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Can you help me understand - in my view the choice to switch doors or keep the same door is irrelevant, because even if you keep the same door you're making a c

Can you help me understand - in my view the choice to switch doors or keep the same door is irrelevant, because even if you keep the same door you're making a choice that is now 2/3 of the right answer. If you switch or keep, you're still choosing from two doors that contain a car and a goat. The other door is no longer relevant and doesn't affect the new state at all. It's your perspective (narrowing the choice down) that changes, not the position of the car. Like getting a run of red in roulette and thinking the next one has gotta be black.

ronyeh 5y ago
Think from the perspective of your first choice. You had a 66.66% chance of choosing the wrong door. So now I just eliminate one of the doors you didn't choose. There are two left, including the one you chose first, which only had a 33.33% chance of being correct. Nothing else has changed about the problem. What if we started with a thousand doors? You chose 1, with a 999/1000 chance of being wrong. Of the remaining doors, I open 499 doors. The prize is behind one of the remaining 500 that you did not choose the first time. Do you want to have a chance to pick from one of the 500?
- ziml77 5y ago
  I like expanding it out to 1000 doors. Even though the logic is the same between 3 and 1000 doors, the chances are more intuitive with more doors. I think the part that really throws the brain off with the 3 door version is that the host opening a door seems like another random event but it's actually dependent on the state of the game at that point. Our natural intuition of the chances is for a version of the game where you pick a door, then the host randomly picks (and does not open as you also haven't) a different door, and then you have to choose if you want to switch to the door that neither of you picked.
TacticalCoder 5y ago
> The other door is no longer relevant and doesn't affect the new state at all. There is no new state, that's the thing. By picking one door you split these in two groups: one group with 1/3 to win, the other with 2/3 to win. When the host reveals the door with a goat behind it from the group that is 2/3 to win, it doesn't change the fact that that group had 2/3 to win to begin with. I mean: that's how I see it.
- bordercases 5y ago
  It changes what you know. States are measured in the mind.
6gvONxR4sf7o 5y ago
I wrote out a hopefully helpful intution down below [0]. Here's a terse part of it: > Or just imagine monty hall with infinite doors. I'm thinking of a number between 1 and infinity (secretly, it's 19083412039102388171230123). You pick a number, and then I'll narrow down your choice to two options. Do you think you just happened to pick my number, or do you switch? You're right that it's absolutely the narrowing down of the choice that's the key. If you happen to have not picked the prize door on your first try (more likely than not), the narrowing down will be your door and the prize door. [0] https://news.ycombinator.com/item?id=27092614 https://news.ycombinator.com/item?id=27092614
- saeranv 5y ago
  This is the best explanation I've read thus far.
- megablast 5y ago
  Ok. But you picked a very very very very small number between 1 and infinity. That can’t be random.
  - 6gvONxR4sf7o 5y ago
    Tons of common distributions have infinite support, but the details don't matter for this bit of intuition. If you want to formalize it in a simpler way, just consider it as an arbitrarily large finite number of doors with a uniform distribution.
  - CRConrad 5y ago
    That it isn't random is precisely the point, isn't it? The only difference that the "very very very very small number" makes is that the odds aren't infinity to one but two to one. Still winning odds.
zeroimpl 5y ago
The other door is not irrelevant. When you first picked a door, there was a 33% chance it was right, and a 67% chance one of the other two doors was right. Once the other door got opened, it is still a 67% chance that the other two doors is right, but you now know which one of those two it would be - the one which wasn’t opened.
- lachstar-x 5y ago
  Thats what I'm stuck on, I suppose. Imagine the entire situation is reversed. The Reverse Monty Hall problem. I'm given a choice between two doors, once of which contains a car and one contains a goat. I have to choose, 1 or 2. Then the game show host reveals that there was also a third door, which contained a goat, which is no longer relevant and never was relevant. I'm then also asked to choose a door (which is also irrelevant since the problem is backwards and the supposed aim is to get the car). Even if that last step repeats 1000 times with 1000 doors and 1 car, each removing a goat-door, the only relevant choice is still the first one as the host appears to be adding new information, but it's always irrelevant information as a new choice is always made at the end.
  - stordoff 5y ago
    In your reverse problem, I don't see how the new door can ever be relevant, as you know the car is behind 1 or 2, so you're only ever interested in picking between those 2. In the original problem, its presence _is_ relevant, as the car could be behind doors 1, 2, or 3. Say you pick door 1 - there's a 1/3 chance you are right. The host is then left with doors 2 and 3. We know there is a 2/3 chance the car is behind _one_ of these doors. When the presenter reveals a goat (say in door 2), he is reveling information about this set of doors - there is still a 2/3 chance that the car is behind one of the doors in this set, but there's only one door left we don't know anything about (3). There is therefore a 2/3 chance that it is behind _this_ door.
  - jarpadat 5y ago
    I think it boils down to the fact that time isn’t reversible, so the “reverse problem” isn’t like the “forward” version. Let’s take some new problems. Suppose you have 100 doors and no switching. Your probability is 1/100, even if the host later opens a goat door, so in that sense the new information is irrelevant. But if we “reverse it” and the host opens the door first and you guess second, your probability improves to 1/99. So now suddenly the same information is relevant. Two things to observe here, one is that the forward and reverse problems are different, the other is that the relevance or irrelevance of the information depends on the direction of time. If you learn the information before you act it is relevant, afterward it is irrelevant. One way to think about Monty Hall is you’re deciding which of these games to play. If you will stick with your first decision, you are sorta turning it into the toy problem above, where you decide the door first and then the goat information is irrelevant. Vs if you will switch, the goat door is opened before you decide, which is relevant. Another way to think about it is with two contestants. Let’s say I pick the door initially, then someone opens the goat door, and finally you decide whether to switch. In this scenario, you don’t have self-preference bias to stick with “my” original door. In fact, my decision was the irrelevant information. It doesn’t matter at all what door I picked, what matters is whether you pick the right door, and involving me at all is a kind of misdirection to anchor you to the 1/3 probability.
