11 ms·

Monty Hall Problem

24 points by lnyan 4y ago

superb-owl 4y ago
The trick that finally got me to accept the conclusion was to imagine there were 100 doors. You pick one, then Monty opens 98 other doors. Then you obviously switch, as the original door has a 1% chance of being right, and the other door has a 99% of being right.
- sdfhdhjdw3 4y ago
  I've hear that before many times, but caution. You're just leaning on your intuition, probably not understanding why you got to the wrong answer in the first place.
- aaronjl 4y ago
  What helped me is the fact that the only way you WON'T win by switching, is if you correctly choose it the first time. And that only happens 1/3 of the time
  - injb 4y ago
    The problem I always had with this (and still have) is that you're not asked about the probability (i.e. what trend you expect to see in a string of repeated experiments), but rather you're asked: "Is it to your advantage to switch your choice?" Well that depends on where the prize is. If the prize is behind the other door, then yes it's to your advantage to switch. Otherwise, it isn't. How can there be any more to it than that? You're not going to be given the opportunity to play the game 100 times. This is your one shot and the location of the prize is already fixed.
    - jjbinx007 4y ago
      Imagine there are 100 doors. You pick one at random. You have a 1 in 100 chance of getting the right answer. If the host opens all the doors that contain the booby prizes bar one (in this case 98 goats) and leaves one door remaining, plus your original guess, it's pretty obvious you are unlikely to have guessed correctly on your first try. Scale it down to 3 doors and the host always revealing a goat and you have the same situation except there's a 1 in 3 chance you'll win by sticking with your first guess and a 2 in 3 chance if you swap.
    - nkrisc 4y ago
      Well if you don't switch then your odds of winning are a simple 1 out of 3. Switching inverts your initial guess, so instead of losing 2 out of 3 times because your initial pick was wrong, switching makes you win 2 out of 3 times when your initial pick is wrong.
    - Jtsummers 4y ago
      Right, it's your one shot. So what gives you better odds (that is, is to your advantage). If you don't switch, you're selecting the lower odds chance of winning, putting yourself at a disadvantage. Since you don't know where it is, but do have a chance to double your chance of winning, then switching is advantageous. Expected value is the thing to consider.
- thehappypm 4y ago
  The problem I have with this is that the original Monty Hall problem has him opening one door out of 2. With the 100 doors analogy, wouldn’t Monty be opening just one of the remaining doors? Or half? Why all 98?
  - injb 4y ago
    Then the question of "switching" wouldn't make sense. Which of the remaining 98 would you be asked to switch to? The whole point is that you end up with a choice of sticking with the door you picked, or switching to the one remaining door.
    - colmanhumphrey 4y ago
      Seems fine to me. Do you want to stick with your original or switch to one of the remaining 98? It’s still worth switching too, just a smaller increase in probability.
  - waynesonfire 4y ago
    i think the idea is to demonstrate more easily the information leak by exaggerating the amount of information being leaked. with 3 doors, there are only so many the host can chose to open. with a 100, there is up to 98. sure, it can be 1 or 2... or 98.
  - jasonswett 4y ago
    The rule is "after the player has guessed a door, open ALL the other doors except the door that was guessed plus one other door". I think your confusion is because if there are 100 doors then the rule is self-evident. But when there are only 3 doors, the rule of "open all doors except the door that was guessed plus one other" is indistinguishable from "open just one door". The key is knowing that the rule is the "open all..." rule.
    - colmanhumphrey 4y ago
      That’s not the rule from the question in the link nor in the original game (the game’s rule isn’t quite the same as the classic problem, it’s even messier). Your proposed rule doesn’t uniquely extend the original, it’s just one of the possible generalizations.
- dahfizz 4y ago
  > Then you obviously switch, as the original door has a 1% chance of being right, and the other door has a 99% of being right. It's not obvious to me that this is true. Why do we update the chance of the "new" door but not the one we already selected? Intuitively, I would expect the odds of both doors to increase to 50% as the other 98 doors are opened. I know that you're correct, I just don't find this to be an intuitive explanation.
  - mcphage 4y ago
    The odds of your original door being right are 1%—that never changes. It’s determined by the options when the game was set up and when you picked it. If your original door was wrong—99% chance—then it’s in one of the 99 other doors. But the host opens 98 of them, so if it was one of the 99 other doors—again, 99% chance—then the only option left is the last door.
antris 4y ago
