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The intuitive way for me to understand the Monte Hall problem is to pretended there are 1000 doors. You pick 1. There’s a 1 in 1000 chance you get it right. The

by gerad 3y ago

The intuitive way for me to understand the Monte Hall problem is to pretended there are 1000 doors. You pick 1. There’s a 1 in 1000 chance you get it right. The host then opens 998 doors that don’t have the prize. Do you keep your original or do you switch? Are the odds 50:50?

Karellen 3y ago
What improvement does your 1000-door variant have over the 100-door variant described in the article?
- michaelcampbell 3y ago
  Bigger numbers make it easier to convince yourself if skeptical. The problem stated with 3 doors is hard, with 5 is still not quite there for many. 100, 100, 10_000_000... becomes almost absurd.
- tshaddox 3y ago
  The most common unintended interpretation seems to be that Monty Hall is only coincidentally opening a door with a goat behind it. I don't quite understand how people arrive at that interpretation, but the 1,000-door variant should make it even more of a stretch to interpret it as Monty Hall coincidentally doing something so extremely improbable.
mithr 3y ago
I think the reason it's so hard to understand is that we try to break the problem down into more digestible problems, and in this case, if you only consider the last stage of the game, it certainly seems like it should be 50:50 odds. In other words, at the final round, you are presented with two doors. Ignore the fact that you already chose one, and assume you were starting at this point: you enter the game, see two doors, and need to pick the one you think hides a car. So intuitively, your odds of choosing which of two doors has the car are 1/2. It's hard to really understand why it matters that the previous rounds occurred at all — why it matters how you narrowed the selection to these two doors. I don't think the 1000 door version makes that easier to understand, because the same thing is true there — if you come into a game, see two doors, and are asked to choose between them, it's very hard to understand why it matters whether in a previous round there were 2, 3, 10, or 1000 doors — there are only two now, when you make your choice.
- rspeele 3y ago
  That's especially tempting since we're used to probability problems often being set up where considering previous state is the sucker's answer. E.g. a fair coin has landed on heads three times in a row, what are the odds of it being tails on the next throw?
- polygotdomain 3y ago
  >It's hard to really understand why it matters that the previous rounds occurred at all It's not hard at all. Humans are not perfectly rational beings. It's not purely about odds, it's about emotions and psychology. In a version where there's no previous selection and it's 50:50, than I pick one and live with it. In the canonical example, the previous selection means that the contestant has already staked their claim, and changing it to the loosing door would have a different emotional response than a simple 50:50 shot with no previous selection. There's a reason why Roulette shows you the last X spins and whether they're odd/even, red/black... because humans make the assumption that the last disconnected data point somehow impacts the current one. You're also glossing over the fact that there was a 1:3 chance you picked the right door to begin with, and 2:3 chance the "other" door was right. The last round isn't 50:50, as you so claim, because there was prior information This is the same reason why the 1000 door example helps explain things; because the math is fundamentally the same, yet significantly more imbalanced. We can also think about how we'd feel about our selection in the 1000 door version, which is likely significantly less confidence, and therefore more likelihood of switching. Whether it's 3 doors, or 1000, the math still say switching is optimal, and our psychology and emotions deal with the choice of switching different in each case.
rspeele 3y ago