marcosdumay 5y ago
It's the presenter sharing knowledge with you, by opening a door that isn't the one you choose nor one that has the prize that changes the odds. And it doesn't help you if you stay with the same door because he never opens your door anyway, so it only adds knowledge about the other doors.
lamontcg 5y ago
Look at it from actual door, first guess door pairs -- which are the random uncorrelated events: 1,1 - switch loses (host may have shown door #2 or #3) 1,2 - switch wins (host showed door #3) 1,3 - switch wins (host showed door #2) 2,1 - switch wins (host showed door #3) 2,2 - switch loses (host may have shown door #1 or #3) 2,3 - switch wins (host showed door #1) 3,1 - switch wins (host showed door #2) 3,2 - switch wins (host showed door #1) 3,3 - switch loses (host may have shown door #1 or #2) The choice of which door to show you is not a random event and is highly dependent upon the events which preceded it, because the host cannot show you the door it is actually behind, or the door that you picked.
- edoceo 5y ago
  incredibly succinct, thanks.
anyonecancode 5y ago
Maybe someone can help me with this -- what if for round two, you flipped a coin: heads you pick the door you picked previously, tails you pick the other door. Does that change things at all, since you're "switching" each time? -- Edited -- Actually @haberman's comment that "By opening a door, Monty lets you cover 100% of the "everything else" makes the most sense to me as I think about this. Imagine there is no Monty hall, just three doors, and I say "you can choose one door and you win if there's a car behind it, or choose two doors and see if there's a car behind it." Clearly you're better off choosing two doors. And that's in fact, functionally what happens by switching after a door is eliminated.
stordoff 5y ago
When you first pick a door, you have a 1/3 chance of it being the right door. There's a 2/3 chance of it being behind a door you didn't pick. When the host then opens a door, there's still a 2/3 chance that it is behind one of the doors you didn't pick. However, there's now only one door in this set, so there's 2/3 chance that it's behind _that_ door. To look at another way, imagine if the host didn't reveal the content of the door, but gave you the option to switch to BOTH of the other doors instead of your door. Your odds of winning clearly go up, as you now have two chances to win (and all are of equal probability). That's equivalent to what's happening here. By showing you the losing door of those two, he doesn't change anything - there's still twice the chance it was behind one of the doors you didn't pick compared to the one you did, and by switching you win if it was behind either of them.
- ledauphin 5y ago
  I've seen this explained ten different ways but this is the way I'm going to explain it from now on. thank you.
rfrey 5y ago
The trick is that the host does not pick a door to open at random... If they did there would be a 1/3 probability they opened your door. If they didn't open yours under that circumstance, there would be a 50/50 chance you already picked the good door. But it's not random... Monty will never open the door you chose. That's what messes up the intuitive probabilities.
- billforsternz 5y ago
  Not only will he not open the door you choose, he won't open the door in front of the car either. I think that's the key point that confuses people. He knows where the car is and he always exposes a goat instead.
admax88q 5y ago
Here's another thought experiment. Forget the opening of the door. Start with picking a door at random, you have a 1/3 chance of having picked the car. On that I think we all agree. Now let's say that the host offers to let you switch from the door you picked, to the other _two_ doors combined. He hasn't opened any doors, they are all closed, you're allowed to stick with your initial guess of one door, or switch to a combined guess of the other two doors. If that's the case it should be fairly obvious that you have a 2/3 chance of getting the car by switching to the combined 2 doors. Now that you've switched, would it really make a difference to your odds if the host opens one of your doors to reveal a goat? Would that lower your probably to 1/2 or would it remain 2/3?
- ikt 5y ago
  Intuitively it makes sense it would be half, because there was a 33% chance of winning but now it's 50/50 because 3-1 = 2, but when you read it logically it makes sense. Yours is the first explanation I 'got', thanks!
  - admax88q 5y ago
    Glad it helped, I just came up with it and it made it click more for me.
- slavik81 5y ago
  It might help to have a contrasting example to clarify the key details. Let's say we buy four scratch-off tickets. The clerk selling the tickets assures us that one of the four is a winner. You choose two tickets, and I take the remaining two. I scratch one of my tickets, revealing that it is a dud. At this point, I suggest we should swap your two tickets for my two tickets. You start to object, but I point out that this is just like the Monty Hall problem. When we made our initial choice, there was a 2/4 chance the winning ticket was in the set you picked, and a 2/4 chance that it was in the set that I picked. The fact that I revealed a ticket didn't change that. So you should be willing to trade, as both sets of tickets are equally likely to contain the winner. You refuse, obviously. The difference is that whether there's two goats behind his doors or one, Monty will always choose a goat. He provides no information about the door you've already chosen, as the probability of him choosing a goat is unchanged based on whether you picked correct or not. I, on the other hand, could have revealed the winning ticket, but the odds of me doing so would be lower if you already had it in your possession. So, my dud updates the probability of all remaining tickets.

