that you have two doors so it's 50/50. A less naive, more interesting analysis, is that there are twelve possible outcomes, of which six are favourable (if the correct option is A, pick A -> presenter picks B is different from pick A -> presenter picks C).
The article hides this away by lumping those two under "Host opens B or C" without further justification, but it's important to notice that this only works because those twelve outcomes have different probabilities.
Edit: In table form,
Car Guest Monty Swap? Probability
A A B No 1/3 * 1/3 * 1/2 = 1/18
A A C No 1/3 * 1/3 * 1/2 = 1/18
A B C Yes 1/3 * 1/3 = 1/9
A C B Yes 1/3 * 1/3 = 1/9
B A C Yes 1/3 * 1/3 = 1/9
B B A No 1/3 * 1/3 * 1/2 = 1/18
B B C No 1/3 * 1/3 * 1/2 = 1/18
B C A Yes 1/3 * 1/3 = 1/9
C A B Yes 1/3 * 1/3 = 1/9
C B A Yes 1/3 * 1/3 = 1/9
C C A No 1/3 * 1/3 * 1/2 = 1/18
C C B No 1/3 * 1/3 * 1/2 = 1/18
Total Yes = 6 * 1/9 = 6/9 = 2/3
Total No = 6 * 1/18 = 6/18 = 1/3
In the Monty Hall problem, you think you are choosing between one door and one other door. But in fact, you are choosing between one door and two doors.
The choice is between "this door" (1/3 probability for winning) and "all other doors" (2/3 probability for winning).
For me the hangup was always the hidden rule: host won’t open a door with a car. That is unstated and remains unstated even in modern discussions of the problem (see Pinker’s “Rationality”). Once explicitly states the outcome becomes intuitive.
Yes, in many formulations the unstated assumptions that the host (a) will always open a door after your initial pick, and (b) that it's always one without a car behind it. Making the assumptions explicitly makes the solution and intuition much simpler.
If your initial choice was a car, the host can open any of the remaining doors, but if your initial choice was a goat this forces the host to reveal extra information to you (namely which of the remaining doors contains the car). Since your initial probability of picking a goat was 2/3, there is 2/3 probability that the host will reveal the prize door for you.
This is why the puzzle is only loosely based on a TV show. No real TV or other iterated games will work like this, since the optimal strategy is pretty simple. In a real TV show, the host would mix up his strategy (never revealing the car door, but only occasionally opening a door after the candidates choice). In that case it's not possible to work out an optimal strategy without additional assumptions or clues wrt the host behavior. E.g. he might be biased to open a remaining door with higher probability when the initial choice was correct, to increase suspension for the viewers, in which the dominant strategy is actually to not switch. But in a real TV show or iterated game, the host behavior is likely not deterministic.
I do not think that when it's explicitly stated it becomes intuitive for everyone and that adds a further wrinkle. Many people will still get it wrong, even when the problem is stated correctly.
However, if the rule is not explicitly stated, how can the player know that the rule exists? Perhaps "Monty" is evil and will not always open a door, "evil Monty" will only open a door when he knows you've chosen correctly.
IOW, without that rule explicitly stated, the answer "Switch" is simply incorrect. Without that rule, the answer is "I don't have enough information to know."
In fact, the Wikipedia Monty Hall article discusses pretty much any aspect of the problem that anyone has ever brought up in any Monty Hall forum thread or blog post.
The easiest way is to convince them is with real money on the line.
Like the article, take a deck of cards. Ask them to pick a card for you without looking. Set it aside.
Tell them if they have the queen of diamonds, you buy lunch, otherwise they buy lunch. Ask if they want to swap decks.
After they inevitably say yes, go through the 51 other cards and turn over 20 cards that aren't the queen of diamonds. Ask if they're sure they'd like to switch.
Remove 20 more cards and repeat. Then 9 more (leaving you with 2 cards.) Ask again, turn over one last card, ask one more time. (This last iteration is the actual Monty Hall problem.)
The key thing is they should understand now that Monty Hall knows where the queen of diamonds/ car is and turns over other cards/goats precisely because he knows they don't change the odds of the original choice, but many people incorrectly believe that it does.
There's actually an easier way to remove the subtleties of this statistical problem by using a hyperbolic example. For kicks and shiggles we'll up the stakes to be a prize of $1B USD.
Consider instead a Monte Hall scenario with 100 boxes (vice just 3), maybe we call this the "deal or no deal" variant of the problem...
The user picks 1 box and has a 1/100 chance of selecting the box with the prize. Now, the host opens 98 boxes that they know do not have keys in them, leaving two unopened boxes (the one the user picked and the one the host left unopened).
Now, pick your box from the remaining selection and claim your prize. You bet your sweet bippy I know which box I'm picking.
Information is gained by observing the winnowing of the field of options.
I don't see how "would you like 1/3rd odds or 2/3rd odds of winning" is less easy to understand that adding more numbers and layers to the problem by adding in words like $1Billion and 100 boxes.
If someone is willing to pick 1/3rds odds of winning, the discussion needs to go somewhere else.
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Imagine a universe with many simultaneous Monty Hall clones playing at once in many studios, where Monty doesn't know and opens another door at random. If that door has the car behind it, Monty and contestant are both shot in the head and the studio burned down and erased from all records. This bloody culling of branches of the probabilities is the same as effected by giving Monty the knowledge and telling him to act on it.
The associated line of reasoning resolved the paradox for me. If I stick with my original choice, it is as if I ignored the new information. If I switch the choice, I react to the new information.
The extreme of this is to pick among countably infinite doors, having the presenter open countably infinite doors and leaving just yours and another closed. Who could reasonable suggest that the chance is still 50/50, assuming you don't flip a coin and base your choice on that?
