Riddle me this

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?

Related General Discussion News on Phys.org
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 [Broken]

Last edited by a moderator:
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
Janitor said:
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
Mentor
confutatis said:
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
Doc Al said:
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.

Last edited by a moderator:
Doc Al
Mentor
confutatis said:
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
Doc Al said:
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
Mentor
confutatis said:
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
Doc Al said:
Point out the errors in the arguments offered.
I can try, but I don't expect you to be convinced.

Doc Al (from the other thread) said:
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
Staff Emeritus
Gold Member
confutatis said:
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
Mentor
confutatis said:
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.

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
Mentor
hypnagogue said:
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
Staff Emeritus
Gold Member
Doc Al said:
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.

Last edited:
confutatis
hypnagogue said:
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
Gold Member
confutatis said:
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
Staff Emeritus
Gold Member
Even better, try a web applet of the game: http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html [Broken]

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.

Last edited by a moderator:
Averagesupernova
Gold Member
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
Staff Emeritus
Gold Member
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.

hypnagogue said:
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 thats 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, im not rejecting the right answer, im jsut not understadning it.

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

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
Gold Member
Dayle Record said:
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
Staff Emeritus
Gold Member
KingNothing said:
I think thats 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, im not rejecting the right answer, im jsut 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
Staff Emeritus