[Oasis]

You are a contestant on the 60s-70s game show Let's Make a Deal.  You are looking at three doors.  Behind one of them is a ticket to the 2014 Rose Bowl, where you will sit (if you choose the right door) on the 50-yard line between John Hennessy, who wants to hear how YOU would allow access to Stanford's Rose Bowl ticket allotment, and Dana Delany deliciously dressed as a 70s Dolly. Behind the other two doors are free Cal football season tickets, without BG.  You choose a door, desperately dreaming of your ultimate Rose Bowl experience.  Monty Hall (the host of the show for those of you who don't remember) opens one of the three doors and reveals Cal season tickets. He then offers you the opportunity to switch to the other unopened door, toward which Carol Merrill (those too young to remember the show can Google her) is pointing seductively.

Should you switch doors?

As Hulk suggested, just figure it out and say if you have an answer.

[Oasis]

You are a contestant on the 60s-70s game show Let's Make a Deal.  You are looking at three doors.  Behind one of them is a ticket to the 2014 Rose Bowl, where you will sit (if you choose the right door) on the 50-yard line between John Hennessy, who wants to hear how YOU would allow access to Stanford's Rose Bowl ticket allotment, and Dana Delany deliciously dressed as a 70s Dolly. Behind the other two doors are free Cal football season tickets, without BG.  You choose a door, desperately dreaming of your ultimate Rose Bowl experience.  Monty Hall (the host of the show for those of you who don't remember) opens one of the three doors and reveals Cal season tickets. He then offers you the opportunity to switch to the other unopened door, toward which Carol Merrill (those too young to remember the show can Google her) is pointing seductively.

Should you switch doors?

As Hulk suggested, just figure it out and say if you have an answer.

[Greyhound]

Definitely switch. Always switch. Unless the answer to Hulk01's brainteaser lies with Carol Merrill.

[Greyhound]

Definitely switch. Always switch. Unless the answer to Hulk01's brainteaser lies with Carol Merrill.

[Leftcoast]

Yup.  This was a problem in intro to statistics back in the day.

[Leftcoast]

Yup.  This was a problem in intro to statistics back in the day.

[CompSci87]

Switching is right under the usual assumption that Monty always opens a door and gives the chance to switch (or that he decides randomly whether to do so -- that works too). IIRC you also need to assume Monty knows what is behind each door.

Monty Hall himself remarked that he might choose to offer you a chance to switch only if your initial choice was the good prize. Then it would always be wrong to switch (when given the chance to)!

[CompSci87]

Switching is right under the usual assumption that Monty always opens a door and gives the chance to switch (or that he decides randomly whether to do so -- that works too). IIRC you also need to assume Monty knows what is behind each door.

Monty Hall himself remarked that he might choose to offer you a chance to switch only if your initial choice was the good prize. Then it would always be wrong to switch (when given the chance to)!

[FarmBoy]

IIRC you also need to assume Monty knows what is behind each door.

I don't believe that is correct. If Monty doesn't know which door holds the prize, you should still switch but 1/3 the time you'll switch to the door he opened because it will have the prize revealed. If you choose door A and Monty randomly opens door B and it's empty, then door C has a 2/3 probability of containing the prize, regardless of Monty's knowledge. If Monty randomly opens door B and it holds the prize, then of course you would choose door B.

[FarmBoy]

IIRC you also need to assume Monty knows what is behind each door.

I don't believe that is correct. If Monty doesn't know which door holds the prize, you should still switch but 1/3 the time you'll switch to the door he opened because it will have the prize revealed. If you choose door A and Monty randomly opens door B and it's empty, then door C has a 2/3 probability of containing the prize, regardless of Monty's knowledge. If Monty randomly opens door B and it holds the prize, then of course you would choose door B.

[Hulk01]

Switch.

When you chose the first time, your odds of being correct were one in three.

Now that Monty has revealed that the tickets can only be behind one of two doors, your odds of being correct are one in two, or 50 percent better.