The fact that the show host knows what's behind the doors and never chooses a door with the prize behind it is crucial. The host leaks information about what's behind, as he chooses a door. I was told about this problem and the person telling it didn't say anything about that, just that you pick a door and the host opens another one which doesn't have a prize behind it. So I thought the probabilities don't change at all because the host just picked a random door. Then everyone who heard of the problem before piled against me like I'm a dumbass, but never really explaining why I was wrong. Yes, I'm still salty about it.
- colinmhayes 4y ago
  Well if the host opens a door and it never has a prize behind it they have to know which door has the prize right?
  - bombcar 4y ago
    It depends on if that's the scenario explained, or you have just one example (not knowing if the host will always show a goat, or if the host will sometimes show you a car either intentionally or by accident).
  - antris 4y ago
    The part about never having a prize behind it was not explained to me. It was just "you pick a door, host opens a door without prize". That doesn't imply that there's never a prize behind the door that the host opens.
    - paxys 4y ago
      What else does "host opens a door without prize" mean?
      - antris 4y ago
        That the host opens a door without prize.
      - 4y ago
VikingCoder 4y ago
I didn't believe the result. So, I had to sit down and decide to write the source code to simulate it, and as I'm typing the code my internal monologue is saying "if you stay, 2/3 of the time you lose, but if you change, 2/3 of the time you win," and I was like, "CRAP!" It was like - my fingers knew how to type the correct solution, and once my brain saw my fingers type it, my brain finally caught up.
- bombcar 4y ago
  Yeah, imagine you pick one door, and Monty says, "You can have what's behind that door, or what's behind these other two doors" - you'd always switch because you're getting two chances (and hey, free goat).
headsupftw 4y ago
To me the key to understand the problem is paying attention to this detail: Monty knows where the car was so he wasn't just randomly opening A door. He always opened a door that didn't have a car behind it. It's this non-randomness that's giving you more information about the remaining door.
- moosinho 4y ago
  Not exactly, if Monty randomly picked a door and happened to open a goat then it would still be beneficial for the player to swap.
PUSH_AX 4y ago
The way I understood this was that if you pick a door, the chances are you didn't pick the prize, given that you probably didn't pick the prize and the host shows you the other non prize door, it makes sense that switching will probably give you the prize more than it doesn't.
nkrisc 4y ago
Here's how I finally understood the problem intuitively: If your initial pick is correct, then switching loses 100% of the time. If your initial pick is incorrect, then switching wins 100% of the time (because you picked wrong, and host will reveal the other incorrect door, leaving only the correct door to switch to). So switching always inverts your initial guess. Well what are the odds of picking incorrectly on your initial guess? 2/3. Switching wins 2/3 times.
- dahfizz 4y ago
  I also find this to be a more intuitive explanation, thank you!
- technothrasher 4y ago
  Yep, this is how I always wrapped my head around it. You’ve got a 1/3 chance of being right, and the host is effectively offering you a chance to bet against yourself. That’s a 2/3 chance.
cupofpython 4y ago
I had this problem beaten to death in my probability and statistics courses in college. To make the problem much more obviously intuitive, simply imagine a thousand or a million doors instead of 3.. and after you make your pick - the host reveals all doors except one to be "failures". Your choice was very obviously 1 in a million, and because the reveal procedure that followed is conditional - if the "success" was behind ANY of the doors you did not pick then it WILL be the only door left in front of you. It is only a 50-50 IF the revealed doors were opened randomly. In this case, sometimes the "success" would be revealed by accident - and so IF you do happen to find yourself in a position of only having 2 doors left THEN it was more likely to happen if you had actually chosen the "success". So arriving at that situation at all is the conditional information that alters the probability of your original choice - but in this case from 1/3 to 1/2 instead of 2/3. (this is how the probabilities in "deal or no deal" work btw) Even knowing the solution, it is a good reminder about the power of information derived from systemic processes. the information leveraged in the processes that lead to circumstances is just as important as the circumstances themselves. The key general point that bypasses most peoples intuition the first time they hear the problem is that the revealed box (or door, etc) is not chosen at random but by a conditional process. The condition is that the host reveals a "failure". In order to pick a failure, he must leverage knowledge about where the "success" is, else he risks revealing it by mistake. Because knowledge of the "success" is embedded into the decision process, more information about the "success" is revealed to the observer which allows them to improve their position.
k3lsi3r 4y ago
The host revealing the door is irrelevant, right? An equivalent offer would be if you had the chance to either: - keep your door OR - choose both of the other doors (since you know at least one of the other doors is a goat/non-prize)
