Win 1 Million Dollars: Pick the Right Door

[KingNothing]

You are a contestant on a game show where you must open the right door of three doors to win 1 million dollars.

You pick the door in the middle the first time, then the host opens a door that you didn't pick to reveal a goat (not 1 million dollars). He gives you one more chance to choose...should you switch your pick, between the remaining two doors?

 

[Doc Al]

This one was beaten to death rather recently: https://www.physicsforums.com/showthread.php?t=17005

 

[confutatis]

And for a really stupid approach to solve a really stupid problem, you can't do much worse than a famous philosopher:

http://jamaica.u.arizona.edu/~chalmers/papers/envelope.html

 

[Janitor]

The contest door problem is something I first heard in the early 1980s. I didn't believe the answer for literally days, then I convinced myself that it really does pay to switch.

 

[confutatis]

The contest door problem is something I first heard in the early 1980s. I didn't believe the answer for literally days, then I convinced myself that it really does pay to switch.

It doesn't, and I can't believe people can actually be convinced of it. This is like being convinced nothing can move because of Zeno's paradox.

Whatever happened to commonsense?

 

[Doc Al]

It doesn't, and I can't believe people can actually be convinced of it. This is like being convinced nothing can move because of Zeno's paradox.

Huh? Here we go again. Did you at least look at the link I posted? And what does this have to do with Zeno?

 

[confutatis]

Huh? Here we go again. Did you at least look at the link I posted?

I did look at the post, but didn't see anyone approaching the problem the right way. I did see some people claiming the odds are increased by switching doors, which is just ridiculous.

And what does this have to do with Zeno?

People who claim that you always choose wrong before the host opens a door, no matter which door you choose, are claiming a paradox. A paradox is a sure sign that your reasoning is faulty.

 

[Doc Al]

I did look at the post, but didn't see anyone approaching the problem the right way. I did see some people claiming the odds are increased by switching doors, which is just ridiculous.

Hope you're not a gambling man... you'd lose your shirt!

Why not share what you think is the "right way"?

Commonsense tells you that switching doors makes no difference. Coming up with a mathematical proof may be tricky, but failure to assert commonsense is in no way proof that commonsense is wrong.

"Commonsense" is of course a very personal--and unreliable--thing. Especially in the area of probability.

 

[confutatis]

Why not share what you think is the "right way"?

Would it make any difference? You seem so sure that the odds change, I'm sure you are mistaken. How can I convince you you are mistaken?

"Commonsense" is of course a very personal--and unreliable--thing. Especially in the area of probability.

Does that mean all logical proofs that offend commonsense are necessarily wrong?

 

[Doc Al]

Would it make any difference? You seem so sure that the odds change, I'm sure you are mistaken. How can I convince you you are mistaken?

In the usual manner: Point out the errors in the arguments offered. Just saying that it's "obviously wrong" or "ridiculous" won't do it.

Does that mean all logical proofs that offend commonsense are necessarily wrong?

Huh? Only to one who holds "commonsense" to be above logic and proof. (Realize that proofs are only as good as the assumptions and premises they are based upon.)

 

[confutatis]

Point out the errors in the arguments offered.

I can try, but I don't expect you to be convinced.

You actually improve your chances from 1/3 to 2/3 by changing your choice.

Not at all. Since the host opened a door without a prize, the prize must be behind one of the two remaining doors, but you are claiming that the prize is more likely to be in the unchosen door regardless of which door was chosen in the first place. This makes absolutely no sense.

If you don't switch, your chances of winning remain at 1/3.

No, they don't. "Chances of winning" means "chances that the prize is behind a particular door". When you have 3 doors, the chance is 1/3; when you have 2 doors, it's 1/2.

Nothing changes when the host opens a door, because you knew from the beginning that, no matter which door you chose, there would always be at least one unchosen door without the prize. You are confused because you think the host gives you information by opening a door - that is definitely not the case. It would be information before you made your choice, but after you do it the host is only confirming what you already knew.

If you flip a coin to decide which remaining curtain to choose, your odds rise to 1/2.

That is correct, but it means that it makes no difference whether you switch doors or not.

But if you take full advantage of the new information given you--and always change your choice--you will increase your odds to 2/3.

Again, there is no new information. You know beforehand that, no matter which door you choose, there will be at least one unchosen door without the prize.

