# The Curious Incident of the Dog in the Night-Time discussion

The Monty Hall problem.

date
newest »

You're wrong.

Here are the three doors;

A B C

The car is behind door A.

There are three things that can happen here.

1. The players chooses door B. The presenter MUST open door C. In this scenario, if the player switches to the other door, door A, he wins the car.

2. The player chooses door C. The presenter MUST open door B. Again, if the player switches to the other door, door A, he wins the car.

3. The player chooses door A. The presenter can open either door B or C. This is the only scenario in which the player shouldn't switch doors.

As you can see, there are two scenarios where the player should switch, and only one where he shouldn't. In other words, switching is right 2/3 times, and sticking is only right 1/3 times - exactly the odds quoted.

This is explained in more depth here; http://montyhallproblem.com/

Here are the three doors;

A B C

The car is behind door A.

There are three things that can happen here.

1. The players chooses door B. The presenter MUST open door C. In this scenario, if the player switches to the other door, door A, he wins the car.

2. The player chooses door C. The presenter MUST open door B. Again, if the player switches to the other door, door A, he wins the car.

3. The player chooses door A. The presenter can open either door B or C. This is the only scenario in which the player shouldn't switch doors.

As you can see, there are two scenarios where the player should switch, and only one where he shouldn't. In other words, switching is right 2/3 times, and sticking is only right 1/3 times - exactly the odds quoted.

This is explained in more depth here; http://montyhallproblem.com/

*Kali wrote: "You're wrong.*

Here are the three doors;

A B C

The car is behind door A.

There are three things that can happen here.

1. The players chooses door B. The presenter MUST open door C. In this s..."

Here are the three doors;

A B C

The car is behind door A.

There are three things that can happen here.

1. The players chooses door B. The presenter MUST open door C. In this s..."

I actually learned about this my Senior year in high school and this is the exact way he described it.

Let's look at the diagram at the end of chapter 101...

The first decision that you make is shown as having three branches. Note that the right hand branch is that "you choose a door with a goat behind it"

The problem states that Monty opens a door and shows us the goat and then asks us to decide to stay or trade.

Let's cover up that whole branch of the outcome matrix. and evaluate our second decision. That leaves us with 4 possible outcomes. In two you get a car, and in two you get a goat. There is no reason to trade (or to NOT trade) to better your odds. The key to understanding this problem is that you're making your second decision on a "pruned" outcome tree.

That's a concept that Chris admits that he has difficulty with. I know that the idea of choices based on subsequent information causes some people difficulty and since

The first decision that you make is shown as having three branches. Note that the right hand branch is that "you choose a door with a goat behind it"

The problem states that Monty opens a door and shows us the goat and then asks us to decide to stay or trade.

Let's cover up that whole branch of the outcome matrix. and evaluate our second decision. That leaves us with 4 possible outcomes. In two you get a car, and in two you get a goat. There is no reason to trade (or to NOT trade) to better your odds. The key to understanding this problem is that you're making your second decision on a "pruned" outcome tree.

That's a concept that Chris admits that he has difficulty with. I know that the idea of choices based on subsequent information causes some people difficulty and since

*my belief*is that there is no harm in swapping, I'll be glad to swap to mollify any of you that still have concerns.
Just like Stephen says:

At first you have a 1:3 chance of being right.

Then a door is KNOWINGLY chosen to be revealed. This is a door that has a goat behind it.

The odds of the door you originally chose to be right now become 1:2. Yes, the odds CHANGE even before you choose if you want to switch or stay. So there really is no difference if you stay or switch. Both doors now have a 50% chance of being the right door.

At first you have a 1:3 chance of being right.

Then a door is KNOWINGLY chosen to be revealed. This is a door that has a goat behind it.

The odds of the door you originally chose to be right now become 1:2. Yes, the odds CHANGE even before you choose if you want to switch or stay. So there really is no difference if you stay or switch. Both doors now have a 50% chance of being the right door.

You're still wrong. And so are you, Ariel.

The second situation is not totally dependent of the first situation. The presenter MUST open a door with a goat behind; his decision is not random. Your "pruned outcome tree" is a totally wrong evaluation of the problem. You are wrong (and extremely arrogant). There is no clearer way to say it.