The explanation that worked for me was: Suppose after you make your initial choice, instead of opening a door, Monty simply asks if you'd like to switch to BOTH the other two doors, such that you win if the prize is behind EITHER of them. That switch is intuitively a great deal, giving you 2/3 odds. The only way you can lose is in the 1/3rd case where you already picked a winner. The original scenario is equivalent to this, since by revealing a goat Monty is allowing you to pick "the best of" the two doors. --- Often people who don't like this problem complain that actually Monty, with his knowledge of the winning door, may be trying to trick you into losing. Maybe he offers this trade conditionally based on whether you've already selected a winner. Of course, this wouldn't support the most common intuitive answer of "it's 50/50", either, making this a bad excuse. If we accept this behavior from Monty -- not stated in the problem -- it becomes a very uninteresting problem based on whether we are dealing with Evil Monty who only offers the trade when it's a bad one (stay wins 100% of the time), Friendly Monty who only offers the trade when it's a good one (switch wins 100% of the time), or somewhere in the middle. It has no analytical answer. So for this to be a solvable problem at all we must assume we are dealing with Fair Monty who offers the trade all the time, or at least, offers it at random times not based on the contestant's initial pick.
- mithr 3y ago
  I do think this modified scenario makes the solution much clearer. At the same time, I also think that it feels like a significantly-enough different scenario from the original that saying the two scenarios have the same probability then becomes the non-intuitive part. The act of revealing what's behind the second door seems like it should change the probability from 1/3 (one door) vs. 2/3 (one of two doors) to 1/2 (one door of the remaining unopened two) vs. 1/2 (one door of the remaining unopened two, since one was "eliminated"). It's amazing how even seeing the probabilities written out, or running simulations, doesn't really make it easier to truly understand the result.
- mcv 3y ago
  I think the important part of the problem is the unstated assumptions around it, and I suspect some people may see different assumptions than others. If Monty always opens a door, and uses his knowledge of which door has the prize to ensure that the door he opens is always empty, then you should switch, because he's providing you with extra information about which door has the prize. However, if Monty changes it up and sometimes doesn't open a door, or opens the door with the prize, the whole problem changes. If he only opens a door and offers you the chance to switch when you've picked the door with the prize, you should never switch of course. If he opens a random door and might reveal the prize (after which he obviously won't let you switch anymore), switching doesn't change your odds. I think a lot of people see it as one of those two situations.
tromp 3y ago
Your formulation is once again ambiguous. Instead of clearly stating that Monty had to open 998 empty doors, you state that he did that. Which could mean he happened to do that. Had to vs happened to makes a big difference. Had he opened them at random and by (extremely small) chance they happened to be empty, the odds are quite different.
- mathstuf 3y ago
  There's a variant where Monty trips and opens one of the doors by accident. Then it is 50/50 again (because he could have tripped into the car door). See also this video where cutting off a question in the middle changes the probability because of the information it provides: https://www.youtube.com/watch?v=bDZieLmya_I https://www.youtube.com/watch?v=bDZieLmya_I
- niel 3y ago
  I'm trying to understand why the odds would be different if Monty opened empty doors by chance versus on purpose. Does it really change anything?
  - Scarblac 3y ago
    Yes, it does. You had a 1/1000 chance of picking the right door at once. Monty had a 1/1000 chance of opening only empty doors. He had a 998/1000 chance of opening a door to a car, but that didn't happen. So it's either of the other two cases, with equal probabilities. If he used his knowledge to never open a car, the 998/1000 case wouldn't exist, and there would be a 999/1000 chance that the door he left unopened had the car.
  - full_monty 3y ago
    Yes. If Monty opens the door by chance there are 3 equally likely cases: There's a 1/3 chance you picked the car and Monty shows you a goat, 1/3 chance you picked a goat and Monty shows you the other goat, 1/3 chance you picked a goat and Monty shows you a car. So if Monty shows you a goat you have equal probability of being in one of the first two cases. If Monty doesn't open the door by chance then he never shows you a car. So 2/3 of the time you picked a goat and Monty shows you the other goat.
topaz0 3y ago
TFA also presents this intuitive argument.
fortran77 3y ago
But this is wrong. The odds become 2/3 in the original problem, not 50:50. Your "intuition" isn't helping you.
- topaz0 3y ago
  I think GP is implying that you would reject 50:50 as absurd in this case (answer that rhetorical question with a "no"), and therefore should also reject it in the original. I agree that was not clear, though.
dynm 3y ago