em500 4y ago
The Month Hall Problem has been discussed many many times on HN and elsewhere. I've read many of the discussions (but of course not nearly all of them. My observations: 1. The linked Wikipedia article is very extensive, and already covers usually all the points and insights that forum posters bring up. And a lot more. 2. Few people will probably bother to read the entire (or even most of the) Wikipedia article, because it's so extensive. 3. I'll copy and highlight the IMO most crucial part the article here: The most famous formulation, a column in Parade magazine in 1990, went thus: Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice? In this formulation the solution is ambiguous. Switching is *not necesarily* the optimal strategy (it depends on the precise strategy/behavior of the host, which is not well formulated). Under the unambigous "standard assumptions": - The host must always open a door that was not picked by the contestant. - The host must always open a door to reveal a goat and never the car. - The host must always offer the chance to switch between the originally chosen door and the remaining closed door. switching is the better strategy. But in this formulation the reason is also a lot more obvious.
chrisbrandow 4y ago
I think the 100 door example is good, but is often not described in a way that was intuitive to me. I've added bit that I found very helpful. The entire trick of the game is that Monty opens 1 (or 98) of the remaining doors. That is because it is completely irrelevant to the odds, but has big impact on the contestants perception. It is not obvious, but Monty actually does two separate and unrelated things: First, a step that basically goes unnoticed, is to propose a new game to the contestant in which the contestant opens either 1 door or 2 (99) doors chosen by Monty, rather than the original game which was to select 1 door out of 3 (100). Obviously, the new game gives them the chance to make a choice that has 2/3 odds (99/100) versus the original game whose odds were 1/3 (1/100). His second step, which is typically the only step noticed and which very effectively misleads the contestant, is to open 1 (98) of the doors he selected. The result is that clearly the 2nd game gives the contestant a definitively better choice, but this is obscured by the fact the contestant only sees two closed doors, so they believe they are comparing 1 door to another, and thus only have 50-50 odds.
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paxys 4y ago
Lots of "intuitive" explanations of the problem shared here, but here's one that works for me. You pick door A. The hosts says you can either keep that door, or instead pick both doors B & C. If either one of them has the prize you get to keep it. You opt to switch. The host opens B, which turns out to be a goat, and then opens C. Regardless of what C contains, it's obvious that switching was the better option.
- cupofpython 4y ago
  This sums up the probabilities cleanly, but doesnt provide any intuition on why the option to pick both doors B&C is the same as the original problem
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ortusdux 4y ago
My favorite Monty Hall explanation: "I have randomly picked a number between 0 and 1 million, and I want you to guess my choice." "Lets go with 408,235." "I will narrow it down for you. Lets eliminate 999,998 incorrect guesses, leaving either 174,999 or your guess of 408,235. Would you like to change your guess?"
dtagames 4y ago
What's missing from this thread is not that Monty always reveals a goat... It's that he never reveals the car! In other words, the "you can switch doors" routine is always part of the show. By removing one door which does not have the car, Monty has narrowed your best choice to one of the other two.
paxys 4y ago
One thing that trips a lot of people up is that the problem is never explicit about the knowledge or strategy of the host. Would he offer you the choice to switch if you initially picked the wrong door? If I'm playing this game in real life, my intuition isn't going to be calculating the correct mathematical odds, but rather thinking "this guy is trying to screw me". So a 1/3 chance that you can be sure of is better than a 2/3 one which is based on trusting your opponent.
ksaj 4y ago
An easy way to word this exact problem in a way that makes clear the reason you should always change your choice is: I have 52 playing cards. I know which one is the Ace of Spades. Point to one of the cards and leave your finger on it. Okay, without telling you whether you are right or not, I'm going to remove every card, leaving only yours and one other on the table, that is the right card if the one you picked is not, or an indifferent card if your choice was right. Do you wish to choose the card I left on the table, or do you still think your 1/52 choice was correct? Notice that I followed the exact rules of the game, but changed the wording and used a lot more "doors." There was a 1/52 chance you would pick the right card, and even after I remove all but one of the remaining cards, it is still a 1/52 chance you originally picked the right card, and a 51/52 chance that the one I left behind is the right card. Now just make it 3 cards, and use the traditional script, performing the exact same steps, and you'll understand why you should always change from your original choice.