You're going to feel extremely stupid one day, when you actually understand the problem.

Here; play the game yourself. Keep track of whether sticking wins you the car 1/2 of the time or 1/3 of the time; http://math.ucsd.edu/~crypto/Monty/mo...

Or, read a more in-depth explanation here; http://mathforum.org/dr.math/faq/faq....

I cannot stand willful ignorance. Stephen, your offer to "mollify" anyone who disagrees by continuing to be ignorant but pretending to agree is downright insulting. Nevermind your statement about your "belief" is something that is easily disprovable!

The second situation is not totally dependent of the first situation. The presenter MUST open a door with a goat behind; his decision is not random. Your "pruned outcome tree" is a totally wrong evaluation of the problem. You are wrong (and extremely arrogant). There is no clearer way to say it.

You're going to feel extremely stupid one day, when you actually understand the problem.

Here; play the game yourself. Keep track of whether sticking wins you the car 1/2 of the time or 1/3 of the time; http://math.ucsd.edu/~crypto/Monty/mo...

Or, read a more in-depth explanation here; http://mathforum.org/dr.math/faq/faq....

I cannot stand willful ignorance. Stephen, your offer to "mollify" anyone who disagrees by continuing to be ignorant but pretending to agree is downright insulting. Nevermind your statement about your "belief" is something that is easily disprovable!

1) I often feel stupid when I realize that I've been wrong.

2) do you mean

3) I don't fancy myself being willfully ignorant so much as fancying myself a peacemaker by offering to swap. Understanding (like faith) is not something one can always command.

4) Can you explain

2) do you mean

*The second situation is not totally***independent**of the first situation?3) I don't fancy myself being willfully ignorant so much as fancying myself a peacemaker by offering to swap. Understanding (like faith) is not something one can always command.

4) Can you explain

**why**the pruned tree metaphor that I used is incorrect?
You already knew that two of the doors hid goats. Knowing which one one of them is doesn't change the fact that you only had a 1/3 chance of choosing correctly in the first place. Look at the three scenarios I described; they are the only three things that can happen. In two out of three cases, switching wins you the car.

Look at the pages I linked to. Google it yourself if you like - the Monty Hall problem is one that a lot of people have trouble with, so there are a great deal of webpages devoted to explaining it.

The pruned tree metaphor you used is incorrect because, as I said, the second situation is not independent of the first. The presenter KNOWS which door hides the car, and makes his decision of which door to open based on that knowledge, leading to the three situations I described in my first post.

I don't know how your mind understands things; mine is visual. It may help you to play around with a pack of cards (ace = car, jack = goat?) if yours is kinetic or visual, or to look at the webpages if yours is verbal.

Look at the pages I linked to. Google it yourself if you like - the Monty Hall problem is one that a lot of people have trouble with, so there are a great deal of webpages devoted to explaining it.

The pruned tree metaphor you used is incorrect because, as I said, the second situation is not independent of the first. The presenter KNOWS which door hides the car, and makes his decision of which door to open based on that knowledge, leading to the three situations I described in my first post.

I don't know how your mind understands things; mine is visual. It may help you to play around with a pack of cards (ace = car, jack = goat?) if yours is kinetic or visual, or to look at the webpages if yours is verbal.

I think my problem lies in considering my second choice as NOT being independent of the first.

However as I said before... I'm willing to always swap. There is an infinity of information in the universe and I can't know or understand everything. So, since my intuition suggests that swapping will cause no HARM I'm willing to swap and admit that I don't understand everything.

On this (and how my catalytic converter works) I'm willing to remain willfully ignorant.

However as I said before... I'm willing to always swap. There is an infinity of information in the universe and I can't know or understand everything. So, since my intuition suggests that swapping will cause no HARM I'm willing to swap and admit that I don't understand everything.

On this (and how my catalytic converter works) I'm willing to remain willfully ignorant.

I apologise for the manner in which I expressed myself the other day. I should not have taken my bad day out on you.

If you're still interested, I suggest another way to visualise the problem.

Imagine there are three cards on a table - an ace and two jacks. You want the ace.

All cards are face down so you can't see them. You pick one at random, and don't look at it.