This assumes that him opening a door is new information. It isn't unless you make an assumption about the game masters intentions.
If the host doesn't want you to win then you will lose 100% of the time if you switch, if the host wants you to win then you could win 100% of the time by switching etc. But people think that these assumptions are "obvious" so of course you must make them, but that makes the riddle bad.
If the riddle doesn't say that the host always opens an empty door after you pick a door, then the typical solution isn't correct. People does the typical "X happened once" and assumes it means "X always happens", that is a typical naive assumption but no, just because it happened once doesn't mean it always happens, you can't make that assumption here unless stated in the riddle.
The host always gives you an opportunity to switch boxes.
And perhaps also, “remember: Monty will always open a door, and the contestant knows it”.
Makes me wonder if there were similar shows where the host can choose not to open a door.
If you choose the door with the car, the host will open another door to tempt you, so then you don't switch, unlike in traditional Monty Hall. And if he didn't open a door, well, that means you choose a goat, so you should switch for a 50% chance. Except the host knows you're thinking this. And you know he knows. And he knows you know he knows. And...
Perhaps, iterated infinitely, this eventually resolves into a limiting expected value for switching and not, but I have no idea how to compute such a scenario.
* if on the 1st try you choose the correct box (33% chance), then the one you can switch to will be wrong
* if on the 1st try you choose the wrong box (66% chance), then the one you can switch to will be correct one
therefore your goal is to pick the wrong box on the 1st try and then switch, and you have 66% chance to do it
"When we pick the original box, we know that the probability that the keys will be in there is 1/3. The probability that the keys will not be in the box you originally chose is 1 - 1/3 = 2/3. Just from this knowledge alone, you could decide that you will always switch, since the probability that the other boxes have the keys is 2/3."
Your sentence the particular way you worded it is not the correct mathematical model.
The player does not get to switch to BOTH OF THE OTHER 2 boxes as an alternative to just the 1st box. Therefore the 2/3rd probability doesn't apply.
Where the non-intuitive 2/3rds probability becomes the answer instead of 50/50 is the host's perfect knowledge of always choosing the door without the car.
It gives a mental image of the host opening 998 boxes, leaving only your selected box and one other. From here it’s easier to see that there must be something special about that one box the host left un-opened!
(Though even then there were people who clung to the “2 boxes means 1-in-2 chance” fallacy, failing to see that the host has revealed information.)
Edit: an other version was to change the hosts proposal: what if he let you choose one box, and then said he would let you switch to having whatever was in the other 999 boxes? Of course you would switch! The crux is understanding that this offer is actually the same as in the first proposal, since the host is not opening the boxes at random.
(I understand the Monty Hall problem, I just don't see how changing the number of doors makes a difference to anyone's intuition.)
By opening the box he knows to be empty, it manufactures an equivalence to if he'd offered to open the box you chose, vs open the other two together.
The article hides this away by lumping those two under "Host opens B or C" without further justification, but it's important to notice that this only works because those twelve outcomes have different probabilities.
Edit: In table form,
In the Monty Hall problem, you think you are choosing between one door and one other door. But in fact, you are choosing between one door and two doors.
The choice is between "this door" (1/3 probability for winning) and "all other doors" (2/3 probability for winning).
If your initial choice was a car, the host can open any of the remaining doors, but if your initial choice was a goat this forces the host to reveal extra information to you (namely which of the remaining doors contains the car). Since your initial probability of picking a goat was 2/3, there is 2/3 probability that the host will reveal the prize door for you.
This is why the puzzle is only loosely based on a TV show. No real TV or other iterated games will work like this, since the optimal strategy is pretty simple. In a real TV show, the host would mix up his strategy (never revealing the car door, but only occasionally opening a door after the candidates choice). In that case it's not possible to work out an optimal strategy without additional assumptions or clues wrt the host behavior. E.g. he might be biased to open a remaining door with higher probability when the initial choice was correct, to increase suspension for the viewers, in which the dominant strategy is actually to not switch. But in a real TV show or iterated game, the host behavior is likely not deterministic.
However, if the rule is not explicitly stated, how can the player know that the rule exists? Perhaps "Monty" is evil and will not always open a door, "evil Monty" will only open a door when he knows you've chosen correctly.
IOW, without that rule explicitly stated, the answer "Switch" is simply incorrect. Without that rule, the answer is "I don't have enough information to know."
In fact, the Wikipedia Monty Hall article discusses pretty much any aspect of the problem that anyone has ever brought up in any Monty Hall forum thread or blog post.
Like the article, take a deck of cards. Ask them to pick a card for you without looking. Set it aside.
Tell them if they have the queen of diamonds, you buy lunch, otherwise they buy lunch. Ask if they want to swap decks.
After they inevitably say yes, go through the 51 other cards and turn over 20 cards that aren't the queen of diamonds. Ask if they're sure they'd like to switch.
Remove 20 more cards and repeat. Then 9 more (leaving you with 2 cards.) Ask again, turn over one last card, ask one more time. (This last iteration is the actual Monty Hall problem.)
The key thing is they should understand now that Monty Hall knows where the queen of diamonds/ car is and turns over other cards/goats precisely because he knows they don't change the odds of the original choice, but many people incorrectly believe that it does.
Consider instead a Monte Hall scenario with 100 boxes (vice just 3), maybe we call this the "deal or no deal" variant of the problem...
The user picks 1 box and has a 1/100 chance of selecting the box with the prize. Now, the host opens 98 boxes that they know do not have keys in them, leaving two unopened boxes (the one the user picked and the one the host left unopened).
Now, pick your box from the remaining selection and claim your prize. You bet your sweet bippy I know which box I'm picking.
Information is gained by observing the winnowing of the field of options.
If someone is willing to pick 1/3rds odds of winning, the discussion needs to go somewhere else.