Great Puzzle

[DistantVoyeur]

After all this arguement I've realized that my logic is valid but that the Monty Hall Priciple does exist, just not for this.

I reasoned on a lottery ticket. one in 14 million. You buy one set of numbers, then eliminate all the other choices but one. The overpowering odds are that the other set left are the winning numbers. Thus the Monty Hall principle is valid on a macro scale. Transposing it onto a micro scale like the 3 doors seems to be a stretch though.

(hoping he has the right set of numbers for the draw though) G'nite all.

 

[Justanormalguy]

This is a fun thread. I'd like to add my two cents.

The way I see it, from the onset, the contestant has a choice of two sets of doors: the door he picks makes up set one, and all the remaining doors make up set two. That'd mean set one has a 1 in X chance of success, and set two has a X-1 in X chance of success.

Opening an empty door in set two is the smoke and mirror thing. It doesn't change the fact that when set one was picked, it only had a 1 in X chance of success.

When the contestant is given a chance to switch, he is really being asked if he believes his initial pick is correct.

Let's stretch it a bit and say X equals 100 doors. When the first door is picked, it has a 1 in 100 chance of success. If Monty opens up 98 empty doors out of the remaining 99, it doesn't change the fact that the first door was picked facing terrible odds. Anyone who doesn't switch believes they were able to pick the right door out of 100 the first time.

Going back to 261252's original query involving only three doors, I'd say the odds aren't terribly bad. I'm not a computer or a casino so I don't have to play the odds. I'll pick whatever door I chose because I feel LUCKY! Hah!

 

[benstt]

Okay, wrap your head around this. Similar theme about information and probabilities.

There are two siblings in a house. (Assume a perfect world, boys and girls are equally likely at birth, no mortality.)

When you knock, one of the two will answer. We're going to make deductions about the other sibling.


1.. You knock, and one of them answers through the door. You can't tell whether the answerer is a boy or girl. What are the odds that the other sibling is a boy?

2. The door is opened and you observe that the answerer is a girl. What are the odds that the other sibling is a boy? Does it change?

3. The girl announces that she is the older sibling of the two. What are the odds that the other, younger sibling is a boy?

To help make sense of this, keep in mind there are at the beginning 4 permutations of older/younger:
Boy/Boy, Boy/Girl, Girl/Boy, Girl/Girl. Each permutation is equally likely.

Many would say the answers are 1. 2/4 2. 2/3 3. 1/2

#2 is the contentious one.

 

[shack]

One more attempt at logic.
When you have 3 choices then you odds can be 1 in 3. By eliminating one choice you no longer have 3 choices. You have your choice and the only other alternative that is now available. You either have the correct one, or it is behind the other door.

I agree with this logic. What was happening prior has no bearing on what the current situation is, i.e. there are two doors and you have one. End of story.

 

[Omnius]

ok...very quickly... and not to prolong this already prolonged topic. The chance of 1/3 would be correct.. if and only if.. monthy also chose one door randomly. And i emphasize... randomly. But monty didnt. he KNEW which door was empty... and chose to show you that. That means NEW information was introduced. Snowleopard had it right when: 1/3 chance u orignally chose right door, 2/3 chance the prize is behind one of the other 2 doors. by eliminating one of those 2 other doors (of which they total a 2/3 chance of winning), the door that is left carries that 2/3 chance of winning (since you have shown the other door empty). again.. THIS ONLY APPLIES cuz monty KNEW to choose the empty door. if he also chose randomly 1 of 3 doors, then the odds would not change. for those that say this is logical and easy to understand... it isnt.. its very counter intuitive. but once you see the logic behind it, then it makes alot of sense.

 

[Omnius]

and to elaborate on the 1 in 100 doors extrapolation... u choose 1 door... its 1 in 100 u chose right, but there is a 99 in 100 chance the prize is in the OTHER doors. if Monty KNOWINGLY eliminated 98 of the other 99 doors, those odds fall on the one door left that Monty didnt choose... ie, the door you originally chose still has the 1/100 chance, but the other door now carries 99/100 chance of winning. (this only works if Monty knowingly chose the empty doors. if Monty RANDOMLY chose those doors, then ALL doors have a 1/100 chance of winning).

 

[shack]

he KNEW which door was empty... and chose to show you that.

However, he did not know which door you were going to choose. Regardless of which one you chose there was still going to be at least one empty one for Monty to show. You had 1 chance in 3 of being right at the beginning. Now there's only 2 left and you have 1 chance of those 2.

 

[DistantVoyeur]

However, he did not know which door you were going to choose. Regardless of which one you chose there was still going to be at least one empty one for Monty to show. You had 1 chance in 3 of being right at the beginning. Now there's only 2 left and you have 1 chance of those 2.

I've struggled with this. They are correct. Initially you have a 66% chance of being wrong, that does not change when empty doors are chosen. It works better with higher numbers because it gets more obvious.
Using 3 choices weakens it because you are looking at a 16% difference from the perceived 50% chance. Take 100 doors and it become more obvious. You have a 1% chance of being correct initially. So you have a 99% chance of being wrong. 99 times out of 100 your choice will be wrong initially. All the host is doing is showing you the other wrong choices, you are still left with a 99% chance of having made the wrong choice, therefore the door that is left has a chance of being the correct door because you only started with a 1% chance of being correct initially.
The concept is harder to grasp because you are using the weakest argument to defend the choice, as the number of choices grows the pattern becomes more obvious.
You have to look at the perspective of the chances of having made the wrong choice initially. If you had a 99% chance of being wrong initially, after all the other wrong choices are eliminated you are still left with a 99% chance of being wrong.
With 3 doors being reduced to 2 you end up with a 50% chance of being right and a 66% chance of being wrong.
With 100 down to 2 you end up with a 50% chance of being right and a 99% chance of being wrong.
It is always the best strategy to choose the other door to maximize your odds.