At this point, there is 2/3 of a chance that you picked up a jack.

To focus our attention back on the ace, we can put it another way; there is 2/3 of a chance that the cards left on the table are an ace and the other jack. This is the most likely scenario.

Your opponent (for lack of a better term) looks at the cards, and removes a jack, leaving one card face-down on the table.

To recap; there is still 2/3 of a chance that the ace never left the table. There is 1/3 of a chance that it is in your hand, and no chance whatsoever that the other person took it, since you know they took a jack.

So, to have the highest chance of getting the ace, you should switch at this point.

The only way the odds could switch to 50-50 at the stick/switch decision is if you put your card back on the table, and shuffled it with the other one till you couldn't tell which was which.

Fortunately, that isn't the case; you know that the card in your hand/door you chose had only 1/3 of a chance of being right, meaning that there is 2/3 of a chance that you left the ace on the table (didn't pick the car). Yo also know that there is at least one jack on the table (goat behind a door you didn't pick). The only information you gain from the other person is which card that is - there's still 2/3 of a chance that you're holding the other jack.

If you're still interested, I suggest another way to visualise the problem.

Imagine there are three cards on a table - an ace and two jacks. You want the ace.

All cards are face down so you can't see them. You pick one at random, and don't look at it.

At this point, there is 2/3 of a chance that you picked up a jack.

To focus our attention back on the ace, we can put it another way; there is 2/3 of a chance that the cards left on the table are an ace and the other jack. This is the most likely scenario.

Your opponent (for lack of a better term) looks at the cards, and removes a jack, leaving one card face-down on the table.

To recap; there is still 2/3 of a chance that the ace never left the table. There is 1/3 of a chance that it is in your hand, and no chance whatsoever that the other person took it, since you know they took a jack.

So, to have the highest chance of getting the ace, you should switch at this point.

The only way the odds could switch to 50-50 at the stick/switch decision is if you put your card back on the table, and shuffled it with the other one till you couldn't tell which was which.

Fortunately, that isn't the case; you know that the card in your hand/door you chose had only 1/3 of a chance of being right, meaning that there is 2/3 of a chance that you left the ace on the table (didn't pick the car). Yo also know that there is at least one jack on the table (goat behind a door you didn't pick). The only information you gain from the other person is which card that is - there's still 2/3 of a chance that you're holding the other jack.

Thanks for trying to explain, but I still get stuck on this. I'm still not entirely convinced that I'm wrong, but As I said before I'll accept your analysis, particularly as there's no perceived downside.

I guess that it's a bit like visualizing a tesseract. There are just some things one's brain is not equipped to do.

I guess that it's a bit like visualizing a tesseract. There are just some things one's brain is not equipped to do.

There's a mythbuster episode that illustrates the results beautifully. You are better off to switch.

OK I'm convinced. The bit in the video where they explained that there's a 2/3 probability that you picked the wrong door and then reducing those two doors down to one made intuitive sense to me.

In this book, our protagonist admitted that he has trouble framing his analysis based on things that MIGHT happen. I guess that was my problem too.

In this book, our protagonist admitted that he has trouble framing his analysis based on things that MIGHT happen. I guess that was my problem too.

This is much better explained if you just make the numbers larger. There is one million dollars behind one door of 1000. You pick one....they open up 998 other doors and then ask you if you want to trade. In effect they have just shown you were the money is.

James I'm afraid that (to me) your explanation doesn't make the problem any clearer. It's a bit of an Occam's Razor thing...

Perception, understanding and belief are strange things. That image of a chalice that other's first perceive as two people kissing, that image of an old hag that other's first perceive as a young woman in an elegant feathered hat, those images that can be perceived as containing three dimensional images (see Magic Eye II: Now You See It)

It's difficult to covince someone to see something in a way that they do not. Sometimes one must just convince them to keep an open mind and keep trying.

Perception, understanding and belief are strange things. That image of a chalice that other's first perceive as two people kissing, that image of an old hag that other's first perceive as a young woman in an elegant feathered hat, those images that can be perceived as containing three dimensional images (see Magic Eye II: Now You See It)

It's difficult to covince someone to see something in a way that they do not. Sometimes one must just convince them to keep an open mind and keep trying.