In addition to the statistical problem that this illustrates, the question also may illuminate a cognitive bias--our risk aversion.  In the situation that you pose, we are afraid to switch doors because we fear the reaction we will have if the Rose Bowl tickets were behind that door that we chose.  Risk looks and feels much more vivid to us than reward, so we will forgo a reward--the increased likelihood of n the ticket--simply to protect us from a risk--in this case, the very good chance of 50 percent-that switching will end up appearing to be the "wrong" decision even though it was the optimum decision.

Or  maybe it's really really early right now and I am dead wrong.

[Hulk01]

Switch.

When you chose the first time, your odds of being correct were one in three.

Now that Monty has revealed that the tickets can only be behind one of two doors, your odds of being correct are one in two, or 50 percent better.

In addition to the statistical problem that this illustrates, the question also may illuminate a cognitive bias--our risk aversion.  In the situation that you pose, we are afraid to switch doors because we fear the reaction we will have if the Rose Bowl tickets were behind that door that we chose.  Risk looks and feels much more vivid to us than reward, so we will forgo a reward--the increased likelihood of n the ticket--simply to protect us from a risk--in this case, the very good chance of 50 percent-that switching will end up appearing to be the "wrong" decision even though it was the optimum decision.

Or  maybe it's really really early right now and I am dead wrong.

[stupac2]

Switch.

When you chose the first time, your odds of being correct were one in three.

Now that Monty has revealed that the tickets can only be behind one of two doors, your odds of being correct are one in two, or 50 percent better.

In addition to the statistical problem that this illustrates, the question also may illuminate a cognitive bias--our risk aversion.  In the situation that you pose, we are afraid to switch doors because we fear the reaction we will have if the Rose Bowl tickets were behind that door that we chose.  Risk looks and feels much more vivid to us than reward, so we will forego a reward--the increased likelihood of gettig the ticket--simply to protect us from a risk--in this case, the very good chance--50 percent-that switching will end up appearing to be the "wrong" decision even though it was the optimum decision.

Or  maybe it's really really early right now and I am dead wrong.

This is the right conclusion but the wrong reasoning. After Monty opens the door, there's still a 1/3 chance that your door has the prize (nothing has changed the fundamental logic of your choice having a 1/3 chance to be correct at the beginning). If it were 1/2 then there'd be no reason to switch. So there's a 1/3 chance your door is right, and a 2/3 chance that the other door is right.

I'm doing this sort shrift, it's really difficult to explain it in a way that's intuitive and whole books have been written on the problem, but the wiki page ought to do it for anyone who's confused: http://en.wikipedia.org/wiki/Monty_hall_problem

[stupac2]

Switch.

When you chose the first time, your odds of being correct were one in three.

Now that Monty has revealed that the tickets can only be behind one of two doors, your odds of being correct are one in two, or 50 percent better.

In addition to the statistical problem that this illustrates, the question also may illuminate a cognitive bias--our risk aversion.  In the situation that you pose, we are afraid to switch doors because we fear the reaction we will have if the Rose Bowl tickets were behind that door that we chose.  Risk looks and feels much more vivid to us than reward, so we will forego a reward--the increased likelihood of gettig the ticket--simply to protect us from a risk--in this case, the very good chance--50 percent-that switching will end up appearing to be the "wrong" decision even though it was the optimum decision.

Or  maybe it's really really early right now and I am dead wrong.

This is the right conclusion but the wrong reasoning. After Monty opens the door, there's still a 1/3 chance that your door has the prize (nothing has changed the fundamental logic of your choice having a 1/3 chance to be correct at the beginning). If it were 1/2 then there'd be no reason to switch. So there's a 1/3 chance your door is right, and a 2/3 chance that the other door is right.

I'm doing this sort shrift, it's really difficult to explain it in a way that's intuitive and whole books have been written on the problem, but the wiki page ought to do it for anyone who's confused: http://en.wikipedia.org/wiki/Monty_hall_problem

[Redrum]

And the reason is that back in the Pleistocene of sports message boards,  occasional visitor Teejers  explained the sudden switch of school preference by recruits, with Stanford "mysteriously" disappearing from a recruit's list of favorites.  Teejers used the Let's Make a Deal analogy to lead us through the possible choices for why a recruit would just disappear... to the conclusion that it was Stanford Admissions... behind Door #1.  And ever since, it has become shorthand code for bad news on the admissions front.

