Monty Hall Problem

[ELD]

I've found sharing this question to be very interesting. 

Before you are 3 doors.  One of them has a fantastic prize behind it, the other two, nothing.  You get to choose one.

After you make your choice, I will show you a loser from the two you did not pick, and then give you the option to switch your choice from your orignal choice to the remaining unknown door. 

Should you switch? 

 

 

[DoubtingThomas]

Yes. The odds will have increased from 33.3% to 50% if you switch.

[ELD]

It is correct to switch, but your numbers are incorrect.    It is not a 50% situation after the reveal.

[Magnus]

Yes, switch. The host's act of opening one door after you pick is not random. It's done with full knowledge. The host will never open (a) the door you picked or (b) the winning door. He always eliminates one of the two losing doors. Therefore, switching to the only remaining unopened door has a 50-50 chance of being the winner. Your initial pick remains only a 1/3 chance of winning. 

[ELD]

As I said before, you do not have a 50/50 chance of being the winner.  You will win twice as much if you switch than if you don't.

[ELD]

 

Yes, switch. The host's act of opening one door after you pick is not random. It's done with full knowledge. The host will never open (a) the door you picked or (b) the winning door. He always eliminates one of the two losing doors. Therefore, switching to the only remaining unopened door has a 50-50 chance of being the winner. Your initial pick remains only a 1/3 chance of winning. 

 

Your logic makes no sense.  How can it switching give you a 50/50 chance of being a winner, when you're saying the door you picked has a 1/3 chance of being the winner

[DoubtingThomas]

 

 

Yes, switch. The host's act of opening one door after you pick is not random. It's done with full knowledge. The host will never open (a) the door you picked or (b) the winning door. He always eliminates one of the two losing doors. Therefore, switching to the only remaining unopened door has a 50-50 chance of being the winner. Your initial pick remains only a 1/3 chance of winning. 

 

Your logic makes no sense.  How can it switching give you a 50/50 chance of being a winner, when you're saying the door you picked has a 1/3 chance of being the winner

 

In the event that the third door is eliminated, it is no longer a 1 in 3 shot. The remaining doors (2) are the only ones considered in the second choice, thus choosing between the remaining two (and the only choice to make is to switch) gives odds of 50%.

Hope that helps.

[Lowe D]

I think that ELD is trying to say that, because probability is conserved, the remaining door has probability 2/3 being the right one ( 1 = 1/3 + 2/3 ).

BTW odds is not the same as probability.  Probability is the ratio of the cardinality of a subset of possible outcomes, to the cardinality of the set of all possible outcomes.  Odds is the ratio of the cardinality of a subset, to the cardinality of the complement of that subset.

In this example, the probabilty of the initial door being right is 1/3.  The odds is 1/2.  The probabilty of the remaining door, after the reveal, being right is 2/3.  The odds is 2/1.

[Roger T]

Yes.  1/3 improves to 2/3. For N doors, your pick has chance 1/N with prize "out there" having chance (N-1)/N. Opening N-2 doors w/o prize reduces "out there" to oneother door (and your pick) :. leaving other door with chance (N-1)/N.

 

[Magnus]

I was thoroughly unclear. Yes, a switching strategy wins, overall, 2/3 times. 

The underlying reason, I believe, is that your initial random guess, when wrong, constrains the host's options. 2/3 of the time, this initial pick will be wrong, which leaves only one door for the host to reveal -- not your pick, and not the winner. Under these circumstances, the remaining one must hold the prize. 

[ELD]

Interesting response.  I wonder why you would say you were unclear when you stated something factually incorrect, instead of just saying you were in error.

[ELD]

 

Yes.  1/3 improves to 2/3. For N doors, your pick has chance 1/N with prize "out there"
 having chance (N-1)/N. Opening N-2 doors w/o prize reduces "out there" to one
other door (and your pick) :. leaving other door with chance (N-1)/N.

 

 

To be clear, this only applies to specifically choosing to reveal doors that are loosers on purpose.  If they are chosen at random, and may reveal the winner, then no information is gained.

[TheRobin]

haha, I remember it taking me hours of explanation on my brother's side until I finally understood why it's 1/3 to 2/3 after the reveal. And just as long when I tried to explain that to some friends of mine. Ah, good times, when talking about goats and doors all night could make for an enjoyable evening

What prompted you to bring the problem up, if I may ask?