I was simply trying to express it in not mathematical terms. In the case of 1000 doors, there is only 0.1% chance that you can pick the right one. Lets assume then that you did not pick it, that leaves 998 doors that have nothing behind them and one with the prize. When the other doors are opened, essentially you are shown with a 99.9% likelihood, exactly where the prize is.

Lets make it more ridiculous. There are 1,000,000,000 doors and one with a prize. You select one and and then all the other doors are opened and one remains shut. Do you change your pick now? You would be stupid not to because you had 1 out of a billion chances to be right, and there are therefore 999,999,999 chances out of a billion that the door being displayed is the one with the prize.

Take that back to the original question. 3 doors each having 33.33% chance of being the right one. No matter where the prize is there is definitely one door that you did not pick that is empty. So by showing you which the empty door is they have shown you where the prize is 2 out of 3 times.

Lets make it more ridiculous. There are 1,000,000,000 doors and one with a prize. You select one and and then all the other doors are opened and one remains shut. Do you change your pick now? You would be stupid not to because you had 1 out of a billion chances to be right, and there are therefore 999,999,999 chances out of a billion that the door being displayed is the one with the prize.

Take that back to the original question. 3 doors each having 33.33% chance of being the right one. No matter where the prize is there is definitely one door that you did not pick that is empty. So by showing you which the empty door is they have shown you where the prize is 2 out of 3 times.

*Ariel wrote: "Just like Stephen says:*

At first you have a 1:3 chance of being right.

Then a door is KNOWINGLY chosen to be revealed. This is a door that has a goat behind it.

The odds of the door you originally..."

At first you have a 1:3 chance of being right.

Then a door is KNOWINGLY chosen to be revealed. This is a door that has a goat behind it.

The odds of the door you originally..."

That is completely incorrect. Your odds never change, and in fact cannot change. They are simply showing you where the car by opening the door where it is not.

I'm guessing that the real genius in including this problem in the book is to give many of us a better understanding of how what we perceive the universe to be is not ALWAYS as obvious to others as we sometimes assume.

*Stephen wrote: "I'm guessing that the real genius in including this problem in the book is to give many of us a better understanding of how what we perceive the universe to be is not ALWAYS as obvious to others as..."*

It is actually extremely interesting how many people do not stop and think this through. Monty Hall did this numerous times over the years and a surprising number of people did not change their pick.

The 1 in a million door analogy never worked for me, either. I understand it, but only because I understand the original question, not vice versa.

The cards on the table work for me, but I'm a tactile thinker.

Edit: I did just try this with a whole pack of cards. Put them facedown on a table, got my boyfriend to pick one, told him the objective was to find the ace of hearts, and he shouldn't look at his card. Then I looked through the rest, put one back down, and said "none of these are the ace of hearts (car)", showing the ones in my hand. Then offered him the choice to stick with the one he'd picked up, or the one I'd put down. It did illustrate the problem better; there was a 1 in 52 chance it was in his hand, 0 chance that it was in my hand, and 51 out of 52 chances that it was the card left.

The cards on the table work for me, but I'm a tactile thinker.

Edit: I did just try this with a whole pack of cards. Put them facedown on a table, got my boyfriend to pick one, told him the objective was to find the ace of hearts, and he shouldn't look at his card. Then I looked through the rest, put one back down, and said "none of these are the ace of hearts (car)", showing the ones in my hand. Then offered him the choice to stick with the one he'd picked up, or the one I'd put down. It did illustrate the problem better; there was a 1 in 52 chance it was in his hand, 0 chance that it was in my hand, and 51 out of 52 chances that it was the card left.

How about this...in Canada we have Lotto 6/49...in which you pick 6 numbers out of 49 and if you are right you win. Now if I select 6 numbers that leaves 13,000,000(ish) possible combinations left. If someone had all the other possible ticket combinations, it can be said with almost complete certainty that they had the winning ticket. Would you trade your one ticket for all the other possible ones? Since all but one of them are wrong it is the same.

If there was 10 cards on the table and you had to choose the one ace, and after you picked one card you could change it for the other 9 would you? Again 8 of those card are also for sure not the ace but one of them most likely is.

Reducing the number you start with make the odds less and less but they are still in your favour to change.