Believe me when I tell you, you are confused. I realize people don't like to be told that, but it's often true nonetheless.

 

[hypnagogue]

Again, there is no new information. You know beforehand that, no matter which door you choose, there will be at least one unchosen door without the prize.

You know beforehand that there will be at least one unchosen door without the prize, true. But if you were to guess beforehand that door number 1 was empty, you would have had a 1/3 chance of being right, whereas if the host reveals door #1 to be empty, you now have certainty that it is empty. Increase in certainty = increase in information.

Think of the problem this way. Suppose that we dispense with the host altogether, and we let you pick 2 doors at once. Now, it is uncontroversial that by picking 2 doors at once, you have a 2/3 chance of being right. But notice that this scenario is essentially equivalent to the original one. In the original scenario, you make an initial choice, which divides the space of possible choices in 2: the one door you picked, and the 2 you did not pick. Prior to the host opening one of the doors, there is a 2/3 chance that the prize lies behind one of the doors you did not pick. When the host opens one of these doors, essentially what he has done is to shunt that 2/3 probability into the remaining door.

Back to picking 2 doors at once. By picking 2 doors at once, you have a priori, absolute certainty that one of the doors will be empty. Does this imply that you if you get to pick 2 doors before any are opened, you only have a 1/2 chance of picking the right one? That is essentially what you are claiming.

 

[Doc Al]

I can try, but I don't expect you to be convinced.

That's the spirit!

Not at all. Since the host opened a door without a prize, the prize must be behind one of the two remaining doors, but you are claiming that the prize is more likely to be in the unchosen door regardless of which door was chosen in the first place. This makes absolutely no sense.

You didn't read the entire thread, did you? This objection is thoroughly discussed there; no point in repeating it.

No, they don't. "Chances of winning" means "chances that the prize is behind a particular door". When you have 3 doors, the chance is 1/3; when you have 2 doors, it's 1/2.

Only if you insist on not taking advantage of the new information provided to you. Again, read the other thread.

Nothing changes when the host opens a door, because you knew from the beginning that, no matter which door you chose, there would always be at least one unchosen door without the prize. You are confused because you think the host gives you information by opening a door - that is definitely not the case. It would be information before you made your choice, but after you do it the host is only confirming what you already knew.

You most certainly are getting new information. Consider the case (discussed in the thread that you apparently haven't read!) where you have 1,000,001 doors. You pick at random. Your odds are pretty low (1/1,000,001) of being right. Now 999,999 doors are revealed to be empty. You'd have to be a bit slow not to realize that the odds are 1,000,000/1,000,001 that the prize is behind the one remaining door.

Believe me when I tell you, you are confused. I realize people don't like to be told that, but it's often true nonetheless.

Sorry, you're still wrong. Read the thread--the truth is out there.

If thinking is not your thing, then perhaps an experiment will convince you. Get someone to hide a coin under one of three cups; you pick at random, then they reveal an empty cup; but don't you dare switch. Bet even money. (And please, bet heavily!) Do this a few hundred times and see what happens.

 

[Doc Al]

Think of the problem this way. Suppose that we dispense with the host altogether, and we let you pick 2 doors at once. Now, it is uncontroversial that by picking 2 doors at once, you have a 2/3 chance of being right. ...

An excellent way of explaining it, hypnagogue!

 

[hypnagogue]

You most certainly are getting new information. Consider the case (discussed in the thread that you apparently haven't read!) where you have 1,000,001 doors. You pick at random. Your odds are pretty low (1/1,000,001) of being right. Now 999,999 doors are revealed to be empty. You'd have to be a bit slow not to realize that the odds are 1,000,000/1,000,001 that the prize is behind the one remaining door.

I'd like to work with this example to really drive the point home.

Scenario 1: There are 1,000,001 doors. Before the contestant does anything, the host proceeds to open 999,999 of the empty ones, leaving only two unopened doors behind for the contestant to choose from.

Scenario 2: There are 1,000,001 doors. The contestant chooses a door. The host proceeds to open 999,999 of the empty remaining ones, leaving the contestant the choice to remain with his original choice or to switch.

The claim is that in scenario 1, the contest has a 1/2 chance of choosing the door with the prize, whereas in scenario 2 he has a 1,000,000/1,000,001 chance of getting the prize if he chooses to switch.

Scenario 1 is uncontroversial; it is just a straightforward reduction of the problem to a simple random choice between two equally likely alternatives. So what has happened in scenario 2 to make it so radically different than scenario 1?