Note, now, that these mysterious changes in school preference have become MUCH less prevalent since the Renaissance began.  There is just more information out there on student athletes in general,  Admissions seems to see itself as less of a slaughterhouse of student athlete dreams.  But, mostly, Stanford and the athletic coaching staffs have become MUCH more adept at evaluating HS student academics and consequently sending more acceptable candidates to the final test.  That there is an administrative post on the football staff that concentrates on: a) finding likely qualified candidates across America and b)  vets and advises them on how to be admissible seems like such a logical step.  You'd think, but only 6 years ago, you would be wrong.  So logical, yet so long in coming.

[Redrum]

And the reason is that back in the Pleistocene of sports message boards,  occasional visitor Teejers  explained the sudden switch of school preference by recruits, with Stanford "mysteriously" disappearing from a recruit's list of favorites.  Teejers used the Let's Make a Deal analogy to lead us through the possible choices for why a recruit would just disappear... to the conclusion that it was Stanford Admissions... behind Door #1.  And ever since, it has become shorthand code for bad news on the admissions front.

Note, now, that these mysterious changes in school preference have become MUCH less prevalent since the Renaissance began.  There is just more information out there on student athletes in general,  Admissions seems to see itself as less of a slaughterhouse of student athlete dreams.  But, mostly, Stanford and the athletic coaching staffs have become MUCH more adept at evaluating HS student academics and consequently sending more acceptable candidates to the final test.  That there is an administrative post on the football staff that concentrates on: a) finding likely qualified candidates across America and b)  vets and advises them on how to be admissible seems like such a logical step.  You'd think, but only 6 years ago, you would be wrong.  So logical, yet so long in coming.

[CompSci87]

IIRC you also need to assume Monty knows what is behind each door.

I don't believe that is correct. If Monty doesn't know which door holds the prize, you should still switch but 1/3 the time you'll switch to the door he opened because it will have the prize revealed. If you choose door A and Monty randomly opens door B and it's empty, then door C has a 2/3 probability of containing the prize, regardless of Monty's knowledge. If Monty randomly opens door B and it holds the prize, then of course you would choose door B.

No, it really does make a difference. I am going to give a slightly handwavy argument that may not convince you, but really *all* arguments on this type of puzzle are handwavy unless you explicitly use Bayes' Theorem. See this book for an excellent treatment: http://www.amazon.com/The-Monty-Hall-Pro...0195367898

Since neither you nor Monty knows which door the prize is behind, assume without loss of generality that you always initially choose A and he always opens B. A priori, there was a 1/3 probability of the prize being behind each door. If it's behind A, you would lose by switching. If it's behind B, Monty shows you the prize -- but that's not what happened here, as we're told that the door he opened did not have the prize. If it's behind C, you would win by switching.

In this variant, we really do have the situation of three alternatives that were equally likely a priori, where one has been eliminated, leaving the two remaining with equal probability. So it doesn't matter if you stick or switch; you have a 1/2 probability either way.

The difference from the original problem is that in the original, if the prize is behind B, Monty doesn't open it; he opens C. (Here I'm assuming WLOG that you always open A.)

[CompSci87]

IIRC you also need to assume Monty knows what is behind each door.

I don't believe that is correct. If Monty doesn't know which door holds the prize, you should still switch but 1/3 the time you'll switch to the door he opened because it will have the prize revealed. If you choose door A and Monty randomly opens door B and it's empty, then door C has a 2/3 probability of containing the prize, regardless of Monty's knowledge. If Monty randomly opens door B and it holds the prize, then of course you would choose door B.

No, it really does make a difference. I am going to give a slightly handwavy argument that may not convince you, but really *all* arguments on this type of puzzle are handwavy unless you explicitly use Bayes' Theorem. See this book for an excellent treatment: http://www.amazon.com/The-Monty-Hall-Pro...0195367898

Since neither you nor Monty knows which door the prize is behind, assume without loss of generality that you always initially choose A and he always opens B. A priori, there was a 1/3 probability of the prize being behind each door. If it's behind A, you would lose by switching. If it's behind B, Monty shows you the prize -- but that's not what happened here, as we're told that the door he opened did not have the prize. If it's behind C, you would win by switching.