If there was 10 cards on the table and you had to choose the one ace, and after you picked one card you could change it for the other 9 would you? Again 8 of those card are also for sure not the ace but one of them most likely is.

Reducing the number you start with make the odds less and less but they are still in your favour to change.

I had a discussion about this with somebody last night and having googled it, have come across this thread. So if I may comment, would somebody care to disprove me...

Taking the 3 doors, 2 goats and 1 car scenario, what if you do not change doors? I understand your odds are only 1/3 compared to 2/3, but what if on that occasion it was the 1/3 and the car was behind your original choice? After all, how many chances is anyone going to give you to win a car?

Taking the 3 doors, 2 goats and 1 car scenario, what if you do not change doors? I understand your odds are only 1/3 compared to 2/3, but what if on that occasion it was the 1/3 and the car was behind your original choice? After all, how many chances is anyone going to give you to win a car?

Can I also add... for those who watch it; Deal or No Deal. From 22 boxes you take 1 and eliminate 20. The 2 prizes left on the board are £50,000 or 10p. The Banker offers you the option to swop boxes.... From the theory explained here you should always swop shouldn't you?

Not necessarily because you choose which ones were opened. I realize it might be very confusing, but the fact that they didn't open the boxes means that the odds are equal about which one you are holding.

Don't worry you're not alone

"Many readers refused to believe that switching is beneficial. After the Monty Hall problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine claiming that vos Savant was wrong. (Tierney 1991) Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy."

It's counter-intuitive, to begin with, but when you actually play it out in your mind suddenly it seems obvious that you have to switch.

"Many readers refused to believe that switching is beneficial. After the Monty Hall problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine claiming that vos Savant was wrong. (Tierney 1991) Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy."

It's counter-intuitive, to begin with, but when you actually play it out in your mind suddenly it seems obvious that you have to switch.

Here's an explanation that works for me, if any of the above is confusing.

Game A: The Monty Hall problem, as described in the book, played by a person who always refuses to change his mind.

Game B: A game where a person can choose one of three doors, opens it immediately, and gets whatever is behind it. Only one of the three doors has a good prize.

It should be obvious, if you pause to think, that Game A is exactly the same as Game B. By choosing not to switch doors, you are choosing to play Game B, pick a door, get whatever is there. The next logical step is to see that a player of Game B only wins 1/3 times. So it should be quite clear that refusing to switch gives exactly a 1/3 chance.

----------------------------------

A wrinkle on the game, which could explain the human instinctive resistance to the solution. Imagine that I, Mr. Devious, offer my own version of the game. I let you pick a door. Then, IF you pick a loser, I immediately open it and shout "You lose, sucker!" IF, on the other hand, you picked a winner, I will then open one of the two doors that you didn't choose. I KNOW that both are losing doors. I then offer you the chance to change your mind. I know that you've never played this game before, you've never seen it before, and I just might trick you into changing your mind, because I want you to lose! In this case, obviously, changing your mind is a bad idea. And this kind of deviousness is how carnival games and three card monty games aim to cheat customers. Our natural distrust tells us to stick to our first choice. The mathematical discussion of probability above can look like further trickery. Any lack of clarity in explaining the exact terms of the game lead to further confusion, suspicion, and resistance. So it's no surprise that even clever people don't readily accept the correct answer. In describing the rules, not enough emphasis is placed on the fact that Monty Hall is consistently compelled to reveal a losing door, and that he does not actively try to cheat or trick the player. Understanding of human motives tends to trump mathematical analysis.

The book, however, is great, and I don't expect Christopher to be so exacting, as he doesn't really understand the very different terms by which other people would approach this problem.

Game A: The Monty Hall problem, as described in the book, played by a person who always refuses to change his mind.

Game B: A game where a person can choose one of three doors, opens it immediately, and gets whatever is behind it. Only one of the three doors has a good prize.

It should be obvious, if you pause to think, that Game A is exactly the same as Game B. By choosing not to switch doors, you are choosing to play Game B, pick a door, get whatever is there. The next logical step is to see that a player of Game B only wins 1/3 times. So it should be quite clear that refusing to switch gives exactly a 1/3 chance.