ronyeh 5y ago

Think from the perspective of your first choice. You had a 66.66% chance of choosing the wrong door. So now I just eliminate one of the doors you didn't choose. There are two left, including the one you chose first, which only had a 33.33% chance of being correct. Nothing else has changed about the problem. What if we started with a thousand doors? You chose 1, with a 999/1000 chance of being wrong. Of the remaining doors, I open 499 doors. The prize is behind one of the remaining 500 that you did not choose the first time. Do you want to have a chance to pick from one of the 500?

ziml77 5y ago
I like expanding it out to 1000 doors. Even though the logic is the same between 3 and 1000 doors, the chances are more intuitive with more doors. I think the part that really throws the brain off with the 3 door version is that the host opening a door seems like another random event but it's actually dependent on the state of the game at that point. Our natural intuition of the chances is for a version of the game where you pick a door, then the host randomly picks (and does not open as you also haven't) a different door, and then you have to choose if you want to switch to the door that neither of you picked.

TacticalCoder 5y ago

> The other door is no longer relevant and doesn't affect the new state at all. There is no new state, that's the thing. By picking one door you split these in two groups: one group with 1/3 to win, the other with 2/3 to win. When the host reveals the door with a goat behind it from the group that is 2/3 to win, it doesn't change the fact that that group had 2/3 to win to begin with. I mean: that's how I see it.

bordercases 5y ago
It changes what you know. States are measured in the mind.