24 points by lnyan 4y ago

The trick that finally got me to accept the conclusion was to imagine there were 100 doors. You pick one, then Monty opens 98 other doors. Then you obviously switch, as the original door has a 1% chance of being right, and the other door has a 99% of being right.

sdfhdhjdw3 4y ago
I've hear that before many times, but caution. You're just leaning on your intuition, probably not understanding why you got to the wrong answer in the first place.
aaronjl 4y ago
What helped me is the fact that the only way you WON'T win by switching, is if you correctly choose it the first time. And that only happens 1/3 of the time
- injb 4y ago
  The problem I always had with this (and still have) is that you're not asked about the probability (i.e. what trend you expect to see in a string of repeated experiments), but rather you're asked: "Is it to your advantage to switch your choice?" Well that depends on where the prize is. If the prize is behind the other door, then yes it's to your advantage to switch. Otherwise, it isn't. How can there be any more to it than that? You're not going to be given the opportunity to play the game 100 times. This is your one shot and the location of the prize is already fixed.
  - jjbinx007 4y ago
    Imagine there are 100 doors. You pick one at random. You have a 1 in 100 chance of getting the right answer. If the host opens all the doors that contain the booby prizes bar one (in this case 98 goats) and leaves one door remaining, plus your original guess, it's pretty obvious you are unlikely to have guessed correctly on your first try. Scale it down to 3 doors and the host always revealing a goat and you have the same situation except there's a 1 in 3 chance you'll win by sticking with your first guess and a 2 in 3 chance if you swap.
  - nkrisc 4y ago
    Well if you don't switch then your odds of winning are a simple 1 out of 3. Switching inverts your initial guess, so instead of losing 2 out of 3 times because your initial pick was wrong, switching makes you win 2 out of 3 times when your initial pick is wrong.
  - Jtsummers 4y ago
    Right, it's your one shot. So what gives you better odds (that is, is to your advantage). If you don't switch, you're selecting the lower odds chance of winning, putting yourself at a disadvantage. Since you don't know where it is, but do have a chance to double your chance of winning, then switching is advantageous. Expected value is the thing to consider.
thehappypm 4y ago
The problem I have with this is that the original Monty Hall problem has him opening one door out of 2. With the 100 doors analogy, wouldn’t Monty be opening just one of the remaining doors? Or half? Why all 98?
- injb 4y ago
  Then the question of "switching" wouldn't make sense. Which of the remaining 98 would you be asked to switch to? The whole point is that you end up with a choice of sticking with the door you picked, or switching to the one remaining door.
  - colmanhumphrey 4y ago
    Seems fine to me. Do you want to stick with your original or switch to one of the remaining 98? It’s still worth switching too, just a smaller increase in probability.
- waynesonfire 4y ago
  i think the idea is to demonstrate more easily the information leak by exaggerating the amount of information being leaked. with 3 doors, there are only so many the host can chose to open. with a 100, there is up to 98. sure, it can be 1 or 2... or 98.
- jasonswett 4y ago
  The rule is "after the player has guessed a door, open ALL the other doors except the door that was guessed plus one other door". I think your confusion is because if there are 100 doors then the rule is self-evident. But when there are only 3 doors, the rule of "open all doors except the door that was guessed plus one other" is indistinguishable from "open just one door". The key is knowing that the rule is the "open all..." rule.
  - colmanhumphrey 4y ago
    That’s not the rule from the question in the link nor in the original game (the game’s rule isn’t quite the same as the classic problem, it’s even messier). Your proposed rule doesn’t uniquely extend the original, it’s just one of the possible generalizations.
dahfizz 4y ago
> Then you obviously switch, as the original door has a 1% chance of being right, and the other door has a 99% of being right. It's not obvious to me that this is true. Why do we update the chance of the "new" door but not the one we already selected? Intuitively, I would expect the odds of both doors to increase to 50% as the other 98 doors are opened. I know that you're correct, I just don't find this to be an intuitive explanation.
- mcphage 4y ago
  The odds of your original door being right are 1%—that never changes. It’s determined by the options when the game was set up and when you picked it. If your original door was wrong—99% chance—then it’s in one of the 99 other doors. But the host opens 98 of them, so if it was one of the 99 other doors—again, 99% chance—then the only option left is the last door.