----------------------------------

A wrinkle on the game, which could explain the human instinctive resistance to the solution. Imagine that I, Mr. Devious, offer my own version of the game. I let you pick a door. Then, IF you pick a loser, I immediately open it and shout "You lose, sucker!" IF, on the other hand, you picked a winner, I will then open one of the two doors that you didn't choose. I KNOW that both are losing doors. I then offer you the chance to change your mind. I know that you've never played this game before, you've never seen it before, and I just might trick you into changing your mind, because I want you to lose! In this case, obviously, changing your mind is a bad idea. And this kind of deviousness is how carnival games and three card monty games aim to cheat customers. Our natural distrust tells us to stick to our first choice. The mathematical discussion of probability above can look like further trickery. Any lack of clarity in explaining the exact terms of the game lead to further confusion, suspicion, and resistance. So it's no surprise that even clever people don't readily accept the correct answer. In describing the rules, not enough emphasis is placed on the fact that Monty Hall is consistently compelled to reveal a losing door, and that he does not actively try to cheat or trick the player. Understanding of human motives tends to trump mathematical analysis.

The book, however, is great, and I don't expect Christopher to be so exacting, as he doesn't really understand the very different terms by which other people would approach this problem.

I completely understand how the obvious 50/50 reasoning is practically implausible. However, I have one question, that I can not explain in mathematical terms... why would you say that the door left unopened by monty hall "inherits" the probability of the door that was opened by MH?

Thanks in advance! :D

Thanks in advance! :D

Purely on the basis that HE opened it. Take the Deal or no Deal scenario. Because the player opens all the boxes the odds of picking the £250K initially are 22:1, and regardless of opening the other boxes those odds remain. If as a player however, you were to choose a box and the banker opened every other box except for the £250K and the box you were holding and offered you the swop option, wouldn't you swop? Would you think that you chose the £250K from 22 boxes or that actually, it's more likely you didn't choose the £250K box and you've just been shown where it is?

*Kieran wrote: "Purely on the basis that HE opened it. Take the Deal or no Deal scenario. Because the player opens all the boxes the odds of picking the £250K initially are 22:1, and regardless of opening the othe..."*

If I understand you correctly, yes that is exactly right.

To me, the magazine's logic makes perfect sense. Unless someone can find me a specific error in it other than "common sense says this:", I will agree with the magazine that the odds are 2/3.

*James wrote: "I was simply trying to express it in not mathematical terms. In the case of 1000 doors, there is only 0.1% chance that you can pick the right one. Lets assume then that you did not pick it, that ..."*This could be a nightmare in three acts. Where the elixer is found behind the one door, but each time the participant finds the right one he has to choose to a new nth power.

hi everyone,

im doing my IB (international baccalaureat) maths higher level coursework about this problem.

i have understood and managed to explain the problem and it is always better to switch.

now i have to questions to make:

1) can anyone help me explain the problem using conditional probability. Cus for tme the problem is so straight forward that i cant see where/how i can use conditional probability.

2) how can i apply the problem/its maths to a day to day basis?

thanks

im doing my IB (international baccalaureat) maths higher level coursework about this problem.

i have understood and managed to explain the problem and it is always better to switch.

now i have to questions to make:

1) can anyone help me explain the problem using conditional probability. Cus for tme the problem is so straight forward that i cant see where/how i can use conditional probability.

2) how can i apply the problem/its maths to a day to day basis?

thanks

*Santiago wrote: "...*

2) how can i apply the problem/its maths to a day to day basis?

"

2) how can i apply the problem/its maths to a day to day basis?

"

I'm not sure that you can apply the lesson of this problem to day to day life. One reason that it's a challenging problem to many people is that it runs counter to our daily experience, thus the intuitive response to the problem is wrong.

In this case, the game operator is constrained by the rules in such a way that he must reveal information that is valuable to us. He's not a competitor, he's an automaton, and the rules of the game are our collaborator.

-----------------------------

In a

*general sense*the lessons we can learn are:

1) Know the "rules of the game." That is, know what constraints exist, and how they determine others' behavior.

2) Always be attentive to the behavior of others who have knowledge that we don't have. It may reveal something.

3) Exploit any information gathered from steps one and two, to the greatest possible advantage.

