I've thought up a new (I think) clear way to explain the 'Monty Hall Problem"

[Little_Nemo]

@thorny_locust

Aaaahhhh.

That’s the only one that makes any sense to me at all.

Increasing the number of doors doesn’t make the problem and its claimed solution make any more sense to me than it does with three doors. Neither does the information that the person who opened a door knows what’s behind it. But being told that one’s getting to pick two out of three –

only appears to help very briefly, until I stop and think about it. Because I’m still picking two out of three if I stick with the door that I started with.

Suppose you start the problem up from scratch at that point: there are three doors, you know something you want is behind one of them, you know that it isn’t behind another, you don’t know which of the two remaining doors it’s behind. I understand that there’s a 50/50 chance of picking the right door out of two. I understand that if I don’t know what’s behind any of three doors there’s a 1 in 3 chance of picking the right door out of three. I understand that 50/50 is better odds than 1 in 3.

What I don’t understand is why it’s not supposed to still be 50/50 between the two remaining doors, giving no advantage to either switching or staying with the earlier choice, just because of a choice made previously. If it would be 50/50 if the problem started there, why is it not still 50/50?

You’re heading off in the wrong direction. Your logical assumptions are not right.

Focus on the simple scenario I described. Don’t think about other versions of this. There are three doors. One has a prize behind it. The other two have nothing. You get to pick one door. Then you’re offered a choice; do you want to get what’s behind that one door or do you want to get everything that’s behind the other two doors?

Just focusing on this scenario, do you see why picking what’s behind two doors gives you a better chance than picking what’s behind one door? If you see this, move on to the next scenario.

Start with the same scenario as above. And when you were given a choice, you chose to take what was behind the two doors because you saw it was better odds for you. A guy shows up to open the doors. He opens each door just a crack, so he can see what’s inside but you can’t. After he’s done this for all three doors, he tells you “I’m going to drag this out as long as possible because I’m annoying. There are three doors and only one door has a prize. You picked two of the doors. So at least one of the doors you picked must have nothing behind it. I’m now going to open an empty door that you picked.” He opens up one of the doors you picked and reveals that there is indeed no prize behind it.

Do you follow this scenario? Do you see how it works regardless of where the prize was? Regardless of how you picked? And do you see that - and this is important - it doesn’t change the odds you figured out above? When you’ve got this, move on.

This one starts out the same. There are three doors. One has a prize behind it. The other two have nothing. You get to pick one door. The guy shows up and looks behind each door so he can see what’s inside but you can’t. After he’s done this for all three doors, he tells you “I’m going to drag this out as long as possible because I’m annoying. There are three doors and only one door has a prize. You picked one door. So at least one of the other two doors you that didn’t picked must have nothing behind it. I’m now going to open an empty door that you didn’t pick.” He opens up one of the doors you didn’t pick and reveals that there is indeed no prize behind it. Now you’re offered the choice. Do you want to get what’s behind the one door that you initially picked or do you want to get everything that’s behind the other two doors, even though you have seen that one of those two doors has nothing behind it?

Do you follow this scenario? Do you see how the odds are still the same here as they were in the first scenario? Do you understand why picking the two doors is still better odds than picking the one door?

[Munch]

I also like the 100 doors - I think it helps people visualize it. I also like to walk through what happens if there’s a contestant for each door.

Initial Guess	Prize In	Choose Between	“I’ll Stay, Monty!”	“I’ll Switch Doors!”
Door 1	Door 3	Door 1 & Door 3	Lose	Win
Door 2	Door 3	Door 2 & Door 3	Lose	Win
Door 3	Door 3	any 2 losing doors	Win	Lose
Door 4	Door 3	Door 4 & Door 3	Lose	Win
Door 5	Door 3	Door 5 & Door 3	Lose	Win
Door 6	Door 3	Door 6 & Door 3	Lose	Win
Door 7	Door 3	Door 7 & Door 3	Lose	Win
Door 8	Door 3	Door 8 & Door 3	Lose	Win
Door 9	Door 3	Door 9 & Door 3	Lose	Win
Door 10	Door 3	Door 10 & Door 3	Lose	Win

I think it makes it easier to see that the only way you lose by switching, is if you pick correctly the first time.

[thorny_locust]

@Riemann

You have to remember that there is information not just in what actually happened, but in the range of things that are ALLOWED to happen under the rules, including things that were allowed but which do not actually occur.

If you pick A initially, Monty is NOT ALLOWED to open A under the rules, simply because you chose it, not because of what it contains.

But door C is ALLOWED to be opened under the rules, unless it contains the car. The fact that Monty actually opens door B, when he was ALLOWED to open door C but did not do so, gives you information about C that you do not have about A.

OK. That does make some sense to me.

@EllisDee

Do this a few times and they will start to see that it is not, in fact, a 50/50 split between the final two cards.