The key point is that in scenario 2, the contestant's initial guess at some random door reduces what is initially a probability space of 1,000,0001 equally likely choices into a drastically different probability space. Having chosen a particular door, and given the host's agreement to open all but one of the remaining doors such that all the opened doors are empty, the contestant is essentially given a binary choice of whether to stay or switch, and so the problem space reduces into two potential choices rather than 1,000,001. But these two potential choices are not equivalent, as in scenario 1. The one choice is to guess that the prize is behind the particular door that was initially picked; the other choice is essentially to guess that the prize is behind one of the 1,000,000 doors that was not picked. Obviously, the better bet is to choose the latter.

Yet another equivalent way to recast the problem is to once again eliminate the host, and simply give the contestant just one negative guess. In this scenario, the contestant guesses that some particular door is not the one with the prize, and if he guesses correctly, he wins the prize. Obviously, in the 3 door case, the contestant has a 2/3 chance of guessing that a certain door does not have the prize, and in the 1,000,001 door case, he has a 1,000,000 / 1,000,001 chance of guessing correctly. But this process of negative guessing is exactly equivalent to what the contestant does in the traditional formulation where he picks a certain door and then switches after the host opens all but one of the remaining doors.

 

[confutatis]

You know beforehand that there will be at least one unchosen door without the prize, true. But if you were to guess beforehand that door number 1 was empty, you would have had a 1/3 chance of being right, whereas if the host reveals door #1 to be empty, you now have certainty that it is empty.

But that only means that if you have chosen #1, you would have lost. But you did not choose it, so the information is irrelevant.

Think of the problem this way. Suppose that we dispense with the host altogether, and we let you pick 2 doors at once. Now, it is uncontroversial that by picking 2 doors at once, you have a 2/3 chance of being right. But notice that this scenario is essentially equivalent to the original one.

Yes in the sense that you have a 1/3 chance of not picking the door with the prize. So what?

In the original scenario, you make an initial choice, which divides the space of possible choices in 2: the one door you picked, and the 2 you did not pick. Prior to the host opening one of the doors, there is a 2/3 chance that the prize lies behind one of the doors you did not pick. When the host opens one of these doors, essentially what he has done is to shunt that 2/3 probability into the remaining door.

All I can say about this is that it's really bad math. This shunting thing would get you in trouble with a good teacher.

Back to picking 2 doors at once. By picking 2 doors at once, you have a priori, absolute certainty that one of the doors will be empty. Does this imply that you if you get to pick 2 doors before any are opened, you only have a 1/2 chance of picking the right one? That is essentially what you are claiming.

I have absolutely no idea how you came up with that.

Can I ask you a question? If the host doesn't open any doors and you just get to choose a door and see if you win or lose, what is the probability that you'll win if you play three times? (the prize may be shifted from door to door between rounds so you can't benefit from the last run)

 

[Averagesupernova]

It doesn't, and I can't believe people can actually be convinced of it. This is like being convinced nothing can move because of Zeno's paradox.

Whatever happened to commonsense?

You and I need to get together. We will use dice and a third party to determine which cup will hide a coin. 1 and 4 are first cup, 2 and 5 are second cup, 3 and 6 are 3rd cup. Obviously each player will not be able to see the roll. He will then choose a cup. The die roller will reveal an empty cup. You stick with your original choice and I will switch each time. $100 a pop. We can play as long as you like as long as it is 10 times or more each. As with anything in stats I believe, one or 2 times can't prove anything.

 

[hypnagogue]

Even better, try a web applet of the game: http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html

Being a good empiricist, I played a quick 500 games with the applet. In 250 of the games I switched, and in the other 250 games I stayed. Here are the stats:

p(win | switch) = 184 / 250 = 73.6%
p(win | stay) = 95 / 250 = 38.0%

I take it for granted that, given the sample sizes involved, I don't need to make any explicit calculations on how unlikely these results would have been if the probability distributions were really 50/50 to assert with high confidence that the probability distributions actually aren't 50/50. All confused theory aside, experiment tells us that the naive 50/50 intuition of this game is just plain wrong.

 

[Averagesupernova]

Ya know, I might be able to get confutatis to see it our way with this:

Suppose you and I sit down and you pick a door and I open one that is NOT the one. If you choose to stay, you still have 1/3 odds at winning.