[ELD]

 

haha, I remember it taking me hours of explanation on my brother's side until I finally understood why it's 1/3 to 2/3 after the reveal. And just as long when I tried to explain that to some friends of mine. Ah, good times, when talking about goats and doors all night could make for an enjoyable evening

What prompted you to bring the problem up, if I may ask?

 

1) It's a really fun problem

2) I've seen people flat out refuse to accept the truth about.  I know two people who've actually written a script to simulate the process, thinking they were going to disprove it.  Even when their program came back to confirm a 1/3 to 2/3 split, they started checking for errors in their program  

3) I'm new here and want to see what the conversations are like.

[Magnus]

 

Interesting response.  I wonder why you would say you were unclear when you stated something factually incorrect, instead of just saying you were in error.

 

I was in error. The reason for my error was that I was typing on my phone, and didn't think it out properly. What I wrote is erroneous. That part was a bit of a brain fart. 

When Monty makes his reveal, it appears (falsely) to be a 50-50 chance that the other door holds the prize, since there are now 2 remaining doors, and one holds a prize. 

But it's not 50-50, because believing it's 50-50 ignores the events that occurred earlier in the game. Your initial guess has influenced Monty's choice about which door to reveal -- he can never reveal the one you have chosen. Furthermore, your initial choice (when it is wrong) dictates Monty's behavior -- he can only open the one remaining wrong door. 

[TheRobin]

 

 

haha, I remember it taking me hours of explanation on my brother's side until I finally understood why it's 1/3 to 2/3 after the reveal. And just as long when I tried to explain that to some friends of mine. Ah, good times, when talking about goats and doors all night could make for an enjoyable evening

What prompted you to bring the problem up, if I may ask?

 

1) It's a really fun problem

2) I've seen people flat out refuse to accept the truth about.  I know two people who've actually written a script to simulate the process, thinking they were going to disprove it.  Even when their program came back to confirm a 1/3 to 2/3 split, they started checking for errors in their program  

3) I'm new here and want to see what the conversations are like.

 

haha, nice. I remember a friend of mine once said (after hours of debating) that he agrees that it is theoretically correct it would practically still be 50-50.

Well, I hope you enjoy the conversations then

[DoubtingThomas]

2/3 as opposed to 1/2

Good catch. It's been a while since I pulled this one up. I assumed ELD was asserting something else, but I now understand the correction.

[ELD]

 

 

Interesting response.  I wonder why you would say you were unclear when you stated something factually incorrect, instead of just saying you were in error.

 

I was in error. The reason for my error was that I was typing on my phone, and didn't think it out properly. What I wrote is erroneous. That part was a bit of a brain fart. 

When Monty makes his reveal, it appears (falsely) to be a 50-50 chance that the other door holds the prize, since there are now 2 remaining doors, and one holds a prize. 

But it's not 50-50, because believing it's 50-50 ignores the events that occurred earlier in the game. Your initial guess has influenced Monty's choice about which door to reveal -- he can never reveal the one you have chosen. Furthermore, your initial choice (when it is wrong) dictates Monty's behavior -- he can only open the one remaining wrong door. 

 

Interesting that it was a phone that caused the error, and not the human.  I wonder why the need to blame something or someone else, instead of just saying you made a simple mistake.  Perhaps something in your past makes it difficult to simply admit when you make an error, without coming up with excuses?

[ELD]

 

 

 

haha, I remember it taking me hours of explanation on my brother's side until I finally understood why it's 1/3 to 2/3 after the reveal. And just as long when I tried to explain that to some friends of mine. Ah, good times, when talking about goats and doors all night could make for an enjoyable evening

What prompted you to bring the problem up, if I may ask?

 

1) It's a really fun problem

2) I've seen people flat out refuse to accept the truth about.  I know two people who've actually written a script to simulate the process, thinking they were going to disprove it.  Even when their program came back to confirm a 1/3 to 2/3 split, they started checking for errors in their program  

3) I'm new here and want to see what the conversations are like.

 

haha, nice. I remember a friend of mine once said (after hours of debating) that he agrees that it is theoretically correct it would practically still be 50-50.

Well, I hope you enjoy the conversations then

 

For the most part it was fun

[ribuck]

Some people may find the following explanation more satisfying:

There are 3 doors, and you choose one of them.

There is a 1-in-3 chance that you have guessed correctly. If you have guessed correctly, you will win if you stick with your original choice, and you lose if you switch.

On the other hand, there is a 2-in-3 chance that you have guessed incorrectly. In that case, the host will open an empty door, and the prize must be behind the other door. You will lose if you stick with your original choice, and you will win if you switch.

