Topic: Episode #280

[Damned Skeptic]

Cryptobovinologist,
You forgot that all of the doors have 1/3 before the host opens a door, so both the contestant's door and the mystery door are 1/3 before the host shows the goat which makes it seem an arbitrary choice to assign the mystery door 2/3 after the reveal. Why not assign it to the contestant door? The fact is it shouldn't get assigned to any door. The game was never between three doors.

[Chew]

It's not arbitrary. You pick one door out of 3. You know ahead of time that one door of the other two doors don't count, so that one other door is really a 2/3's choice. It's the same as asking, "Would you like to choose door 1? Or doors 2 and 3?"

[Damned Skeptic]

It's not arbitrary. You pick one door out of 3. You know ahead of time that one door of the other two doors don't count, so that one other door is really a 2/3's choice. It's the same as asking, "Would you like to choose door 1? Or doors 2 and 3?"

And how do you know you didn't pick the 2/3 door?

[Chew]

It's impossible to pick the 2/3's door; the 2/3's door is always the door you didn't pick.

[werecow]

Cryptobovinologist,
You forgot that all of the doors have 1/3 before the host opens a door, so both the contestant's door and the mystery door are 1/3 before the host shows the goat which makes it seem an arbitrary choice to assign the mystery door 2/3 after the reveal. Why not assign it to the contestant door? The fact is it shouldn't get assigned to any door. The game was never between three doors.

No, that's not what I'm saying, and I think you're making the mistake I highlighted in the 1000000 doors reductio ad absurdum. I'll try to make it clearer:

 Assume for the moment that you're a computer, and you don't care what anyone says, either before or after you choose your initial door (at random).

Each door has a 1/3rd chance of hiding the car a priori. This means that any initial guess, has that same chance of 1/3rd of getting the car - it doesn't matter which door you pick.

So now you've made your guess. Your door has a 1 in 3 chance of being right, and there is a 2 in 3 chance that the car is behind one of the other doors. Do we agree so far?

OK, now one of the remaining two doors, which taken together have a 2/3rd chance of being correct a priori, is opened, to reveal a goat.

Being the computer that you are, you infer that this means that, whereas the two doors together have a 2/3 chance of containing a car, the one that was just opened has a 0 out of 3 chance.

Still. the remaining two doors MUST have a shared probability of 2/3rds, because your initial guess cannot have had a chance of more than 1/3rd of containing a car (assuming the car is behind a random door). This is because you don't change your initial guess after the fact; you made your choice while there were still three doors, so it had 1/3rd probability, and this cannot change unless you have a time machine.

That means that the remaining 3rd must shift to the remaining door, giving that door a 2/3rds probability of hiding the car.  You still have that 1/3rd chance of being right that you had all along, but the odds are against you.

Try doing the 1000000 door substitution on this narration, or even a 4 door substitution (remember that Monte opens all but one door, rather than just one extra door. That cognitive flip helped me a lot when I was struggling to understand it).

[werecow]

BTW, all of this assumes that you have a fixed strategy, of course. If you randomly switch between sticking to your guns and switching, you'll have a 0.5 chance on average of getting the car, as that's the average of the expected gains for the two strategies (1/3rd stick + 2/3rds switch / 2 = 0.5). The point is, it's better to stick with switching.

[Damned Skeptic]

BTW, all of this assumes that you have a fixed strategy, of course. If you randomly switch between sticking to your guns and switching, you'll have a 0.5 chance on average of getting the car, as that's the average of the expected gains for the two strategies (1/3rd stick + 2/3rds switch / 2 = 0.5). The point is, it's better to stick with switching.

After spending way too much time reading and thinking about this problem I have little confidence in my original position without gaining any confidence in the solutions I've seen. It was interesting to see how various people approached the problem, but of the ones I could understand, I always felt like there was something not quite right about all of them. Now it seems likely that this is a problem that I will never grasp. That's kind of disappointing, but it isn't the first time and I doubt it will be the last.

[Chew]

My step-mom has a masters in math. When I told her about the MH problem she refused to believe it. It wasn't until she played with the applets that she accepted it.

Podcast 142 describes how this problem has screwed up some reseach.
Show notes for 142: http://www.theskepticsguide.org/archive/podcastinfo.aspx?mid=1&pid=142

[Moloch]

I've found it useful for people struggling with the Monty Hall problem just to simply spell out the results of an "always switch" strategy to highlight the odds. No formula or anything needed.

Let's say the car is behind door 1 and you'll always switch.

You choose door 1, Monty opens door 2, you switch to door 3 and lose.

You choose door 2, Monty opens door 3, you switch to door 1 and win.

You choose door 3, Monty opens door 2, you switch to door 1 and win.

There, simple! All 3 outcomes of the switching strategy and you win 2/3!

[Citizen Wolf]

Perhaps part of the problem that some people have with the Monty Hall problem is not realising that the host knows where the prize is, from the outset, and this influences the door that the host reveals. The intent of the host is important.