6gvONxR4sf7o 5y ago

I wrote out a hopefully helpful intution down below [0]. Here's a terse part of it: > Or just imagine monty hall with infinite doors. I'm thinking of a number between 1 and infinity (secretly, it's 19083412039102388171230123). You pick a number, and then I'll narrow down your choice to two options. Do you think you just happened to pick my number, or do you switch? You're right that it's absolutely the narrowing down of the choice that's the key. If you happen to have not picked the prize door on your first try (more likely than not), the narrowing down will be your door and the prize door. [0] https://news.ycombinator.com/item?id=27092614 https://news.ycombinator.com/item?id=27092614

saeranv 5y ago
This is the best explanation I've read thus far.
megablast 5y ago
Ok. But you picked a very very very very small number between 1 and infinity. That can’t be random.
- 6gvONxR4sf7o 5y ago
  Tons of common distributions have infinite support, but the details don't matter for this bit of intuition. If you want to formalize it in a simpler way, just consider it as an arbitrarily large finite number of doors with a uniform distribution.
- CRConrad 5y ago
  That it isn't random is precisely the point, isn't it? The only difference that the "very very very very small number" makes is that the odds aren't infinity to one but two to one. Still winning odds.

zeroimpl 5y ago

The other door is not irrelevant. When you first picked a door, there was a 33% chance it was right, and a 67% chance one of the other two doors was right. Once the other door got opened, it is still a 67% chance that the other two doors is right, but you now know which one of those two it would be - the one which wasn’t opened.

lachstar-x 5y ago
Thats what I'm stuck on, I suppose. Imagine the entire situation is reversed. The Reverse Monty Hall problem. I'm given a choice between two doors, once of which contains a car and one contains a goat. I have to choose, 1 or 2. Then the game show host reveals that there was also a third door, which contained a goat, which is no longer relevant and never was relevant. I'm then also asked to choose a door (which is also irrelevant since the problem is backwards and the supposed aim is to get the car). Even if that last step repeats 1000 times with 1000 doors and 1 car, each removing a goat-door, the only relevant choice is still the first one as the host appears to be adding new information, but it's always irrelevant information as a new choice is always made at the end.
- stordoff 5y ago
  In your reverse problem, I don't see how the new door can ever be relevant, as you know the car is behind 1 or 2, so you're only ever interested in picking between those 2. In the original problem, its presence _is_ relevant, as the car could be behind doors 1, 2, or 3. Say you pick door 1 - there's a 1/3 chance you are right. The host is then left with doors 2 and 3. We know there is a 2/3 chance the car is behind _one_ of these doors. When the presenter reveals a goat (say in door 2), he is reveling information about this set of doors - there is still a 2/3 chance that the car is behind one of the doors in this set, but there's only one door left we don't know anything about (3). There is therefore a 2/3 chance that it is behind _this_ door.
- jarpadat 5y ago
  I think it boils down to the fact that time isn’t reversible, so the “reverse problem” isn’t like the “forward” version. Let’s take some new problems. Suppose you have 100 doors and no switching. Your probability is 1/100, even if the host later opens a goat door, so in that sense the new information is irrelevant. But if we “reverse it” and the host opens the door first and you guess second, your probability improves to 1/99. So now suddenly the same information is relevant. Two things to observe here, one is that the forward and reverse problems are different, the other is that the relevance or irrelevance of the information depends on the direction of time. If you learn the information before you act it is relevant, afterward it is irrelevant. One way to think about Monty Hall is you’re deciding which of these games to play. If you will stick with your first decision, you are sorta turning it into the toy problem above, where you decide the door first and then the goat information is irrelevant. Vs if you will switch, the goat door is opened before you decide, which is relevant. Another way to think about it is with two contestants. Let’s say I pick the door initially, then someone opens the goat door, and finally you decide whether to switch. In this scenario, you don’t have self-preference bias to stick with “my” original door. In fact, my decision was the irrelevant information. It doesn’t matter at all what door I picked, what matters is whether you pick the right door, and involving me at all is a kind of misdirection to anchor you to the 1/3 probability.

marcosdumay 5y ago

It's the presenter sharing knowledge with you, by opening a door that isn't the one you choose nor one that has the prize that changes the odds. And it doesn't help you if you stay with the same door because he never opens your door anyway, so it only adds knowledge about the other doors.

lamontcg 5y ago

Look at it from actual door, first guess door pairs -- which are the random uncorrelated events: 1,1 - switch loses (host may have shown door #2 or #3) 1,2 - switch wins (host showed door #3) 1,3 - switch wins (host showed door #2) 2,1 - switch wins (host showed door #3) 2,2 - switch loses (host may have shown door #1 or #3) 2,3 - switch wins (host showed door #1) 3,1 - switch wins (host showed door #2) 3,2 - switch wins (host showed door #1) 3,3 - switch loses (host may have shown door #1 or #2) The choice of which door to show you is not a random event and is highly dependent upon the events which preceded it, because the host cannot show you the door it is actually behind, or the door that you picked.