antris 4y ago

The fact that the show host knows what's behind the doors and never chooses a door with the prize behind it is crucial. The host leaks information about what's behind, as he chooses a door. I was told about this problem and the person telling it didn't say anything about that, just that you pick a door and the host opens another one which doesn't have a prize behind it. So I thought the probabilities don't change at all because the host just picked a random door. Then everyone who heard of the problem before piled against me like I'm a dumbass, but never really explaining why I was wrong. Yes, I'm still salty about it.

colinmhayes 4y ago
Well if the host opens a door and it never has a prize behind it they have to know which door has the prize right?
- bombcar 4y ago
  It depends on if that's the scenario explained, or you have just one example (not knowing if the host will always show a goat, or if the host will sometimes show you a car either intentionally or by accident).
- antris 4y ago
  The part about never having a prize behind it was not explained to me. It was just "you pick a door, host opens a door without prize". That doesn't imply that there's never a prize behind the door that the host opens.
  - paxys 4y ago
    What else does "host opens a door without prize" mean?
    - antris 4y ago
      That the host opens a door without prize.
    - 4y ago

VikingCoder 4y ago

I didn't believe the result. So, I had to sit down and decide to write the source code to simulate it, and as I'm typing the code my internal monologue is saying "if you stay, 2/3 of the time you lose, but if you change, 2/3 of the time you win," and I was like, "CRAP!" It was like - my fingers knew how to type the correct solution, and once my brain saw my fingers type it, my brain finally caught up.

bombcar 4y ago
Yeah, imagine you pick one door, and Monty says, "You can have what's behind that door, or what's behind these other two doors" - you'd always switch because you're getting two chances (and hey, free goat).

headsupftw 4y ago

To me the key to understand the problem is paying attention to this detail: Monty knows where the car was so he wasn't just randomly opening A door. He always opened a door that didn't have a car behind it. It's this non-randomness that's giving you more information about the remaining door.

moosinho 4y ago
Not exactly, if Monty randomly picked a door and happened to open a goat then it would still be beneficial for the player to swap.

PUSH_AX 4y ago

The way I understood this was that if you pick a door, the chances are you didn't pick the prize, given that you probably didn't pick the prize and the host shows you the other non prize door, it makes sense that switching will probably give you the prize more than it doesn't.

nkrisc 4y ago

Here's how I finally understood the problem intuitively: If your initial pick is correct, then switching loses 100% of the time. If your initial pick is incorrect, then switching wins 100% of the time (because you picked wrong, and host will reveal the other incorrect door, leaving only the correct door to switch to). So switching always inverts your initial guess. Well what are the odds of picking incorrectly on your initial guess? 2/3. Switching wins 2/3 times.

dahfizz 4y ago
I also find this to be a more intuitive explanation, thank you!
technothrasher 4y ago
Yep, this is how I always wrapped my head around it. You’ve got a 1/3 chance of being right, and the host is effectively offering you a chance to bet against yourself. That’s a 2/3 chance.

cupofpython 4y ago

I had this problem beaten to death in my probability and statistics courses in college. To make the problem much more obviously intuitive, simply imagine a thousand or a million doors instead of 3.. and after you make your pick - the host reveals all doors except one to be "failures". Your choice was very obviously 1 in a million, and because the reveal procedure that followed is conditional - if the "success" was behind ANY of the doors you did not pick then it WILL be the only door left in front of you. It is only a 50-50 IF the revealed doors were opened randomly. In this case, sometimes the "success" would be revealed by accident - and so IF you do happen to find yourself in a position of only having 2 doors left THEN it was more likely to happen if you had actually chosen the "success". So arriving at that situation at all is the conditional information that alters the probability of your original choice - but in this case from 1/3 to 1/2 instead of 2/3. (this is how the probabilities in "deal or no deal" work btw) Even knowing the solution, it is a good reminder about the power of information derived from systemic processes. the information leveraged in the processes that lead to circumstances is just as important as the circumstances themselves. The key general point that bypasses most peoples intuition the first time they hear the problem is that the revealed box (or door, etc) is not chosen at random but by a conditional process. The condition is that the host reveals a "failure". In order to pick a failure, he must leverage knowledge about where the "success" is, else he risks revealing it by mistake. Because knowledge of the "success" is embedded into the decision process, more information about the "success" is revealed to the observer which allows them to improve their position.