-------------------------------------------

Now, I'm not sure if I said this before, but I'll risk repetition. The reason that this problem does not reflect ordinary daily conditions is the special way the game was contrived. Imagine, instead, a more ordinary human style experience:

-You go to a carnival. A carny is running a game which appears similar to the game we've analyzed here (but the rules governing the carny are not the same). For $1 you can choose one of three boxes. One of the boxes has a prize of $2 in it, the others have discarded banana peels. You choose any box you like, and win whatever is in it.

-In this one time experience, having never played the game before, you choose box 'C.' The carny opens box 'B,' which contains banana peels, then offers you the opportunity to change your mind. Should you? No, of course not.

-Why not? Because the carny is unconstrained. He's never obligated to show you the contents of a box, he's never obligated to let you change your mind, he's just doing it now. Or at least, you have no knowledge of any rule that forces him to let you change your mind.

-What does game theory dictate that the carny should do? Well, if you're wrong on your first choice, he should open your choice immediately and not offer you a chance to change your mind. Haha sucker! But if you're right in your first choice, he should always give you a chance to change your mind, and to give you the impression that he's helping you out, he should always show you one of the other boxes, which contains banana peels. If you never change your mind, this practice doesn't hurt him... you're only going to win this game, at most, 1/3 times. But if you sometimes change your mind, then the carny improves his chances/expected profit.

-With this in mind, game theory dictates your choice. Never change your mind. But then, game theory also dictates your earlier choice. You should previously have chosen not to play this game at all, as it has guaranteed negative expectation.

This is the intuitive thinking which subconsciously dominates in most people's minds, which is why they stubbornly refuse to understand the Monty Hall problem... that, plus the people who present the problem rarely give adequate emphasis to the rules that absolutely dictate that the game operator must always give away information! Then they scoff when mere mortals bring their real world understanding into the game and fail to solve the purely mathematical problem which is disguised as a carnival game.

Thanks! I think i need to come up With a regular day today example But your answer will help! Thanks

If anyone knows Derren Brown, he did this in one of his live shows (I think it was Svengali). He put his shoe in one of three boxes and got an audience member to choose a box she thought the shoe was in. Then he opened an empty box and asked if she wanted to switch. In my head I was screaming, "switch! switch!" but of course she didn't. Obviously she didn't know about Monty Hall. :)

Looks like Stephen is gone, I'm hoping that means you all managed to convince him! I mean if you don't understand the Monty Hall problem, there are plenty of people willing to teach you, if you open your ears and your mind. Just don't decide Monty Hall is wrong because you don't get it.

Looks like Stephen is gone, I'm hoping that means you all managed to convince him! I mean if you don't understand the Monty Hall problem, there are plenty of people willing to teach you, if you open your ears and your mind. Just don't decide Monty Hall is wrong because you don't get it.

I'm not gone. I've just been convinced. It is still interesting though to see people who aren't argue.

I still don't understand how it's more likely to win a car. Yes there is a 50% chance you will get it, but it's 50% whether you change it or not.

*Gabby wrote: "I still don't understand how it's more likely to win a car. Yes there is a 50% chance you will get it, but it's 50% whether you change it or not."*

Watch the link in post 13 and see if you still feel that way. That's the argument that convinced me. But then again, we all approach understanding these things differently.

*Gabby wrote: "I still don't understand how it's more likely to win a car. Yes there is a 50% chance you will get it, but it's 50% whether you change it or not."*

Because there is a 2/3 (~66.67%) chance you'll get the car if you switch, not 50%.

This has all been discussed already but I'll make an attempt to make it simple:

**If the door you pick originally is a goat and you switch, then you win a car because the other door has to be a car.**

There is a 2/3 chance of picking a goat, so if you switch every time after the host shows you the goat, then you have a 2/3 chance of winning a car.

Some people prefer to view it this way:

**You will lose when you switch only if you originally pick a car, which happens 1/3 of the time.**Therefore, you win a car 2/3 of the time if you switch.

I hope that made sense.

Remember, this is all contingent upon the constraints and rules of the aforementioned problem.

Just in case you might be sincere:

If you buy a lottery ticket, you might win. That doesn't mean you have a 50% chance of winning. The fact that there are two possible outcomes does not mean that they are equally probable.