In this variant, we really do have the situation of three alternatives that were equally likely a priori, where one has been eliminated, leaving the two remaining with equal probability. So it doesn't matter if you stick or switch; you have a 1/2 probability either way.

The difference from the original problem is that in the original, if the prize is behind B, Monty doesn't open it; he opens C. (Here I'm assuming WLOG that you always open A.)

[FarmBoy]

Ahhhh, I got it. Essentially, Monty has eliminated a door at random meaning the remaining doors are equal in proability and the chance of being right is 50/50. But interestingly, the probabiliy of winning this game is the same as the one where Monty knows where the prize is and reveals an empty door every time (assuming that in your version, if Monty mistakenly reveals the prize you get to keep it). In both games, you have a 2/3 chance of winning, I think.

[FarmBoy]

Ahhhh, I got it. Essentially, Monty has eliminated a door at random meaning the remaining doors are equal in proability and the chance of being right is 50/50. But interestingly, the probabiliy of winning this game is the same as the one where Monty knows where the prize is and reveals an empty door every time (assuming that in your version, if Monty mistakenly reveals the prize you get to keep it). In both games, you have a 2/3 chance of winning, I think.

[Extra Point]

The solution to this problem has nothing to do with statistics. The objective of the producers is to increase viewership and in turn advertising revenue. Millions of dollars are at stake. To do this, they want the show to be as as emotionally engaging as possible. Since the majority of viewers are rooting for the participants, and the majority of the participants are reluctant to switch, it is better to stick with the inital choice. While Monte may not know which door has the better prize, the producers certanily do, and the order to offer the option can easily be given with a cue card.

[Extra Point]

The solution to this problem has nothing to do with statistics. The objective of the producers is to increase viewership and in turn advertising revenue. Millions of dollars are at stake. To do this, they want the show to be as as emotionally engaging as possible. Since the majority of viewers are rooting for the participants, and the majority of the participants are reluctant to switch, it is better to stick with the inital choice. While Monte may not know which door has the better prize, the producers certanily do, and the order to offer the option can easily be given with a cue card.

[card_alum_11]

[quote author=FarmBoy link=topic=9125.msg78005#msg78005 date=1386851373]

IIRC you also need to assume Monty knows what is behind each door.

I don't believe that is correct. If Monty doesn't know which door holds the prize, you should still switch but 1/3 the time you'll switch to the door he opened because it will have the prize revealed. If you choose door A and Monty randomly opens door B and it's empty, then door C has a 2/3 probability of containing the prize, regardless of Monty's knowledge. If Monty randomly opens door B and it holds the prize, then of course you would choose door B.

No, it really does make a difference. I am going to give a slightly handwavy argument that may not convince you, but really *all* arguments on this type of puzzle are handwavy unless you explicitly use Bayes' Theorem. See this book for an excellent treatment: http://www.amazon.com/The-Monty-Hall-Pro...0195367898

Since neither you nor Monty knows which door the prize is behind, assume without loss of generality that you always initially choose A and he always opens B. A priori, there was a 1/3 probability of the prize being behind each door. If it's behind A, you would lose by switching. If it's behind B, Monty shows you the prize -- but that's not what happened here, as we're told that the door he opened did not have the prize. If it's behind C, you would win by switching.

In this variant, we really do have the situation of three alternatives that were equally likely a priori, where one has been eliminated, leaving the two remaining with equal probability. So it doesn't matter if you stick or switch; you have a 1/2 probability either way.

The difference from the original problem is that in the original, if the prize is behind B, Monty doesn't open it; he opens C. (Here I'm assuming WLOG that you always open A.)
[/quote]

Interesting.  At first, I thought that Monty wouldn't know what was behind each door and thus you've gone from having a 1/3 chance of getting the tickets to having a 1/2 change of getting the RB tickets so switch or not makes no difference.