I actually like to back up even further and start with this problem: There are 1000 doors. You pick one. There's a 1 in 1000 chance you get it right. Now, do you want what's behind your door, or what's behind the other 999 doors? https://dynomight.net/2020/09/17/making-the-monty-hall-problem-weirder-but-obvious/ https://dynomight.net/2020/09/17/making-the-monty-hall-probl...
- mihaic 3y ago
  I think this is the clearest explanation I've heard of the principle, kudos!
  - czzr 3y ago
    Ironically this is exactly the explanation vos Savant gave in her original column.
- cynix 3y ago
  Do you really want 998 goats though? It must cost a fortune to transport them home from the studio, finding a space for them to live in, and feed them, etc.
HumblyTossed 3y ago
The way I think about it, in general, is that Monty, by purposely not opening the correct door, is giving you a huge hint and thus changing the game which changes the odds. Think of it similarly to those people who play the same lotto #s all the time because they think eventually it has to hit. Well, no. It's a new game every time. If the lotto commission retired each combination that won each week, then sure. And this is essentially what Monty is doing. He's retiring some doors and changing the odds.
- lelanthran 3y ago
  > And this is essentially what Monty is doing. He's retiring some doors and changing the odds. That still leads to the wrong answer, because he is retiring one of the losers, leaving a 50:50 chance of getting a prize.
  - jakelazaroff 3y ago
    You mean 1/3:2/3, right?
    - lelanthran 3y ago
      That's the correct answer. The wrong answer is 50:50. The act of retiring leads to the wrong answer, so it does not aid in understanding.
hinkley 3y ago
No explanation worked for me. I did a truth table, and said, "...huh." You don't need to model all three door choices. Just say you pick Door A, not Door 1. The first door you didn't pick is B, and C is the other door. Then map the prize behind each of 3 doors, and whether you stay or change your answer. Then count the number of successes for change versus stay. Spoilers: It's 3/6 vs 2/6.
- shadowgovt 3y ago
  My experience was I coded a simulation in Python and got 50/50. ... this revealed a major flaw in my intuition that helped me to grasp it. I had subconsciously been assuming Monty sometimes opens the door revealing the prize instead of always opening a goat door. When I coded that into the simulation, of course it broke (because the player was now playing foolishly, looking at the prize and intentionally choosing the closed door that definitely still had a goat behind it).
  - bemusedthrow75 3y ago
    This is an even better learning experience than coding it right the first time, and I am so happy you had it!
lelanthran 3y ago
I drew a decision tree once to show all possible states, with the 1/3 and 2/3 probabilities along each edge, which finally made it click for the coworkers I was trying to convince. In the tree it becomes blindingly obvious that the probabilities are.
full_monty 3y ago
I never understood why the 1000 door version was more intuitive. Here's how I understand it. There's 1/3 chance of picking the car. 1/3 of the time you pick the car, switch, and lose. 2/3 of the time you pick a goat, switch, and win. Why? Because 2/3 of the time you picked a goat, Monty shows you the other goat, so if you switch you definitely get the car.
- michaelcampbell 3y ago
  username checks out =D I never liked the 10^x other doors version either. My thought was, what say me and a friend play a the same time, and pick the same initial door? If I NEVER change, and he ALWAYS does... I'm still going to win 1/3 of the time, and if Monty always shows the goat, ONE OF US has to win, so if I'm 1/3 by not switching, he HAS to be 2/3 by switching.
michaelcampbell 3y ago
When first confronted with this, I wrote a simulation in GW-Basic on a PC with an orange monochrome monitor and ran it a bunch of times to convince myself switching was the right thing to do, statistically. I came up with another way to visualize it that made more sense than increasing the number of doors, but it requires that one accept the fact that your chances of winning are 1/3 if you don't switch, and you know Monty always shows a losing door.
andrewla 3y ago
Did the host have to open 998 doors, or could he have chosen to just let you open your selected door? If he's "out to get you" then you get very little information here, because he could have just let you open your door. The only reason he'd open the other doors is to try to get you to switch, which means that it's pretty close to 50-50.
DerekL 3y ago
*Monty Hall
mhb 3y ago
Yes. Just like in TFA.