If you buy a lottery ticket, you might win. That doesn't mean you have a 50% chance of winning. The fact that there are two possible outcomes does not mean that they are equally probable.

It does when one of them IS a goat and one of them IS a car. When you are syre there is one in each, it becomes 50%

That, unfortunately, is also not true.

To take an extreme example: There are two rooms, there is a goat in one, and there is a car in the other. You are SURE that one of them IS a goat, and one of them IS a car. Now, your best friend peeks into the room and shouts at you "THE CAR IS IN ROOM B!!!!"

Again, you are sure that one is a goat, and one is a car, but the odds are not 50% each, because you've been given information to guide your choice.

The rules of the Monty Hall problem force the emcee to reveal information, and that's the key to the solution.

As has been said before in this thread, if you take a shuffled deck of cards, and your job is to guess which one is the queen of spades, imagine this scenario: You pick at random, and before I reveal your choice, I look through the remaining deck, throw 50 of the cards in the trash can, and say "none of those were the queen of spades!" I even show them to you to assure you that I'm not cheating. Now you have to choose between your first choice, or the one remaining card that I didn't throw in the trash.

You know that one IS a queen of spades, and one ISN'T, but the odds are no longer 50/50 from the point of view of a person who has been given this extra information.

-----------------------------

Final, most extreme example: There are two rooms, one with a car, one with a goat. You have to choose one and accept the prize. The doors are OPEN and you can SEE the car! Are the odds still 50/50 when you make your choice?

To take an extreme example: There are two rooms, there is a goat in one, and there is a car in the other. You are SURE that one of them IS a goat, and one of them IS a car. Now, your best friend peeks into the room and shouts at you "THE CAR IS IN ROOM B!!!!"

Again, you are sure that one is a goat, and one is a car, but the odds are not 50% each, because you've been given information to guide your choice.

The rules of the Monty Hall problem force the emcee to reveal information, and that's the key to the solution.

As has been said before in this thread, if you take a shuffled deck of cards, and your job is to guess which one is the queen of spades, imagine this scenario: You pick at random, and before I reveal your choice, I look through the remaining deck, throw 50 of the cards in the trash can, and say "none of those were the queen of spades!" I even show them to you to assure you that I'm not cheating. Now you have to choose between your first choice, or the one remaining card that I didn't throw in the trash.

You know that one IS a queen of spades, and one ISN'T, but the odds are no longer 50/50 from the point of view of a person who has been given this extra information.

-----------------------------

Final, most extreme example: There are two rooms, one with a car, one with a goat. You have to choose one and accept the prize. The doors are OPEN and you can SEE the car! Are the odds still 50/50 when you make your choice?

I've given up on this discussion. My opinion is here is 50% whether or whether or not you change, because there's no garuntee there's a car behind the door if you swap. And really, you could change, and then change back and it would be 50% anyway...

So that's it, I don't want to talk about this anymore, so i'm leaving.

So that's it, I don't want to talk about this anymore, so i'm leaving.

The million doors I think makes it clearer. Suppose I let you pick a number from 1 to one million. Now after you pick I tell you that you can either keep the number you picked or you can have every other number. What would you do? That is basically the Monty Hall problem reduced to 3 numbers. Pick a door. Now you can either keep your door or have the other two doors. What will it be?

all discussions on this book |
post a new topic

Magic Eye II: Now You See It (other topics)

### Books mentioned in this topic

The Curious Incident of the Dog in the Night-Time (other topics)Magic Eye II: Now You See It (other topics)

message 1:by Stephen (last edited Jan 01, 2013 05:43AM) (new) - rated it5starsThe columnist actually WAS wrong. There is NO benefit to changing the door you select. What this problem entails is that we have subsequent information. The odds of either door having the car are 50 50 once we KNOW that one of the doors had a goat. (Despite the diagrams in the book)

On the other hand, our protagonist has admitted that he has trouble with subjunctive/conditional logic. That is he has trouble framing his analysis based on things that MIGHT happen. So it's understandable that he can't see WHY his answer is not correct.

[I've since been convinced that my initial opinion was wrong. Read on for more details if you like. But keep an open mind.]