So, "not switching" wins 1-in-3 times, and "switching" wins 2-in-3 times.

[ELD]

 

Some people may find the following explanation more satisfying:

There are 3 doors, and you choose one of them.

There is a 1-in-3 chance that you have guessed correctly. If you have guessed correctly, you will win if you stick with your original choice, and you lose if you switch.

On the other hand, there is a 2-in-3 chance that you have guessed incorrectly. In that case, the host will open an empty door, and the prize must be behind the other door. You will lose if you stick with your original choice, and you will win if you switch.

So, "not switching" wins 1-in-3 times, and "switching" wins 2-in-3 times.

 

A fine explaination.  This board grasps the Monty Hall Problem better than many other places I've posted it.  The funny thing about this problem, is many people are unable to grasp it, even with a clear, concise explaination like yours.  They hear it, follow it, then abandon it and say it's 50/50.  Very fun to see how different people approach problems.

[Waster]

A similar problem: There are three boxes. In each box there are two balls. One box has two black balls, one box has two white balls, one box has a white and a black ball in it. If the first ball you get out of the box is white, what is the chance the other ball is also white?

 

[ritherz]

 

Derren brown has a very good way of explaining it.  The explenation starts around 6 mins in.

[Lowe D]

@ Waster

Unless I'm mistaken, that problem is different from the MHP in that the first draw is always random: it affects the probablity of the desired event (both white).

The probability of both whites is 1/3.  When the first white is drawn, we condition on that outcome, the probability of which is 1/2.

P(both white, first white) / P(first white) = P(both white) / P(first white) = (1/3) / (1/2) = 2/3

[ELD]

 

A similar problem: There are three boxes. In each box there are two balls. One box has two black balls, one box has two white balls, one box has a white and a black ball in it. If the first ball you get out of the box is white, what is the chance the other ball is also white?

 

 

Love it.  Never heard of that one.  Let's see if I understand.

So you pick a ball at random.  It's white.  You have a 2/3 chance of having pulled it out of the box with two white, and a 1/3 chance of having pulled it out of the 50/50 box.  The chances that the next ball is white would therefore be 2/3. 

Is that correct?

[Magnus]

 

Interesting that it was a phone that caused the error, and not the human.  I wonder why the need to blame something or someone else, instead of just saying you made a simple mistake.  Perhaps something in your past makes it difficult to simply admit when you make an error, without coming up with excuses?

 

Phones can't cause errors of thought.  They're inanimate.  It was my sloppy thinking that produced my error.

But it's good to see that you're beginning to at least consider why people say what they say, and what they hope to accomplish by their manner and mode of communication.  I hope you find that line of inquiry to be more productive than trying to force people to stay on your preferred topic, and becoming irritated when they don't.

You can be sure, however, that I wasn't testing you -- posting in a way just to see how you respond.  My error was sincere.

[NotDarkYet]

Here is how I explan the answer to people who refuse to believe the answer.  

 

 

Imagine you have 100 doors.  You stand in front of door 1.   Now you have TWO SETS of doors.  

 

SET 1 = Your door.  

Set 2 = The 99 other doors.

 

Of those two sets.  Where is the money more likely to be?

 

In the large set, of course.

 

 

Now the host OPENS 98 of the 99 doors.  Leaving 1 remaining.

 

It becomes very very clear that the money is probably in that other door.

 

 

The 3 door problem is the same, but on the smallest possible scale.

[Waster]

 

 

A similar problem: There are three boxes. In each box there are two balls. One box has two black balls, one box has two white balls, one box has a white and a black ball in it. If the first ball you get out of the box is white, what is the chance the other ball is also white?

 

 

Love it.  Never heard of that one.  Let's see if I understand.

So you pick a ball at random.  It's white.  You have a 2/3 chance of having pulled it out of the box with two white, and a 1/3 chance of having pulled it out of the 50/50 box.  The chances that the next ball is white would therefore be 2/3. 

Is that correct?

 

This is correct yes. Maybe this problem is harder because it involves conditional probability and most people dont even know what it is. The subtelity is in the part 'given that the first ball is white'. 

[Waster]

 

@ Waster

Unless I'm mistaken, that problem is different from the MHP in that the first draw is always random: it affects the probablity of the desired event (both white).

The probability of both whites is 1/3.  When the first white is drawn, we condition on that outcome, the probability of which is 1/2.