A few times doesn’t mean much, because a short run’s not necessarily going to give a 50/50 split, or anything like it. You’d have to be there for quite a while for it to start to look like it’s really showing something.

@Riemann

There’s a 1/3 chance the car is behind door A when you pick it. You always know that there is at least one goat among B and C, so if someone who knows what’s behind all the doors DELIBERATELY (not just by picking at random) shows you a goat among one of B and C, you have no new information. You still have a 1/3 chance of winning if you stick with A

That’s actually more convincing than telling me that I am being given new information – except that you argued in your previous post that I do have new information.

@Little_Nemo

You get to pick one door. Then you’re offered a choice; do you want to get what’s behind that one door or do you want to get everything that’s behind the other two doors?

That one actually might be convincing.

@Little_Nemo

Do you want to get what’s behind the one door that you initially picked or do you want to get everything that’s behind the other two doors, even though you have seen that one of those two doors has nothing behind it?

– I think what’s going on in my head is that when I look at that one, I still see 50/50, because the reason for choosing two-out-of-three in the previous round would be because either door two or door three might have the prize; while I’d know one of them was a blank, I wouldn’t know which. I think I’m imagining the doors as specific doors – did Monty hide it in the back hall, or in the basement, or on the porch? Ah, now I know it’s not the porch, so it must be either the back hall or the basement, but it could be either – whereas all the logicians are seeing it as an abstraction.

On the abstract level I can see, I think, that you’re making sense. But in practice I still feel like I’d have an equal chance heading for the basement or the back hall, no matter which one I’d picked first.

Oh. OK. Maybe I am seeing it. If I picked the basement, and it’s actually in the basement, Monty might pick either the porch or the back hall. If I picked the basement, and it’s in the back hall, he’s got to pick the porch.

Turn it around. If I picked the back hall, and it’s in the basement, he has to pick the porch; if I picked the back hall, and it’s in the back hall, he could pick either the porch or the basement.

What needs to be looked at, I think, isn’t my chances of having guessed right. It’s Monty’s chances of picking a particular door.

– If I picked the porch, and it’s in the basement, he has to pick the back hall; or, if it’s in the back hall, he has to pick the basement. If it’s in the porch, he can pick either one. But he’s going to be limited if it’s not in the one that I guessed, in a way in which he’s not limited if it is in the one that I guessed.

[Little_Nemo]

You can see the advantage of picking two doors instead of one door. Obviously two doors doubles your chances of picking the door with the prize behind it.

You know that only one door has a prize. So when you pick two doors, you know you’re picking at least one door that doesn’t have a prize. Monty knows what’s behind all of the doors; when he opens one of the two doors and shows you there’s no prize behind it, he isn’t telling you anything you didn’t already know.

So the fact that your odds are better picking two doors instead of one isn’t changed when one of the two doors is opened.

[EllisDee]

@thorny_locust

A few times doesn’t mean much, because a short run’s not necessarily going to give a 50/50 split, or anything like it. You’d have to be there for quite a while for it to start to look like it’s really showing something.

It’s not a 50/50 split, it’s a 98/2 split, and it will become incredibly obvious why after the second or third round.

The part you quoted was describing what will happen when you do the whole deck variant, where the skeptic picks a card, then you show them 50 cards that aren’t the ace of spades and ask them to pick between their original choice and the ace of spades you pulled out of the remainder of the deck.

Go grab a deck of cards right now and try it 3 times and you’ll see. Here’s a quote for the specifics without having to scroll up:

Have them pick a card, any card. Now fan the remaining 51 cards (not showing them) and pull out the ace spades, but keep it covered. Now flip over the 50 loser cards, then say “You can stick with your first choice, or switch to this covered card here.”

[TheBoltEater]

Others have chimed in, but to answer your reply to me -

Yes, you DO get 2 other doors - that’s why it’s 2 out of 3 to switch. And you already know that one of those doors has a goat, so when Monty shows you a goat he’s telling you nothing. Maybe think about it this way - what if he offered you to switch before he opened the door? You can have Door A or Doors B and C. Of course you choose B and C. Then he says, “But wait! What if I told you that at least one of those two doors has a goat?!” Well, yeah, of course it does - they can’t both have the car.

Opening the door changes nothing, period. It just feels like it does.

[thorny_locust]

I haven’t the faintest idea what will convince other people who have difficulty with this problem.

But what convinced me was turning the thing around to look not at my choices, but at Monty’s choices. If I didn’t pick the right door the first time, Monty has no choice; there’s only one door he can pick. If I did pick the right door the first time, Monty has a choice – but, presuming he chooses randomly, there’s a 50% chance that he’ll pick the same door that he will if he hasn’t got a choice. Therefore, overall there’s a greater chance that he’ll pick that door than the other one. I was picking at random: but Monty isn’t.