by gerad 3y ago

Karellen 3y ago

What improvement does your 1000-door variant have over the 100-door variant described in the article?

michaelcampbell 3y ago
Bigger numbers make it easier to convince yourself if skeptical. The problem stated with 3 doors is hard, with 5 is still not quite there for many. 100, 100, 10_000_000... becomes almost absurd.
tshaddox 3y ago
The most common unintended interpretation seems to be that Monty Hall is only coincidentally opening a door with a goat behind it. I don't quite understand how people arrive at that interpretation, but the 1,000-door variant should make it even more of a stretch to interpret it as Monty Hall coincidentally doing something so extremely improbable.

mithr 3y ago

I think the reason it's so hard to understand is that we try to break the problem down into more digestible problems, and in this case, if you only consider the last stage of the game, it certainly seems like it should be 50:50 odds. In other words, at the final round, you are presented with two doors. Ignore the fact that you already chose one, and assume you were starting at this point: you enter the game, see two doors, and need to pick the one you think hides a car. So intuitively, your odds of choosing which of two doors has the car are 1/2. It's hard to really understand why it matters that the previous rounds occurred at all — why it matters how you narrowed the selection to these two doors. I don't think the 1000 door version makes that easier to understand, because the same thing is true there — if you come into a game, see two doors, and are asked to choose between them, it's very hard to understand why it matters whether in a previous round there were 2, 3, 10, or 1000 doors — there are only two now, when you make your choice.

rspeele 3y ago
That's especially tempting since we're used to probability problems often being set up where considering previous state is the sucker's answer. E.g. a fair coin has landed on heads three times in a row, what are the odds of it being tails on the next throw?
polygotdomain 3y ago
>It's hard to really understand why it matters that the previous rounds occurred at all It's not hard at all. Humans are not perfectly rational beings. It's not purely about odds, it's about emotions and psychology. In a version where there's no previous selection and it's 50:50, than I pick one and live with it. In the canonical example, the previous selection means that the contestant has already staked their claim, and changing it to the loosing door would have a different emotional response than a simple 50:50 shot with no previous selection. There's a reason why Roulette shows you the last X spins and whether they're odd/even, red/black... because humans make the assumption that the last disconnected data point somehow impacts the current one. You're also glossing over the fact that there was a 1:3 chance you picked the right door to begin with, and 2:3 chance the "other" door was right. The last round isn't 50:50, as you so claim, because there was prior information This is the same reason why the 1000 door example helps explain things; because the math is fundamentally the same, yet significantly more imbalanced. We can also think about how we'd feel about our selection in the 1000 door version, which is likely significantly less confidence, and therefore more likelihood of switching. Whether it's 3 doors, or 1000, the math still say switching is optimal, and our psychology and emotions deal with the choice of switching different in each case.

rspeele 3y ago

The explanation that worked for me was: Suppose after you make your initial choice, instead of opening a door, Monty simply asks if you'd like to switch to BOTH the other two doors, such that you win if the prize is behind EITHER of them. That switch is intuitively a great deal, giving you 2/3 odds. The only way you can lose is in the 1/3rd case where you already picked a winner. The original scenario is equivalent to this, since by revealing a goat Monty is allowing you to pick "the best of" the two doors. --- Often people who don't like this problem complain that actually Monty, with his knowledge of the winning door, may be trying to trick you into losing. Maybe he offers this trade conditionally based on whether you've already selected a winner. Of course, this wouldn't support the most common intuitive answer of "it's 50/50", either, making this a bad excuse. If we accept this behavior from Monty -- not stated in the problem -- it becomes a very uninteresting problem based on whether we are dealing with Evil Monty who only offers the trade when it's a bad one (stay wins 100% of the time), Friendly Monty who only offers the trade when it's a good one (switch wins 100% of the time), or somewhere in the middle. It has no analytical answer. So for this to be a solvable problem at all we must assume we are dealing with Fair Monty who offers the trade all the time, or at least, offers it at random times not based on the contestant's initial pick.

