Einstein's puzzle: Can you sovle it?

The gimmick of this riddle is the fact that the door that is opened is not randomly chosen. It will never reveal the car, it is guaranteed to reveal one of the two goats.

This tells you nothing you didn't already know, except for which of the three doors does not contain the car.

Let's say you chose door #3 to begin with.
The original odds of the car being behind:
either door #1 or door #2 is 66.6%
either door #1 or door #3 is 66.6%.
either door #2 or door #3 is 66.6%.

While it is true that door #1 and door #2 together constitute a group which has 66.6% probability of containing the car in the first place, the fact is that #1+#3 and #2+#3 also constitute groupings with 66.6% probability of containing the car. Thus your original selection necessarily belongs to two out of three of the 66.6% probability groups. The odds are 2 out of 3 that your door will belong to a pairing of doors containing the car. The odds of the car being behind the door you selected is exactly equal to the odds of it being behind any individual door from any pair of doors.

Revealing the goat from one of the two unselected doors alters the probability of the car being behind all three pairings of doors.

Let's say the inevitable goat is shown behind door #1.
The odds of the car being behind:
either door #1 or door #2 is now 50%
either door #1 or door #3 is now 50%
either door #2 or door #3 is now 100%

The revealed goat alters the relative probabilities of the three pairings of doors, but not of the remaining individual unopened doors. They are still equal. The odds of your door now containing the car is still 50%, not 33.3%, even though you know the door which revealed the goat was not chosen randomly in 2/3 of the cases. This is because although door #2 belongs to the group #1+#2 which originally had the 66.6% probability of containing the car, door #3 also belongs to two pairings that originally had a 66.6% probability of containing the car. After door #1 has been shown to contain a goat, door #2 and door #3 both belong to the only pair guaranteed to contain the car. Since they both belong to the same winning group, their odds of containing the car are exactly equal.

Should I switch my original selection expecting to double my odds of winning? Naaaaaaaaaaaaaaah! :)

You should switch from your first choice to the remaining closed door, and by doing so you increase your probability of choosing the auto from 33% to 67%. Initially this may seem to be counter-intuitive, because there are two remaining doors that ORIGINALLY had the same probability of hiding the car (33% for each of the three).

Consider however that after your first choice there are two groups, your choice which has 33% likelyhood, and the group of the other two which have combined 67% likelyhood. Since there is always at least one goat in the second group of two doors, the game show host does not change the relative likelyhood by showing you one goat (no new info about the two groups). What he does allow you to do is capture the entire 67% likelyhood by picking the remaining door of that second group.
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The illusion of logic that makes this riddle is the assumption that "after your first choice there are two groups", one with 33% odds and one with 66% odds. The fact is, that after your first choice there are three groups, all with equal probabilities. Comparing probabilities of groups vs. individuals is comparing apples to oranges. It creates a numerical illusion, but not a statistical fact.
 
Quote from hii a_ooiioo_a:


The illusion of logic that makes this riddle is the assumption that "after your first choice there are two groups", one with 33% odds and one with 66% odds. The fact is, that after your first choice there are three groups, all with equal probabilities. Comparing probabilities of groups vs. individuals is comparing apples to oranges. It creates a numerical illusion, but not a statistical fact.
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That is an excellent summation of the problem, hiia_ooiioo.

Basically, from the way I viewed it is from an information theory standpoint. You are presented with no knowledge about three conditions, one of which is the correct decision. The odds of you selecting correctly is 1/3, since you have no information.

As soon as Mr. Monty Hall shows you a door that does not possess a prize, you are given a new piece of information that you originally did not have.

You can always list all possibilities to see how it plays out (this is sometimes the best way to solve an information theory problem)

Three doors represent several combinations:

I = Incorrect Door
C = Correct Door

The possible combinations are:

I I C
I C I
C I I

From each one of those combinations, there are three ways you can choose INITIALLY.

X = Your choice

X I C I X C I I X

X C I I X I I C X

X I I C X I C I X


As you can see, for each group (Row), you would have selected the correct door once and the incorrect door twice.

This means that if Monty Hall shows you an incorrect door, the odds of you switching from a correct door to an incorrect door are 1 and 3.

It makes sense to switch given the new information.
 
The odds are 2 out of 3 that you will switch from an incorrect door to a correct door, when none of the doors have been opened.

Once a door has been opened, that eliminates 1/3 of the possible switches (unless your intention is to go home with a goat).

So after seeing one door opened, the odds become 50/50 of switching from car to goat, or from goat to car.