But if Monty has knowledge and we know he won't select the door with the RB tickets, then you're gaining information with his selection.

There are three combinations when you select. 

Door 1, Door 2, Door 3
RB, LAU, LAU
LAU,RB,LAU
LAU,LAU,RB

(*LAU = Kal, RB = Grand Daddy of them all)

Once you select, Monty's combinations look like this.

LAU,LAU
RB,LAU
LAU,RB

So Monty has a 2 in 3 chance of having the Rose Bowl tickets in his two doors.  Which means that when he made his selection there was a 2/3 chance that he deliberately picked that door in order to avoid picking the Rose Bowl tickets.  So you have a 2/3 chance of knowing that Monty was communicating to you the actual location of the Rose Bowl tickets. 

But if he randomly picked and happened to expose some Kal tickets.  There was a 2/3 chance he got lucky.  The two possible combinations assuming he picked door two would be {LAU, LAU} or {LAU, RB}.  Because you know he picked randomly, then these are the combinations you are picking from and deciding if you want to switch from door 1 to door 3.  You're at 50/50.  You lose the possibility that he selected door 3 due to seeing {RB,LAU} and that's why the pick isn't giving you more information.

[card_alum_11]

[quote author=FarmBoy link=topic=9125.msg78005#msg78005 date=1386851373]

IIRC you also need to assume Monty knows what is behind each door.

I don't believe that is correct. If Monty doesn't know which door holds the prize, you should still switch but 1/3 the time you'll switch to the door he opened because it will have the prize revealed. If you choose door A and Monty randomly opens door B and it's empty, then door C has a 2/3 probability of containing the prize, regardless of Monty's knowledge. If Monty randomly opens door B and it holds the prize, then of course you would choose door B.

No, it really does make a difference. I am going to give a slightly handwavy argument that may not convince you, but really *all* arguments on this type of puzzle are handwavy unless you explicitly use Bayes' Theorem. See this book for an excellent treatment: http://www.amazon.com/The-Monty-Hall-Pro...0195367898

Since neither you nor Monty knows which door the prize is behind, assume without loss of generality that you always initially choose A and he always opens B. A priori, there was a 1/3 probability of the prize being behind each door. If it's behind A, you would lose by switching. If it's behind B, Monty shows you the prize -- but that's not what happened here, as we're told that the door he opened did not have the prize. If it's behind C, you would win by switching.

In this variant, we really do have the situation of three alternatives that were equally likely a priori, where one has been eliminated, leaving the two remaining with equal probability. So it doesn't matter if you stick or switch; you have a 1/2 probability either way.

The difference from the original problem is that in the original, if the prize is behind B, Monty doesn't open it; he opens C. (Here I'm assuming WLOG that you always open A.)
[/quote]

Interesting.  At first, I thought that Monty wouldn't know what was behind each door and thus you've gone from having a 1/3 chance of getting the tickets to having a 1/2 change of getting the RB tickets so switch or not makes no difference.

But if Monty has knowledge and we know he won't select the door with the RB tickets, then you're gaining information with his selection.

There are three combinations when you select. 

Door 1, Door 2, Door 3
RB, LAU, LAU
LAU,RB,LAU
LAU,LAU,RB

(*LAU = Kal, RB = Grand Daddy of them all)

Once you select, Monty's combinations look like this.

LAU,LAU
RB,LAU
LAU,RB

So Monty has a 2 in 3 chance of having the Rose Bowl tickets in his two doors.  Which means that when he made his selection there was a 2/3 chance that he deliberately picked that door in order to avoid picking the Rose Bowl tickets.  So you have a 2/3 chance of knowing that Monty was communicating to you the actual location of the Rose Bowl tickets. 

But if he randomly picked and happened to expose some Kal tickets.  There was a 2/3 chance he got lucky.  The two possible combinations assuming he picked door two would be {LAU, LAU} or {LAU, RB}.  Because you know he picked randomly, then these are the combinations you are picking from and deciding if you want to switch from door 1 to door 3.  You're at 50/50.  You lose the possibility that he selected door 3 due to seeing {RB,LAU} and that's why the pick isn't giving you more information.