Now, suppose your friend walks by AFTER the empty door is opened and is given the oppurtunity to pick and win also . Your friend has a 50/50 chance. You however, only have a 1 in 3 chance if you choose to stay. You are GUARANTEED a 1 in 3 chance if you don't switch. Since you cannot get any worse than a 1/3 odds, doesn't it make sense to switch? Hypnagogue showed 38% when not switching. A little above 1/3, but I'll still take the switch any day ahead of slightly above 1/3.

 

[hypnagogue]

The actual probabilities are 1/3 and 2/3. My own results with a finite number of trials yielded higher results on both counts, but would have converged to the actual probablities with a higher number of trials. In any case, the important point is that if staying or switching really was a 50/50 proposition, it would be exceedingly unlikely to get the skewed results that I recorded with such a high number of trials.

 

[KingNothing]

Think of the problem this way. Suppose that we dispense with the host altogether, and we let you pick 2 doors at once. Now, it is uncontroversial that by picking 2 doors at once, you have a 2/3 chance of being right. But notice that this scenario is essentially equivalent to the original one.

I think that's a good way of looking at it...unfortunately I just can't seem to bend my mind around to getting what everyone else got. I mean, I am not rejecting the right answer, I am just not understadning it.

Can you continue with how they are exactly the same scenario?

 

[Dayle Record]

I don't think there is a difference mathematically, but go with your gut. You chose a door. Your choice prompted the host to open another door upping your probabilities of winning; but synthetically adding a new element of probability, and tension. My gut would say he made a distracting move, because you had chosen the correct door to begin with.

 

[Averagesupernova]

I don't think there is a difference mathematically, but go with your gut. You chose a door. Your choice prompted the host to open another door upping your probabilities of winning; but synthetically adding a new element of probability, and tension. My gut would say he made a distracting move, because you had chosen the correct door to begin with.

You don't think there is a difference mathematically? C'mon! Let's go back to the 1,000,000 doors scenario. It's simply an exaggerated scenario of the 3 doors. So you pick a door out of 1,000,000. The host opens all doors except the one you picked and the 'correct' one. If you stay, there is a million to one shot you will get it right. Do you honestly think you are that lucky? If you switch, you are almost guaranteed to get the correct door since it is QUITE unlikely you will pick the right one out of a million. The same thing happens with 3 doors. You are GUARANTEED that one of the 2 doors left will be the right one. The difference is that you are more likely to pick the correct door the first time because the odds are 3:1 instead of 1,000,000:1. The million to one odds virtually guarantees you a wrong first pick. So wouldn't it be smart to pick the one that the host left covered? Same thing with 3 doors, it's just that it takes more tries to show the results. Someone correct me if I'm wrong because I'm no expert in stats, but it appears to me that what happens is the odds are inverted.

 

[hypnagogue]

I think that's a good way of looking at it...unfortunately I just can't seem to bend my mind around to getting what everyone else got. I mean, I am not rejecting the right answer, I am just not understadning it.

Can you continue with how they are exactly the same scenario?

Let's go with the negative choice scenario again. Recall that in the negative choice scenario, in order to win the prize the contestant must correctly guess that a certain door does not contain the prize. So the contestant has a 2/3 chance of winning the prize in the negative choice scenario with 3 doors.

What happens in the original version of the game if the contestant picks a certain door and then switches? Essentially, the contestant has done exactly the same as above; he's picked out one particular door, and by switching, he's betting that it's not the one with the prize.

If the contestant gets to pick 2 doors at once without the host being involved, that is also an equivalent scenario in the 3 door problem. By picking 2 out of the 3 doors, the contestant is implicitly guessing that the unpicked door doesn't have the prize behind it; hence this is also a sort of negative choice scenario.

If your intuition is still working against you, try considering the 1,000,001 door problem again. In the original version of the game, once you pick a certain door, the host opens 999,999 of the remaining empty doors and gives you the option to switch to the one remaining one that is closed. Notice that if you choose to switch, all the doors that you did not pick initially will have been opened. Thus, the situation we have when we switch is exactly equivalent to the situation we would have if we picked out one certain door initially, and then proceeded to open all the remaining ones. This in turn is equivalent to betting that that initially chosen door did not have the prize behind it.