The most important thing you need to know to increase your probability of winning on Let's Make A Deal is to wear an outrageous or eye-catching costume. That will increase your odds of being chosen by Monty Hall by about 1,000%.
 
This whole riddle is a mental illusion. It's like looking at an Escher drawing. It really looks like things are going one way, even though you know it's physically impossible. You have to look at it a long time before you start to see how the illusion was done.

The illusion of the goats vs. cars is that you know the opened door was not opened totally at random. Only a goat will be shown. But regardless of that, your original choice was chosen at random by you. You could just as easily have chosen the other unopened door. If you had chosen #2 instead of #3, you would now be thinking you have twice the odds of getting the car by switching to #3 after you see the goat behind #1.

Your odds of getting the car from #2 or #3 is still exactly equal, regardless of the goat behind #1. The lure to switch for statistical advantage is an Escher illusion.
 
Quote from hii a_ooiioo_a:

The odds are 2 out of 3 that you will switch from an incorrect door to a correct door, when none of the doors have been opened.
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This is incorrect. The odds are not 2 out of 3 that you will switch from an incorrect door to a correct door.

If you are switching FROM an incorrect door, then you already know it is incorrect, so you can't ask that question without changing the parameters of the question.

There are several possibilities:

Take the following scenerio:

We are god and know that door number 2 is the correct door. That leaves 1 and 3 the incorrect door.

If the person chooses door #1, they have a 50/50 chance of selecting the correct door IF they switch.

If the person chooses door #2, they have a 0 chance of selecting the correct door IF they switch.

If the person chooses door #3, they have a 50/50 chance of selecting the correct door IF they switch.

Once they select a door, regardless of whether it was the correct one or not, and Monty Hall plays god and shows another door that was incorrect, it is always advantegous for you to switch.
 
Quote from hii a_ooiioo_a:

The lure to switch for statistical advantage is an Escher illusion.
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It is not an illusion. If you pick a door and Monty shows you another door that is wrong, you better switch.
 
He is NEVER going to show you the "right" door! He will ALWAYS show you a goat. Once he has shown you the goat, it is no longer a 33% vs. 66%, or 33% vs. 33% vs. 33%. One of the 33% choices has been eliminated. It is now 50% vs. 50%.

It looks like half the Escher people are walking upside-down down the staircase. But gravity will not allow it. It may look like you have 66% odds of getting the car by switching, but it's just a very clever illusion. The odds are still 50/50.
 
hii a_ooiioo_a,

This is a branch of information theory. Here is a complete example that I am taking the time out to help you understand this. Believe me, it took me a long time to understand it because it is not intuitive (meaning, it goes against what you would consider normal logic -- yet it is logical).

The first thing I will do is select one of the possible scenarios.

The prize is behind DOOR #2. That leaves a goat behind DOOR #1 and DOOR #3

Here are all the possibilities ...

You initially select DOOR #1. Monty shows you DOOR #3. You switch to DOOR #2. YOU WIN

You initially select DOOR #2. Monty shows you DOOR #3. You switch to DOOR #1. YOU LOSE

You initially select DOOR #3. Monty shows you DOOR #1. You switch to DOOR #2. YOU WIN

The previous 3 are IF YOU SWITCH DOORS. You win 2/3 times.

Now, what happens if you don't switch?

You initially select DOOR #1. Monty shows you DOOR #3. You stay with DOOR #1. YOU LOSE

You initially select DOOR #2. Monty shows you DOOR #3. You stay with DOOR #2. YOU WIN

You initially select DOOR #3. Monty shows you DOOR #1. You stay with DOOR #3. YOU LOSE

In the previous examples of all combinations, you win 1/3 times.

Therefore, if you switch, you stand a greater chance of winning.
 
The odds are not 2 out of 3 that you will switch from an incorrect door to a correct door.

If you are switching FROM an incorrect door, then you already know it is incorrect, so you can't ask that question without changing the parameters of the question.
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The way the riddle is stated, you never know whether your door was incorrect. If you knew that, there would be no riddle.
 
Quote from hii a_ooiioo_a:

The way the riddle is stated, you never know whether your door was incorrect. If you knew that, there would be no riddle.
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You don't need to know what door is incorrect. That's part of the game, obviously.

The point I am trying to make is that this non-intuitive riddle is best solved by switching doors given the condition that Mr. Hall will always show you an incorrect door after your first choice.

I know it sounds wacked, but that is just how it is -- it is one of those profound mysteries of logic that we don't grasp immediately but it is true.

If you want verification, just work out every possible scenerio. Then find out your odds for success if you don't switch vs. your odds of success if you do switch -- GIVEN that Mr. Hall shows you an incorrect door after your choice.
 
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