I thought I said that already; but as people seem to still be trying to convince me using techniques that didn’t work for me (the card trick version might convince me that it works, but would still leave me not understanding why), maybe I need to try again.

[Ynnad]

@Little_Nemo

You know that only one door has a prize.

I thought all three doors had a prize. It’s just that one door has nothing but a stupid car.

[Riemann]

@Riemann

There’s a 1/3 chance the car is behind door A when you pick it. You always know that there is at least one goat among B and C, so if someone who knows what’s behind all the doors DELIBERATELY (not just by picking at random) shows you a goat among one of B and C, you have no new information. You still have a 1/3 chance of winning if you stick with A.

@thorny_locust

That’s actually more convincing than telling me that I am being given new information – except that you argued in your previous post that I do have new information.

There was a typo/omission in what I wrote there - it should have read as follows, note the critical addition in bold:

There’s a 1/3 chance the car is behind door A when you pick it. You always know that there is at least one goat among B and C, so if someone who knows what’s behind all the doors DELIBERATELY (not just by picking at random) shows you a goat among one of B and C, you have no new information ABOUT DOOR A. You still have a 1/3 chance of winning if you stick with A.

(Whereas you certainly DO have information about the relative merits of doors B and C that you did not have before.)

[Ynnad]

• I want the car.
• I’d rather have a goat.

[Little_Nemo]

A variant of the classic game.

There are four doors. There is a prize randomly hidden behind one of the doors.

You pick one of the doors. Monty Hall (who is aware of where the prize is) opens one of the remaining three doors and show you there is nothing behind it. He then offers you the choice of staying with the door you initially picked or switching to one of the other two unopened doors. (There are no further rounds of opening empty doors and giving you choices.)

Should you switch to one of the other doors or stay with your first pick? Unlike the classic game, there’s the possibility in this game of switching from one losing door to another, which changes the odds.

[thorny_locust]

@Riemann

There was a typo/omission in what I wrote there [ . . . ]
you have no new information ABOUT DOOR A . [ . . . ]
(Whereas you certainly DO have information about the relative merits of doors B and C that you did not have before.)

Ah. Makes sense.

[Ann_Hedonia]

Here’s my elaboration of the 100 doors variation. I can’t do this without constructively talking about doors, because the problem centers around revealing a hidden variable. I can call them shades or windows or boxes but I can’t get around it. But I like boxes.

There are two rounds to the game. In round one you have 100 identical boxes. One contains a prize, the rest are empty The rules are simple - pick a box. That box has a 1 in 100 chance of containing a prize.

The rules for round two are more elaborate and are the same as used on Let’s Make a Deal.

Every box except for the box containing the prize and one other box are removed. If you happened to pick the prize in round one, the second box will be a random empty. If you failed to pick the prize in round one, the second box will contain the prize.

Stay or switch?

[Ann_Hedonia]

I made an error in the above and missed the edit window. The first line in the last paragraph should be:

Every box except the box containing the prize and THE BOX YOU PICKED are removed. If you happened to pick the prize in round one, the second box will be a random empty. If you failed to pick the prize in round one, the second box will contain the prize.

Stay or switch?

[Author_Balk]

Once upon a time… I was put on a new team at work. I mentioned the Monty Hall program to a new team member (I’ll call him John). John insisted that it was 50/50 no matter how I explained it. I understand now that he would never admit he was wrong. He eventually said that he knows because he had advanced math classes. And I think he held this against me because he saw this as questioning his knowledge.
A little later they made some organizational changes and he became my manager. I doubt I would have got along with him, but I’m sure it made things worse.
So… Maybe it would have been better if I would have left him in his ignorance.

[EllisDee]

@thorny_locust

(the card trick version might convince me that it works, but would still leave me not understanding why)

That’s fair. (Though did you physically try it?)

The main benefit of the deck of cards approach compared to the 100 doors approach proposed in the OP is that a deck of cards is a ubiquitous physical item you can use to physically try it out. No imagining a thought experiment required.

EDIT: I didn’t post the cards method to say it’s the be-all end-all. It was in direct response to the OP, advocating a 100-door thought experiment. People only reach the correct conclusion in thought experiments if they understand it, and Monty Hall resists understanding. But once you see it with your own eyes and can physically touch the “doors” and “prizes” with your own hands, it strikes a chord in a more visceral way than a thought experiment can. (That’s my thinking, anyway.)

[Telemark]

@Little_Nemo

Should you switch to one of the other doors or stay with your first pick? Unlike the classic game, there’s the possibility in this game of switching from one losing door to another, which changes the odds.

Your initial pick has a 1/4 chance of being right. Switching to one of the remaining 2 doors should give you half of 3/4 chance of being right, or 3/8. So you should switch, you’re 50% more likely to win regardless of which door you choose.

[pjd]

@Munch

…
I think it makes it easier to see that the only way you lose by switching, is if you pick correctly the first time.

This !