Imagine the following. You are asked to pick the ace-of-spades out of a deck of cards. You don’t get to see the card you picked. You have a 1/52 chance that you picked the right card, and thus there’s a 51/52 chance that the ace is in the group of cards you didn’t pick.  Simple enough so far.

What would you do if, after you pick your card, the host shows you one other card from his side of the deck (which won’t be the ace-of-spades) then asks you if you want to keep your original card or take ALL of the remaining cards. Well, you’d be a fool not to go with all of the rest, because you know from the outset that the chance that you picked the right card was only 1/52, and the odds of it being somewhere amongst all the others was 51/52. Just because the host showed you one other card doesn’t change that.

This is effectively what the host is doing in the original 3-door problem. The fact that he showed you one of the non-prize doors makes no difference to the chances that the odds of the prize being in the door you picked was only 1/3.

To help understand the problem better, imagine that the host DOESN’T know where the ace-of-spades is. This changes the game completely. You pick your card at random. You still only have a 1/52 chance of getting the right card. Now the host turns over another card at random. It’s also unlikely to be the ace-of-spades.  Now, if the host asked you at this point whether you wanted to switch from your original card to ALL of the others, the choice would still be clear. However, for the moment we won’t give you the option to switch; just follow through the scenario. The host continues to turn over all the cards one at a time, and let’s suppose that by chance, he gets all the way through the deck without finding the ace-of-spades, until there are only 2 cards left; your card and the last card left in his side of the deck.

He then asks you if you want to switch. Well, in this case there is NO benefit in switching. This is because the chances that he managed to go through all of the remaining 51 cards, at random, not knowing where the ace-of-spades was, and not manage to pick it, is about the same as you picking the correct card right at the beginning. 

So remember, it is important what the host knows, and how the host acts with that information. That is what should influence your decisions.  In the original 3-door problem, the host will never show you the prize. This is information that’s useful to you. 

[Johnny Slick]

Let's face it: the biggest issue with the Monty Hall problem isn't the sample space or the trials or not getting that Hall knows where the car is ahead of time, it's that something - and I can't quite put my finger on what - about this trick is counter to human intuition. The only reason it makes sense is that everything works out mathematically and logically.

[lonely moa]

Let's face it: the biggest issue with the Monty Hall problem isn't the sample space or the trials or not getting that Hall knows where the car is ahead of time, it's that something - and I can't quite put my finger on what - about this trick is counter to human intuition. The only reason it makes sense is that everything works out mathematically and logically.

That's why I use it for a prelude to the lecture and discussion on the role of heuristics and avalanche involvement. 

[James]

BTW, all of this assumes that you have a fixed strategy, of course. If you randomly switch between sticking to your guns and switching, you'll have a 0.5 chance on average of getting the car, as that's the average of the expected gains for the two strategies (1/3rd stick + 2/3rds switch / 2 = 0.5). The point is, it's better to stick with switching.

After spending way too much time reading and thinking about this problem I have little confidence in my original position without gaining any confidence in the solutions I've seen. It was interesting to see how various people approached the problem, but of the ones I could understand, I always felt like there was something not quite right about all of them. Now it seems likely that this is a problem that I will never grasp. That's kind of disappointing, but it isn't the first time and I doubt it will be the last.

This riddle is ver counter-intuitive and hence its popularity.  I've explained it enough times that I think I'vegot down to its simplest understanding.

K, First off, Monty knows where the car and the goats are, and since there are two goats, he always reveals a goat after you pick.

If you always stay with your first pick then your odds of winning is 1/3  you have to pick the car, which will happen 1/3 times,  you stay with the car, you win the car!  If you pick a goat, which happens 2/3 times, you lose!

If you always switch, if you happen to pick the car, Monty reveals a goat, you switch and now you lose.   As stated above, picking car first happens 1/3, but this time you lose cuz you switched.

If you always switch and happen to pick a goat, Monty reveals the other goat, you are switching, so now you won the car!  As stated above, picking a goat first happens 2/3 of the time, so by always switching you will win 2/3 of the time!

To summarise, if you stay with initial pick, you have to pick correctly and that happens 1/3!

If you always switch then you DONT want to pick the car first and that happens 2/3 times! If you pick  a goat when always switching-you win and there's TWO goats.

Hope that clears it up

[James]

If that doesn't clear it up, then we can play for money, two pennies and a one hundred dollar bill.  We each pick 20 times, you sticking to your picks and me switching everytime!

[Trinoc]

The difficulties people have with the Monty Hall problem are symptoms of a fundamental misunderstanding about probability, namely:-

An event does not, in general, have a fixed, a priori, probability, but the probability depends on what prior knowledge you have of the outcome.

In this case, Monty Hall knows where the car is, and betrays part of that knowledge by opening the door. This makes the probability of winning different from what it would be if choosing with no prior knowledge at all.