mithr 3y ago
I do think this modified scenario makes the solution much clearer. At the same time, I also think that it feels like a significantly-enough different scenario from the original that saying the two scenarios have the same probability then becomes the non-intuitive part. The act of revealing what's behind the second door seems like it should change the probability from 1/3 (one door) vs. 2/3 (one of two doors) to 1/2 (one door of the remaining unopened two) vs. 1/2 (one door of the remaining unopened two, since one was "eliminated"). It's amazing how even seeing the probabilities written out, or running simulations, doesn't really make it easier to truly understand the result.
mcv 3y ago
I think the important part of the problem is the unstated assumptions around it, and I suspect some people may see different assumptions than others. If Monty always opens a door, and uses his knowledge of which door has the prize to ensure that the door he opens is always empty, then you should switch, because he's providing you with extra information about which door has the prize. However, if Monty changes it up and sometimes doesn't open a door, or opens the door with the prize, the whole problem changes. If he only opens a door and offers you the chance to switch when you've picked the door with the prize, you should never switch of course. If he opens a random door and might reveal the prize (after which he obviously won't let you switch anymore), switching doesn't change your odds. I think a lot of people see it as one of those two situations.

tromp 3y ago

Your formulation is once again ambiguous. Instead of clearly stating that Monty had to open 998 empty doors, you state that he did that. Which could mean he happened to do that. Had to vs happened to makes a big difference. Had he opened them at random and by (extremely small) chance they happened to be empty, the odds are quite different.

mathstuf 3y ago
There's a variant where Monty trips and opens one of the doors by accident. Then it is 50/50 again (because he could have tripped into the car door). See also this video where cutting off a question in the middle changes the probability because of the information it provides: https://www.youtube.com/watch?v=bDZieLmya_I https://www.youtube.com/watch?v=bDZieLmya_I
niel 3y ago
I'm trying to understand why the odds would be different if Monty opened empty doors by chance versus on purpose. Does it really change anything?
- Scarblac 3y ago
  Yes, it does. You had a 1/1000 chance of picking the right door at once. Monty had a 1/1000 chance of opening only empty doors. He had a 998/1000 chance of opening a door to a car, but that didn't happen. So it's either of the other two cases, with equal probabilities. If he used his knowledge to never open a car, the 998/1000 case wouldn't exist, and there would be a 999/1000 chance that the door he left unopened had the car.
- full_monty 3y ago
  Yes. If Monty opens the door by chance there are 3 equally likely cases: There's a 1/3 chance you picked the car and Monty shows you a goat, 1/3 chance you picked a goat and Monty shows you the other goat, 1/3 chance you picked a goat and Monty shows you a car. So if Monty shows you a goat you have equal probability of being in one of the first two cases. If Monty doesn't open the door by chance then he never shows you a car. So 2/3 of the time you picked a goat and Monty shows you the other goat.

topaz0 3y ago

TFA also presents this intuitive argument.

fortran77 3y ago

But this is wrong. The odds become 2/3 in the original problem, not 50:50. Your "intuition" isn't helping you.

topaz0 3y ago
I think GP is implying that you would reject 50:50 as absurd in this case (answer that rhetorical question with a "no"), and therefore should also reject it in the original. I agree that was not clear, though.

dynm 3y ago

I actually like to back up even further and start with this problem: There are 1000 doors. You pick one. There's a 1 in 1000 chance you get it right. Now, do you want what's behind your door, or what's behind the other 999 doors? https://dynomight.net/2020/09/17/making-the-monty-hall-problem-weirder-but-obvious/ https://dynomight.net/2020/09/17/making-the-monty-hall-probl...

mihaic 3y ago
I think this is the clearest explanation I've heard of the principle, kudos!
- czzr 3y ago
  Ironically this is exactly the explanation vos Savant gave in her original column.
cynix 3y ago
Do you really want 998 goats though? It must cost a fortune to transport them home from the studio, finding a space for them to live in, and feed them, etc.