edoceo 5y ago
incredibly succinct, thanks.

anyonecancode 5y ago

Maybe someone can help me with this -- what if for round two, you flipped a coin: heads you pick the door you picked previously, tails you pick the other door. Does that change things at all, since you're "switching" each time? -- Edited -- Actually @haberman's comment that "By opening a door, Monty lets you cover 100% of the "everything else" makes the most sense to me as I think about this. Imagine there is no Monty hall, just three doors, and I say "you can choose one door and you win if there's a car behind it, or choose two doors and see if there's a car behind it." Clearly you're better off choosing two doors. And that's in fact, functionally what happens by switching after a door is eliminated.

stordoff 5y ago

When you first pick a door, you have a 1/3 chance of it being the right door. There's a 2/3 chance of it being behind a door you didn't pick. When the host then opens a door, there's still a 2/3 chance that it is behind one of the doors you didn't pick. However, there's now only one door in this set, so there's 2/3 chance that it's behind _that_ door. To look at another way, imagine if the host didn't reveal the content of the door, but gave you the option to switch to BOTH of the other doors instead of your door. Your odds of winning clearly go up, as you now have two chances to win (and all are of equal probability). That's equivalent to what's happening here. By showing you the losing door of those two, he doesn't change anything - there's still twice the chance it was behind one of the doors you didn't pick compared to the one you did, and by switching you win if it was behind either of them.

ledauphin 5y ago
I've seen this explained ten different ways but this is the way I'm going to explain it from now on. thank you.

rfrey 5y ago

The trick is that the host does not pick a door to open at random... If they did there would be a 1/3 probability they opened your door. If they didn't open yours under that circumstance, there would be a 50/50 chance you already picked the good door. But it's not random... Monty will never open the door you chose. That's what messes up the intuitive probabilities.

billforsternz 5y ago
Not only will he not open the door you choose, he won't open the door in front of the car either. I think that's the key point that confuses people. He knows where the car is and he always exposes a goat instead.

admax88q 5y ago

Here's another thought experiment. Forget the opening of the door. Start with picking a door at random, you have a 1/3 chance of having picked the car. On that I think we all agree. Now let's say that the host offers to let you switch from the door you picked, to the other _two_ doors combined. He hasn't opened any doors, they are all closed, you're allowed to stick with your initial guess of one door, or switch to a combined guess of the other two doors. If that's the case it should be fairly obvious that you have a 2/3 chance of getting the car by switching to the combined 2 doors. Now that you've switched, would it really make a difference to your odds if the host opens one of your doors to reveal a goat? Would that lower your probably to 1/2 or would it remain 2/3?

ikt 5y ago
Intuitively it makes sense it would be half, because there was a 33% chance of winning but now it's 50/50 because 3-1 = 2, but when you read it logically it makes sense. Yours is the first explanation I 'got', thanks!
- admax88q 5y ago
  Glad it helped, I just came up with it and it made it click more for me.
slavik81 5y ago
It might help to have a contrasting example to clarify the key details. Let's say we buy four scratch-off tickets. The clerk selling the tickets assures us that one of the four is a winner. You choose two tickets, and I take the remaining two. I scratch one of my tickets, revealing that it is a dud. At this point, I suggest we should swap your two tickets for my two tickets. You start to object, but I point out that this is just like the Monty Hall problem. When we made our initial choice, there was a 2/4 chance the winning ticket was in the set you picked, and a 2/4 chance that it was in the set that I picked. The fact that I revealed a ticket didn't change that. So you should be willing to trade, as both sets of tickets are equally likely to contain the winner. You refuse, obviously. The difference is that whether there's two goats behind his doors or one, Monty will always choose a goat. He provides no information about the door you've already chosen, as the probability of him choosing a goat is unchanged based on whether you picked correct or not. I, on the other hand, could have revealed the winning ticket, but the odds of me doing so would be lower if you already had it in your possession. So, my dud updates the probability of all remaining tickets.