3 Door Logic Problem

[Overman]

Articulett started this in the Atracttive thread...

"Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors?"

Most people think the odds ar 50/50 because two choices are left--and so they stick with their first "intuition". But the odds for the original door being correct is still 1/3 as it was in the beginning (the host can always turn over a door with a booby prize because there are two--it doesn't change your odds as to your first choice being correct--it does, however, eliminate the door with the booby prize from the equation). This means you will win the good prize 2/3 of the time by switching.

The ways human make errors in logic is a fascinating study--and a necessity for any magician or person who desires to exploit this behavior. It would serve you well to learn more about them--lest you be exploited by someone who understands how your thinking can go wrong.

Remember--2 options does not mean 50-50 probability. It might for heads or tells it is not the same when it comes to the existence of fairies. A women may or may not be pregnant, right? But the odds are greatly in the favor on non pregnant if you tested women at random--moreso if they are over 50. Learn the logic skills others here can teach you, and pass them on.

 

[Overman]

I do not understand....

If you have a 1 in 3 chance, then one of the guesses is eliminated, wouldn't that give you a new start, a new equation with a 1 in 2 chance?

 

[Lamuella]

I do not understand....

If you have a 1 in 3 chance, then one of the guesses is eliminated, wouldn't that give you a new start, a new equation with a 1 in 2 chance?

no, because what is eliminated is one of the WRONG guesses.

This is called the Monty Hall Problem (after the game show host of the same name). It ALWAYS pays to switch.

Look at it this way. There are two possible scenarios here. Scenario 1 involves you getting the right door on your first go. This has a one out of three probability. In this case, if you switch, you will lose.

Scenartio two involves you getting the WRONG dooron your first go. It has a two out of three probability. After you have picked your door, the other wrong door is revealed. Thus the other door must be the RIGHT door.

THus two things can happen:

You get the right door and switch to a wrong door (1/3 probability)
You get the wrong door and switch to a right door (2/3 probability)

 

[tkingdoll]

The forum's millionth Monty Hall thread! Still, it's always a good one. Overman, the solution is counter-intuitive but Lamuella is correct. Google for Monty Hall problem or Marilyn Vos Savant.

There's a simulation here: http://www.userpages.de/monty_hall_problem/

 

[Ersby]

The easiest way to understand this problem (for me, anyway) is to forget three doors. Instead imagine someone with a deck of cards. They say you have to find the ace of hearts and ask you to pick one, without seeing it, and they put that face down on the table. They then go through the rest of the pack, and seperate one card.

So now you have three piles: the card you chose, the card they chose, and a pile of 50 cards. They then discard the pile of 50 cards, saying the ace of hearts isn't amongst them. Now you have two cards. Yours and theirs. The odds clearly aren't 50-50, so it would make sense to change.

Now imagine this with one less card each time, until you get down to three cards. The reasoning stays the same.

 

[Stellafane]

I do not understand....

If you have a 1 in 3 chance, then one of the guesses is eliminated, wouldn't that give you a new start, a new equation with a 1 in 2 chance?

It took me a long time to wrap my brain around this one, but the following thought problem did it for me: Instead of 3 doors, imagine there were 1000. You pick number 10. The host then opens every other door except number 777. You'd probably switch your choice to 777 pretty quick in that situation, wouldn't you?

Once I accepted that scenario, I just substituted 3 for 1000. Voila! The scales fell from my eyes.

 

[Overman]

Look at it this way. There are two possible scenarios here. Scenario 1 involves you getting the right door on your first go. This has a one out of three probability. In this case, if you switch, you will lose.

Scenartio two involves you getting the WRONG dooron your first go. It has a two out of three probability. After you have picked your door, the other wrong door is revealed. Thus the other door must be the RIGHT door.

OK, I can see it as long as you are assuming that The Game Show Host would not open the third door if you picked correctly the first time.

Say if you pick correctly the first time or if you pick incorrectly the first time and the GSH still opens a goat door, then it would be 50/50 correct?

 

[Lamuella]

OK, I can see it as long as you are assuming that The Game Show Host would not open the third door if you picked correctly the first time.

Say if you pick correctly the first time or if you pick incorrectly the first time and the GSH still opens a goat door, then it would be 50/50 correct?

NO.

THe game show host ALWAYS opens a door.

But (and this is the crucial point) there is always a door to open that will contain a wrong answer.

Let's run this again as three scenarios, as the two scenarios clearly confused you. THere are three doors: A, B, and C In all these scenarios, the prize is behind door A.

Scenario 1: You pick door A. The host opens one of door B or C, it honestly doesn't matter because there is nothing behind either of them. You change doors and lose.
Scenario 2: You pick door B. The host opens door C. You change to door A and win.
Scenario 3: You pick door C. The host opens door B. You change to door A and win.

So in two out of these three equally likely scenarios, you win by changing doors.

 

[TheChadd]

Makes sense to me, once I had a bit of a think about it. Thanks for bringing this one up guys

 

[Overman]

NO.

THe game show host ALWAYS opens a door.

But (and this is the crucial point) there is always a door to open that will contain a wrong answer.

Let's run this again as three scenarios, as the two scenarios clearly confused you. THere are three doors: A, B, and C In all these scenarios, the prize is behind door A.

Scenario 1: You pick door A. The host opens one of door B or C, it honestly doesn't matter because there is nothing behind either of them. You change doors and lose.
Scenario 2: You pick door B. The host opens door C. You change to door A and win.
Scenario 3: You pick door C. The host opens door B. You change to door A and win.

