I think it all comes down to I can’t comprehend how the odds don’t change after I have more information . . .

I’m not sure what you mean by the odds not changing. Maybe you mean the probability that we picked the correct door initially. Then, you’re right, the probability that that door is the correct door does not change after Monty opens one of the doors.

When Monty opens another door revealing a goat, that tells us nothing about whether the car is behind our door or not, because Monty will never open a door revealing a car. If we initially picked the correct door, Monty will reveal a goat. If we initially picked the wrong door, Monty will reveal a goat. Therefore, Monty’s revealing a goat tells us absolutely nothing about whether we initially picked the correct door. Before Monty opened a door, the probability that we picked the correct door was 1/3. Since Monty’s opening a door tells us nothing about whether our initial pick was correct, the probability that it is is still 1/3 after Monty opens a door.

However, if we initially picked the wrong door—and there is a 2/3 probability that we did—then by Monty revealing another losing door, Monty is telling us exactly where the car is, namely, under the remaining door.

So, there is a 1/3 chance that we picked the correct door initially, but a 2/3 chance that we did not, in which case the car *is* behind the last remaining door. Therefore, we should switch.

Jay

]]>No. Your wording leaves out a key point in my solution, that the probability of winning by switching, given that the door you first picked was wrong, is 1.

Now why would that be useful thing to communicate when one is talking about a strategy for working with uncertainty?

The revised wording was specifically intended to remove that rather obvious statement of fact, namely that if you know what the right answer is, pick it! In the MTP or any like it, the contestant never *knows* the answer until the game is *over*.

I wrote:

There is a 1/3 probability that your initial choice was correct, but there is a 2/3 probability that it was wrong, in which case you will win by switching doors.

I know it is a slight nit of wording, but maybe a more accurate way to phrase it might be “…but there is a 2/3 probability that it was wrong, hence you will have better chances of winning by switching doors. So switch”

No. Your wording leaves out a key point in my solution, that the probability of winning by switching, given that the door you first picked was wrong, is 1.

]]>The problem bases itself off of the “don’t double-count” aspect of sets, which is why the more generic version (not specifying day of week for either child, just probability of both boys given 1 is a boy) is 1/3 rather than 1/2.

]]>There is a 1/3 probability that your initial choice was correct, but there is a 2/3 probability that it was wrong, in which case you will win by switching doors.

I know it is a slight nit of wording, but maybe a more accurate way to phrase it might be “…but there is a 2/3 probability that it was wrong, hence you will *have better chances of winning* by switching doors. So switch”.

This way the explanation doesn’t rhetorically discount the feature that “always switching” actually looses 1/3 of the time.

]]>But, unlike Barbara’s reluctance to think more choices initially doesn’t clarify things by making being wrong initially quite obviously likely, I think this helps (like in my twisted lottery example) but it is *better than mine* because nearly everyone can relate to picking a card from a deck especially when using the “ace of spades” as the psychologically attractive prize marker. No strange alternatives of familiar processes are required unlike in my lottery style game. Nice one!

At the other posting on OddsMustBeCrazy, I put this comment as an alternate way of explaining the problem:

Incidentally, I have a method for explaining the basis of the MHP to those who still seem to have trouble after a direct explanation. The key, I’ve found, is to exaggerate the probabilities without making the numbers too large.

Here’s what I do: Take a full deck of 52 standard playing cards. Shuffle them thoroughly, then hold them fanned out so you can see their values but the person you’re trying to convince cannot. Tell them the object is for them to get the Ace of Spades in order to win. Have them pick one card and place it face-down on the table without letting them see what it is. Then pick one card of your own and place it face down. Be sure that between your two cards, one of them was the Ace of Spades.

Then take the rest of the fanned-out deck and spread them out face-up and tell them to confirm that the Ace of Spades is not present in the remaining 50 that have been revealed. Ask them which is more likely: that they picked the Ace on their initial pick or that your card is the Ace. Ask them if they want to switch before turning over the cards to see if they’ve won. Ask them to explain their choice.

There is still another issue with the over-simplification of the Monty Hall Problem (without actually stating the assumptions used to simplify) which I also addressed at OddsMustBeCrazy, but assuming the simplistic version it really does give better odds to switch.

BTW, MythBusters looked at the Monty Hall Problem in an episode that aired last November. Watch It Here!

]]>**Here is how it maps**

Note: As a garden path like strategy, you *first* present the 20 digit example so they can see the initial pick is almost certainly wrong, once switching in *that* game is ‘accepted’ you then repeat the game restricted to a lottery number that’s 1,2 or 3 (1-3) in order to map it *directly* into Monty Hall:

**specifically**

The lottery number (1,2 or 3) maps into doors 1,2 or 3.

The commissioner knows the winning number and hides it in the envelope maps into: Monty Hall knows which door has the car.

Your ticket maps into your chosen door.

The commissioner declares that he holds the only remaining potentially winning ticket leaving only his and your ticket: that maps into the revealing the “wrong” door leaving only 2 doors.

You have a choice between the 2 remaining doors, maps into the choice between commissioner’s ticket (the remaining unopened door) or your ticket (your door)

Your offer to swap ticket ownership, maps into the choice to “switch” doors.

The envelope is opened and the winning number is revealed, maps into the winning door is revealed.

The door you end with maps into the ticket you own.

**Finally**

I agree with you that the essence of the game is the choice between two things and not between more than two things.

I acknowledge you made that point in your reply to Tim, but in my follow ups I only tried to make the same worded differently. Based on Tim’s post my attempts were unsuccessful as well, since he states he still doesn’t get it.

The challenge as I see it is to present and describe an equivalent game that people would accept *easily*, free of the cognitive misdirection the wording of the initial problem creates. Then you just map the new game into Monty Hall, which I think I did above.

If I did the mapping incorrectly, please help me by pointing out where I went off the rails.

]]>There is a 1/3 probability that your initial choice was correct, but there is a 2/3 probability that it was wrong, in which case you will win by switching doors. So switch.

For a real mind f***, try the Tuesday Birthday Problem: A man walks up to you on the street and says, “I have two children, one of whom is a son who was born on a Tuesday. What is the probability that my other child is also a son?”

Jay

]]>