 

[261252]

u got it distantvoyeur congrats.

It occurs to me that there are 2 more ways to solve this

This is the easiest way of all

If door ones' chance is always 33.3%, and it is , this never ,ever changes if, and only if, Monty deliberately choses the empty door than the remaining door (2) has a 100 - 33.3 % chance as the odds of it being behind one of the doors is 100%.


This is the hardest way of all

What has occurred here is called the variable change effect. As new info is added your odds change. Use the information Monty has given and set up an algebraic equation

Firstly, when Monty deliberately chose an empty door, there were 2 possible scenarios.

Secondly, the information these scenarios gave us is

If the car is behind door one then it does not matter which door he opens first as the other two doors are both empty and no new information has been added. The odds of this occurring is 50%

If door one is empty then Monty has to open the remaining door that is also empty thus revealing with certainty which door has the car ( door two.)
The oods of this occurring is also 50%

Assign these chances letters and make your equation equal 100%

 

[RandyAndy2]

I've struggled with this. They are correct. Initially you have a 66% chance of being wrong, that does not change when empty doors are chosen. It works better with higher numbers because it gets more obvious.
Using 3 choices weakens it because you are looking at a 16% difference from the perceived 50% chance. Take 100 doors and it become more obvious. You have a 1% chance of being correct initially. So you have a 99% chance of being wrong. 99 times out of 100 your choice will be wrong initially. All the host is doing is showing you the other wrong choices, you are still left with a 99% chance of having made the wrong choice, therefore the door that is left has a chance of being the correct door because you only started with a 1% chance of being correct initially.
The concept is harder to grasp because you are using the weakest argument to defend the choice, as the number of choices grows the pattern becomes more obvious.
You have to look at the perspective of the chances of having made the wrong choice initially. If you had a 99% chance of being wrong initially, after all the other wrong choices are eliminated you are still left with a 99% chance of being wrong.
With 3 doors being reduced to 2 you end up with a 50% chance of being right and a 66% chance of being wrong.
With 100 down to 2 you end up with a 50% chance of being right and a 99% chance of being wrong.
It is always the best strategy to choose the other door to maximize your odds.

DV, I'm really disappointed in you. You had it right and you now choose to be wrong. Sigh. Tell me you were joking with your post.
Take your 100 door example. I agree that initially your door has a 1% chance of being correct. After you have eliminated 98 doors you have a 50% chance of being correct. Your liklihood of being wrong does not stay the same - it drops to 50% so that the chance of being right + the chance of being wrong (the only two possibilities) = 100%.
In the 3 door Monty Hall example, the fact that the first selection has a 2/3 chance of being wrong has NO EFFECT on the second selection. Probabilities of success are calculated seperately each time a selection is made.

 

[RandyAndy2]

Here's another way of looking at the MH example:
Assuming the game requires Monty to remove one door from consideration, the contestant really onlyhas two choices, and so his/her probability of success is 50% from the start. Changing the selection after a door has been eliminated would still provide a 50% chance. It just doesn't matter which door is chosen.

 

[benstt]

DV, I'm really disappointed in you. You had it right and you now choose to be wrong. Sigh. Tell me you were joking with your post.
Take your 100 door example. I agree that initially your door has a 1% chance of being correct. After you have eliminated 98 doors you have a 50% chance of being correct. Your liklihood of being wrong does not stay the same - it drops to 50% so that the chance of being right + the chance of being wrong (the only two possibilities) = 100%.
In the 3 door Monty Hall example, the fact that the first selection has a 2/3 chance of being wrong has NO EFFECT on the second selection. Probabilities of success are calculated seperately each time a selection is made.

Probabilities are calculated separately for independent events. That's the whole issue; Monty's reveals add information to the second selection, so it is not independent from the first.

This really messes up many people, especially those who studied probability rules.

 

[261252]

Here's another way of looking at the MH example:
Assuming the game requires Monty to remove one door from consideration, the contestant really onlyhas two choices, and so his/her probability of success is 50% from the start. Changing the selection after a door has been eliminated would still provide a 50% chance. It just doesn't matter which door is chosen.

Daer Randy,

Yet another disbeliever. The key here is that Monty is deliberately choosing empty doors.

My thread just above yours provides the logic.

 

[DistantVoyeur]

To me the confusion came in that you assume if one probability is 50%, the other has to be as well because it must add up to 100%. It doesn't. Look at the 100 door scenario and it comes a bit clearer.
I resolved it with thoughts of the lottery as sets of doors. initially you have 14 million chances, you pick your one set of numbers. There is a one in 14 million chance you have the correct set/door. Remove all other chances of winning because they did not win, only the winning number set has the prize behind the door. There are now only 2 alternatives left, yours and the winning numbers. Do you still think you have a 50% chance of winning the prize.
In reality the other number sets are removed by not being drawn, just as Monty would remove a goat.

 

[261252]

benstt is correct.

Monty has given you information because his pick of an empty door was deliberate. This info is what he was thinking, so, look at the problem from montys point of view.

you can use this info to deduce your probabilities.

If the choose of an empty door was random then Monty has given you no info because he had nothing to consider in such an act of non deliberation.