So in two out of these three equally likely scenarios, you win by changing doors.

Got ya!

Thanks!

 

[brettDbass]

This scenario depends on the Host having prior knowledge of what's behind each door though.

Remove that, make it 25 red boxes instead of 3 doors and you've got Deal or No Deal.

Notice how people have been so eager to grasp the WOO by getting crowds to shout support, systems of lucky numbers and all sorts of other claptrap as soon as the element of pure chance has become involved.
Notice also that many players go on way beyond the point of maximising returns for minimum risk and most of them end up winning very small amounts.

You now have an example of logic completely falling apart in the presence of greed.

 

[PopeTom]

This scenario depends on the Host having prior knowledge of what's behind each door though.

Remove that, make it 25 red boxes instead of 3 doors and you've got Deal or No Deal.

Notice how people have been so eager to grasp the WOO by getting crowds to shout support, systems of lucky numbers and all sorts of other claptrap as soon as the element of pure chance has become involved.
Notice also that many players go on way beyond the point of maximising returns for minimum risk and most of them end up winning very small amounts.

You now have an example of logic completely falling apart in the presence of greed.

Though does anyone know the locations of the various prizes in Deal or No Deal? Or does the deal offer get based off the possible values still left in the cases?

-PopeTom

 

[pgwenthold]

THe game show host ALWAYS opens a door.

I just re-read the initial post in this thread, and I do not see any statement that could be interpreted to indicate that the game show host ALWAYS opens a door.

This is your assumption, and it is necessary in order to answer the question. It may or may not be true, but the question itself does not provide the information.

If it is true that the host always opens a door, then it is indeed a 2/3 chance of winning by switching. However, consider the following:

1) The host ONLY opens a door if your initial guess is correct. In that case, you have a 0% chance of getting it right by switching.

2) The host only opens a door if your initial guess is wrong. In that case, you have a 100% chance of winning by switching.

3) The host randomly elects to open a non-winning door for you. In this case, it is still better to switch, with a prob of 2/3 in those cases.

The premise only states that he has opened the door in this case. It does not say what the host did the last time this happened, or if he will do it again next time. Your claim that he does it all the time is an assumption and not given.

 

[Jaggy Bunnet]

This scenario depends on the Host having prior knowledge of what's behind each door though.

Remove that, make it 25 red boxes instead of 3 doors and you've got Deal or No Deal.

Notice how people have been so eager to grasp the WOO by getting crowds to shout support, systems of lucky numbers and all sorts of other claptrap as soon as the element of pure chance has become involved.
Notice also that many players go on way beyond the point of maximising returns for minimum risk and most of them end up winning very small amounts.

You now have an example of logic completely falling apart in the presence of greed.

I was under the impression that the offer given in Deal or No Deal was always less than the expected return from playing on (certainly has been in the couple of shows I have seen).

What do you mean by playing beyond the point of maximising returns for minimum risk?

 

[Meffy]

Wrote my own program to test it. Got 1/2-1/2 results. Checked my code -- oops, error in logic; fixed it. Got 1/3-2/3 results. Sure 'nuff. Marilyn makes some pretty awful mistakes* but this ain't one of them.
________________________
*F'rex, "Q: What part of a car travels the farthest? A: The keys! Tee hee hee!" I think for the average case this ain't so. My vote would be for any point on the tires' tread, or possibly on one of the engine belts. If you travel by air or rail a lot, and carry your car keys along, then Marilyn's right.)

 

[pgwenthold]

Wrote my own program to test it. Got 1/2-1/2 results. Checked my code -- oops, error in logic; fixed it. Got 1/3-2/3 results. Sure 'nuff.

Which you will, if you assume premises not stated.

The short answer is that the situation _as given in the original post_ does not provide enough information to solve the problem.

 

[sphenisc]

Which you will, if you assume premises not stated.

The short answer is that the situation _as given in the original post_ does not provide enough information to solve the problem.

Given the OP is a one-off situation then the answer is " That depends on which door the car's behind."

 

[brettDbass]

I was under the impression that the offer given in Deal or No Deal was always less than the expected return from playing on (certainly has been in the couple of shows I have seen).

What do you mean by playing beyond the point of maximising returns for minimum risk?

There comes a point in the show when you've say halved the number of boxes, taken out some low and some high values.
As the offers made are usually just below the rough average of values remaining there comes a point when you run a high risk of eliminating the remaining boxes which contain a higher value than the offer received one or two rounds ago.

 

[tsg]

I just re-read the initial post in this thread, and I do not see any statement that could be interpreted to indicate that the game show host ALWAYS opens a door.

This is a common source of confusion with this problem. There are some initial assumptions which need to be explicitly stated:

1. The host knows where the prize is.
2. The host will always offer you a choice.
3. The host will always open another door and never show you the prize.
4. The contestant is aware of 1-3.

Only if all four of these conditions are met does switching to the remaining unopened door give the contestant an advantage. (Technically, #4 doesn't have to be true, but if it isn't he can't make the determination that switching is better.)

ET clarify #3.

 

[tsg]

I was under the impression that the offer given in Deal or No Deal was always less than the expected return from playing on (certainly has been in the couple of shows I have seen).

http://www.internationalskeptics.com/forums/showthread.php?t=49473

The bank offer is a calculation based on the expected value of the remaining cases. It is fairly complicated, but in general, you are offered a fraction of the average of the large (> $100,000) cases and the average of the small cases, as if there was one fewer case remaining.