k3lsi3r 4y ago

The host revealing the door is irrelevant, right? An equivalent offer would be if you had the chance to either: - keep your door OR - choose both of the other doors (since you know at least one of the other doors is a goat/non-prize)

em500 4y ago

The Month Hall Problem has been discussed many many times on HN and elsewhere. I've read many of the discussions (but of course not nearly all of them. My observations: 1. The linked Wikipedia article is very extensive, and already covers usually all the points and insights that forum posters bring up. And a lot more. 2. Few people will probably bother to read the entire (or even most of the) Wikipedia article, because it's so extensive. 3. I'll copy and highlight the IMO most crucial part the article here: The most famous formulation, a column in Parade magazine in 1990, went thus: Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice? In this formulation the solution is ambiguous. Switching is *not necesarily* the optimal strategy (it depends on the precise strategy/behavior of the host, which is not well formulated). Under the unambigous "standard assumptions": - The host must always open a door that was not picked by the contestant. - The host must always open a door to reveal a goat and never the car. - The host must always offer the chance to switch between the originally chosen door and the remaining closed door. switching is the better strategy. But in this formulation the reason is also a lot more obvious.

chrisbrandow 4y ago

I think the 100 door example is good, but is often not described in a way that was intuitive to me. I've added bit that I found very helpful. The entire trick of the game is that Monty opens 1 (or 98) of the remaining doors. That is because it is completely irrelevant to the odds, but has big impact on the contestants perception. It is not obvious, but Monty actually does two separate and unrelated things: First, a step that basically goes unnoticed, is to propose a new game to the contestant in which the contestant opens either 1 door or 2 (99) doors chosen by Monty, rather than the original game which was to select 1 door out of 3 (100). Obviously, the new game gives them the chance to make a choice that has 2/3 odds (99/100) versus the original game whose odds were 1/3 (1/100). His second step, which is typically the only step noticed and which very effectively misleads the contestant, is to open 1 (98) of the doors he selected. The result is that clearly the 2nd game gives the contestant a definitively better choice, but this is obscured by the fact the contestant only sees two closed doors, so they believe they are comparing 1 door to another, and thus only have 50-50 odds.

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paxys 4y ago

Lots of "intuitive" explanations of the problem shared here, but here's one that works for me. You pick door A. The hosts says you can either keep that door, or instead pick both doors B & C. If either one of them has the prize you get to keep it. You opt to switch. The host opens B, which turns out to be a goat, and then opens C. Regardless of what C contains, it's obvious that switching was the better option.

cupofpython 4y ago
This sums up the probabilities cleanly, but doesnt provide any intuition on why the option to pick both doors B&C is the same as the original problem

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ortusdux 4y ago

My favorite Monty Hall explanation: "I have randomly picked a number between 0 and 1 million, and I want you to guess my choice." "Lets go with 408,235." "I will narrow it down for you. Lets eliminate 999,998 incorrect guesses, leaving either 174,999 or your guess of 408,235. Would you like to change your guess?"

dtagames 4y ago

What's missing from this thread is not that Monty always reveals a goat... It's that he never reveals the car! In other words, the "you can switch doors" routine is always part of the show. By removing one door which does not have the car, Monty has narrowed your best choice to one of the other two.

paxys 4y ago

One thing that trips a lot of people up is that the problem is never explicit about the knowledge or strategy of the host. Would he offer you the choice to switch if you initially picked the wrong door? If I'm playing this game in real life, my intuition isn't going to be calculating the correct mathematical odds, but rather thinking "this guy is trying to screw me". So a 1/3 chance that you can be sure of is better than a 2/3 one which is based on trusting your opponent.

ksaj 4y ago

An easy way to word this exact problem in a way that makes clear the reason you should always change your choice is: I have 52 playing cards. I know which one is the Ace of Spades. Point to one of the cards and leave your finger on it. Okay, without telling you whether you are right or not, I'm going to remove every card, leaving only yours and one other on the table, that is the right card if the one you picked is not, or an indifferent card if your choice was right. Do you wish to choose the card I left on the table, or do you still think your 1/52 choice was correct? Notice that I followed the exact rules of the game, but changed the wording and used a lot more "doors." There was a 1/52 chance you would pick the right card, and even after I remove all but one of the remaining cards, it is still a 1/52 chance you originally picked the right card, and a 51/52 chance that the one I left behind is the right card. Now just make it 3 cards, and use the traditional script, performing the exact same steps, and you'll understand why you should always change from your original choice.