 

[Integral]

consider this:
Under the given conditions you will pick the correct door 1 out of 3 times. This, of course, means that you will select the INCORRECT door 2 out of 3 times. So 2 out of 3 times the prize will be behind one of the doors you DID NOT SELECT. Now when the host exposes on of the 2 remaining doors as not containing the prize, 2 out of 3 times the prize will be behind the remaining door.

 

[Doc Al]

No it won't. You are erroneously assuming that the revelation that there is no prize behind the opened door does not affect your chance of winning.

Njorl, I think you are misinterpreting Integral's comment. I believe Integral would agree that revealing the door without a prize provides essential information.

 

[Njorl]

Njorl, I think you are misinterpreting Integral's comment. I believe Integral would agree that revealing the door without a prize provides essential information.

Sorry, I deleted my message, wasn't fast enough.

"I think I better think it out again!" (Fagan from Oliver)

 

[Njorl]

I was thinking about the problem incorrectly.

If the host opens a door that is guaranteed to be empty, you are better off switching. If the host opens a door that may or may not be empty, it does not matter.

It becomes easier to visualize with many doors.

We will play the game two ways.

The first way, the host opens random doors that are not your choice. In this game, it is anti-climactic because the prize is almost always revealed well before it comes down to two. It is never beneficial (or harmful) to switch choices in this game. Even if we are reduced to two doors, it does not matter, but there is only a 1 in 500 chance to get to this point.

Essentially, you and the host are making 1 in a 1000 chance guesses. Trading your guess for his is pointless.

In the other game, which simulates our logic problem, you make a 1 in a thousand guess. The host, by eliminating all possibilities but two is then asking you "Do you think your guess is right or do you think your guess is wrong?" Since your guess was 1/1000, it was probably wrong.

 

[Integral]

I didn't see Norjls post, but it is my understanding of this riddle, that the host has knowledge of where the prize is and always exposes an empty door.

Otherwise the host will expose the prize 1 out of 3 times.

 

[Njorl]

Here is a good way to think of it. Choose a door. The host then asks you if you think you guessed right or wrong. Since you only had a 1/3 chance to be right, you should say that you were wrong. Saying you answered wrong is equivalent to switching doors after one was revealed. While the final choice between two doors might appear to be a standard 50/50 chance, it is not. It is a choice between whether your original guess was right or wrong.

If there was a chance that the host would reveal the prize when he opened that first door, then you would be justified in saying it doesn't matter

I must admit, my convictions were strong in the opposite direction at the onset. So many people are bad at probability that my first instinct was to show those ignoramuses how they just don't know math. Oops

Njorl

 

[gurl4god77]

the door on the RIGHT!

 

[e(ho0n3]

When I first encountered this problem, I did't seem right to me that switching would help. Many posters before me gave other excellent ways of looking at the problem that might convince one that switching is good. I will propose another way of looking at the problem here that doesn't rely much on the probability of it all, but on intuition.

Consider these two games:

Game 1: You get to pick one of the curtains, the host doesn't reveal anything to you, and you are given the option to switch.

Game 2: This is the original game. You pick one curtain, the host reveals an empty curtain, and you are given the option to switch.

What happens if you don't switch in either of the two games? Your intuition should tell you here that your chances of winning will be the same in both games.

Now what happens if you do switch? Which game will you win more in if you play the game a considerable amount of times? (Note that in Game 1, your chances of winning will not change whether you switch or not).

Before I leave, I would like to propose a variation of the original game. Suppose a contestant picks one of the three curtains and leaves. The host lifts one of the 'bad' curtains leaving two curtains (one of them being the one the contestant picked). In comes another contestant (who doesn't know what has happened) and is asked to picked a curtain. The probability that this contestant will choose the winning curtain is 1/2 right? Now what if the contestant chose the the same curtain as previous one. Would his/her chances of winning increase by switching? What if before choosing one of the two curtains, the current contestant was told the whole story of what happened. Will this information allow the current contestant to choose a curtain that will increase his chances of winning?

e(ho0n3

 

[jcsd]

I think you've forgiot to mentiomn the fact that there are 3 curtains:

Game 1: the probabilty remains 1/3 whether or or not you decide to change.

Game 2: the proabbilty increases to 2/3 if you change curtains

In your final paragraph, the proabilty of choosing the correct curtain is indeed 1/2 if he doesn't have any prior knowledge, BUT this will increase to 2/3 if he chooses the curtain that the last player DIDN'T pick