HumblyTossed 3y ago

The way I think about it, in general, is that Monty, by purposely not opening the correct door, is giving you a huge hint and thus changing the game which changes the odds. Think of it similarly to those people who play the same lotto #s all the time because they think eventually it has to hit. Well, no. It's a new game every time. If the lotto commission retired each combination that won each week, then sure. And this is essentially what Monty is doing. He's retiring some doors and changing the odds.

lelanthran 3y ago
> And this is essentially what Monty is doing. He's retiring some doors and changing the odds. That still leads to the wrong answer, because he is retiring one of the losers, leaving a 50:50 chance of getting a prize.
- jakelazaroff 3y ago
  You mean 1/3:2/3, right?
  - lelanthran 3y ago
    That's the correct answer. The wrong answer is 50:50. The act of retiring leads to the wrong answer, so it does not aid in understanding.

hinkley 3y ago

No explanation worked for me. I did a truth table, and said, "...huh." You don't need to model all three door choices. Just say you pick Door A, not Door 1. The first door you didn't pick is B, and C is the other door. Then map the prize behind each of 3 doors, and whether you stay or change your answer. Then count the number of successes for change versus stay. Spoilers: It's 3/6 vs 2/6.

shadowgovt 3y ago
My experience was I coded a simulation in Python and got 50/50. ... this revealed a major flaw in my intuition that helped me to grasp it. I had subconsciously been assuming Monty sometimes opens the door revealing the prize instead of always opening a goat door. When I coded that into the simulation, of course it broke (because the player was now playing foolishly, looking at the prize and intentionally choosing the closed door that definitely still had a goat behind it).
- bemusedthrow75 3y ago
  This is an even better learning experience than coding it right the first time, and I am so happy you had it!

lelanthran 3y ago

I drew a decision tree once to show all possible states, with the 1/3 and 2/3 probabilities along each edge, which finally made it click for the coworkers I was trying to convince. In the tree it becomes blindingly obvious that the probabilities are.

full_monty 3y ago

I never understood why the 1000 door version was more intuitive. Here's how I understand it. There's 1/3 chance of picking the car. 1/3 of the time you pick the car, switch, and lose. 2/3 of the time you pick a goat, switch, and win. Why? Because 2/3 of the time you picked a goat, Monty shows you the other goat, so if you switch you definitely get the car.

michaelcampbell 3y ago
username checks out =D I never liked the 10^x other doors version either. My thought was, what say me and a friend play a the same time, and pick the same initial door? If I NEVER change, and he ALWAYS does... I'm still going to win 1/3 of the time, and if Monty always shows the goat, ONE OF US has to win, so if I'm 1/3 by not switching, he HAS to be 2/3 by switching.

michaelcampbell 3y ago

When first confronted with this, I wrote a simulation in GW-Basic on a PC with an orange monochrome monitor and ran it a bunch of times to convince myself switching was the right thing to do, statistically. I came up with another way to visualize it that made more sense than increasing the number of doors, but it requires that one accept the fact that your chances of winning are 1/3 if you don't switch, and you know Monty always shows a losing door.

andrewla 3y ago

Did the host have to open 998 doors, or could he have chosen to just let you open your selected door? If he's "out to get you" then you get very little information here, because he could have just let you open your door. The only reason he'd open the other doors is to try to get you to switch, which means that it's pretty close to 50-50.

DerekL 3y ago

*Monty Hall

mhb 3y ago

Yes. Just like in TFA.