P(both white, first white) / P(first white) = P(both white) / P(first white) = (1/3) / (1/2) = 2/3

 

I think you mean this:

P(second white | first white) = P(both white) / P(first white) = (1/3) / (1/2) = 2/3

 

[Lowe D]

What I wrote is the expression for the conditional probability P(both white | first white).

The MHP does involve conditional probability.  It's more difficult to evaluate than in your problem, because the host's action is not random.

[AnarcoB]

 

I've found sharing this question to be very interesting. 

Before you are 3 doors.  One of them has a fantastic prize behind it, the other two, nothing.  You get to choose one.

After you make your choice, I will show you a loser from the two you did not pick, and then give you the option to switch your choice from your orignal choice to the remaining unknown door. 

Should you switch? 

 

One could switch, but there would be no empirical reason to do so.  Why switch at all if there is no evidence tipping the odds of 50:50?

[ELD]

 

 

I've found sharing this question to be very interesting. 

Before you are 3 doors.  One of them has a fantastic prize behind it, the other two, nothing.  You get to choose one.

After you make your choice, I will show you a loser from the two you did not pick, and then give you the option to switch your choice from your orignal choice to the remaining unknown door. 

Should you switch? 

 

One could switch, but there would be no empirical reason to do so.  Why switch at all if there is no evidence tipping the odds of 50:50?

 

Hi,  If you read the previous posts, you'll see your logic is incorrect.  You have twice as high a chance of winning if you switch.  Empyrically speaking

[AnarcoB]

 

 

 

I've found sharing this question to be very interesting. 

Before you are 3 doors.  One of them has a fantastic prize behind it, the other two, nothing.  You get to choose one.

After you make your choice, I will show you a loser from the two you did not pick, and then give you the option to switch your choice from your orignal choice to the remaining unknown door. 

Should you switch? 

 

One could switch, but there would be no empirical reason to do so.  Why switch at all if there is no evidence tipping the odds of 50:50?

 

Hi,  If you read the previous posts, you'll see your logic is incorrect.  You have twice as high a chance of winning if you switch.  Empyrically speaking

 

I suppose if the initial question were worded differently I would be incorrect, but I still don't see it.  For instance, if I make a second guess, the odds of being correct will be better than the odds of being correct with the first guess.  If however we look at whether I "should" make a second guess because it increases my odds... I don't see it.

[Nathan T_ Freeman]

 

 

 

 

I've found sharing this question to be very interesting. 

Before you are 3 doors.  One of them has a fantastic prize behind it, the other two, nothing.  You get to choose one.

After you make your choice, I will show you a loser from the two you did not pick, and then give you the option to switch your choice from your orignal choice to the remaining unknown door. 

Should you switch? 

 

One could switch, but there would be no empirical reason to do so.  Why switch at all if there is no evidence tipping the odds of 50:50?

 

Hi,  If you read the previous posts, you'll see your logic is incorrect.  You have twice as high a chance of winning if you switch.  Empyrically speaking

 

I suppose if the initial question were worded differently I would be incorrect, but I still don't see it.  For instance, if I make a second guess, the odds of being correct will be better than the odds of being correct with the first guess.  If however we look at whether I "should" make a second guess because it increases my odds... I don't see it.

 

Since this is a philosophy board, the meaning of "should you" is certainly open for debate. But the basic assumption of the Monty Hall Problem is that you want to win a good prize.

If you can accept that premise, then it follows that you should take whatever actions maximize your chance of winning. It has been demonstrated quite clearly in this thread that switching doors increases your chance of winning from 1 in 3 to 2 in 3. If you're struggling to accept that (and you'd hardly be the first person to argue against this counter-intuitive-but-true argument) then read over all the replies. The explanation has been presented in full.

The MHP and people's reactions to it are an excellent measure of cognitive bias. A less mathematically intense but similarly challenging concept is the economic Law of Comparative Advantage.

A similarly interesting measure can be extrapolated from how people deal with the answers to the question that are correct, but backed by imprecise math. So when someone asks "should you switch?" and another person says "yes because your odds of winning improve by X%" and the reply is "no, they improve by Y%" and then a debate ignites about who admits when they're wrong and how they do it. To me, that is a far more interesting signal about the participants in a conversation.

Of course, the internet is pretty much the perfect feeding ground for the pedantic...

 

[AnarcoB]

I give up.  This must be a joke, or Asch part two.

(Correction, Asch not Milgram)