If you haven’t seen it yet, you will need to read part 1 first to understand what is going on here.

Let’s start with Blythman’s early paragraph about the merit of scientific inquiry;

Like many scientists, he thinks he deals in absolute fact, offering a rock-solid “evidence-based” scientific “truth” that trumps all other perspectives – social, moral, political – and which is superior to other types of knowledge we might bring to bear on our decisions, such as intuition, experience, observation, or even common sense.

Actually, it is. Intuition, experience, observation and common sense are all very well, but all of them are inevitably coloured by the subjectivity of the individual. Scientific enquiry is all about eliminating subjective elements to uncover objective truths.

Common sense would tell you that the World is flat, because (barring the odd lumpy bit such as the Himalayas or the Cotswolds) it looks flat, especially at the coast. Inductive reasoning would lead you to think, “well, hang on… if I see a ship appear over the horizon the mast is visible before the hull. So maybe the sea isn’t flat but very slightly curved.” Common sense would certainly lead you to suppose the Sun travels round the Earth.

Many of the findings of mathematics or science *are* counter-intuitive. A classic example is the Monty Hall problem.

Suppose you’re on a game show and you’re given the choice of three doors. Behind one door is a car; behind the others, goats [that is, booby prizes]. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you “Do you want to switch to Door Number 2?” Is it to your advantage to change your choice?

As the player cannot be certain which of the two remaining unopened doors is the winning door, most people assume that each of these doors has an equal probability and conclude that switching does not matter. In fact, the player should switch—doing so doubles the probability of winning the car from 1/3 to 2/3.

*Thanks to Wikipedia for that*

To emphasise the counter-intuitive nature of this paradox it is worth pointing out that when Parade magazine published the problem and solution (admittedly in a truncated form which was slightly ambiguous), many graduates including mathematicians wrote in complaining that the solution was wrong.

You might think about this if you’re ever a contestant on Deal or No Deal. With all the variables of so many prizes involved intuition is bound to trip you up.

Quantum mechanics is more or less an affront to common sense, but without it there would be no transistors, and eventually no digital computers. Without digital computers there would of course be no World Wide Web. So we have the delicious irony that you can view the Sunday Herald’s website via a medium which no knowledge system other than science could produce, where you can read Joanna Blythman denying the special status of science ahead of other knowledge systems.

November 14, 2009 at 10:40 pm |

I might be wrong, but I am not convinced with the given solution: my intuition is that in the beginning the player had 1/3 chance to win and then after one door is opened, it becomes a 50% chance because then behind one door is a goat whether behind the other is the car; I would say as well that switching would not double that chance: for the player it remains the same from the beginning and the game is just to get him/her believe that he/she get more chances to succeed as the time goes by. Could you please explain to why of the solution you posted here? Thanks in advance!

Here is another paradox:

I am nobody and would like to remain so, so please don’t identify me as nobody!

November 15, 2009 at 7:27 am |

Hi Ghizlane (or is it nobody? I’m not quite sure) – thank you for your comment.

I feel your pain. There is an explanation at http://montyhallproblem.com/ which I hope will clarify this.

Also, Ion Salinou has written out all the possible cases at http://saliu.com/monty-paradox.html. See section 3.

Mark Haddon’s 2003 novel “The Curious Incident of the Dog in the Night” had an account of this problem with a detailed explanation.

An important part of this is that the game show host knows what’s behind the doors. When he shows you behind a second door with a booby prize he hasn’t picked it at random – prolonging the tension makes for better TV after all. Now when you selected a door it had a 1 in 3 chance of being the car, and it stays that way. Monty has revealed a booby prize, so the third door has a 2 in 3 chance of being a prize, ie the balance of the unity after your choice was made.

November 15, 2009 at 10:19 am |

Hi quantsuff!

Thanks for the links to the explanation: I went through the first one and understood the maths behind it but still think it’s a 50% chance and switching does not change anything.

Why? Because when calculations are made to have the 2/3 chance solution (if switching), time is not taken into consideration; whereas in the 50% chance the problem actually changes as time goes by. You may argue that time does not exist according to Einstein, but the space-time does!

Somehow, the second part of this problem is like opening Schrodinger’s box: once you do, there is no going back (either the cat is alive or dead)! so once a losing door is opened, the player has before him two remaining doors and can not go back and make calculations taking into consideration that losing one.

I hope I made myself clear (sorry I don’t have any specific links to support my reasoning).

P.S: it’s Ghizlane by the way!

November 15, 2009 at 10:56 am |

Hi Ghizlane,

Let’s look at all the permutations;

... Stay Change

---- ------

1 2 3 W L

1 3 2 W L

2 1 3 L W

2 3 1 L W

3 1 2 L W

3 2 1 L W

Column 1 is the player’s choice; if column 1 shows 1, the player guessed right. So that was 2/6 of the time. If the 1st column is not 1, it’s a booby prize, so that’s 4/6 of the time. The host eliminates a losing choice which leads you to think you face a 1 in 2 chance with what’s left. But if you map out all the permutations like this, you see that by changing his initial choice the player loses 2 out of 6 times (rows 1 & 2) and wins 4 out of 6.

It feels unutterably weird, but writing the permutations convinces me at least.

I still don’t know how to win Deal or no Deal though!

November 15, 2009 at 5:08 pm |

Hi Quantstuff!

I am still not convinced because you didn’t give me an answer to the variable of time. I also got the time to go through the second link you gave me: it is about a program calculating the probabilities; which means it does calculate by taking into consideration the three doors all the time (from the beginning to the end of the game), while the number of doors to chose from changes in the second stage of the game.

So unless there is a program which can take those changes into consideration (a program with the ability to travel through time, add the new variables and cancel those which are no longer relevant), I still think it’s a 50% chance and switching does not change anything.

I thought of another way to explain my view of it though:

There are 3 doors: number 1, number 2 and number 3. Only one has a price behind it, so the player has a 1/3 chance of making the right choice. If he picks the winning door, without knowing of course (1/3 chance), then he would have to choose between it and a losing one; If he picks the losing door (2/3 chances) he will have to chose between it and the winning one.

So somehow, no matter what door the player chooses in the beginning, he will have to make a choice eventually between a winning and a losing door and no matter what were his chances in the beginning, he will have to make a choice between two possibilities. It’s a 50% chance.

So I guess, the elimination of a losing door only makes the player feel better and/or closer to the price; and the first choice from 3 only meant to lower his chances of picking the winning door from the very beginning. At stage two (after one losing door is opened), he has still the choice and it is between one losing door and the winning one: the first door does no longer count in calculating his probabilities of winning.

Somehow, from the beginning, eventually he was going to make a 50% choice no matter what door he chose the first time. He just feels his chances are getting higher as the time goes by… you know, to make a show, keep the suspense until the end, give the spectator (and player) a sensation that he is getting closer to his goal….

November 16, 2009 at 1:48 am |

Hi quantsuff,

I like what you’re doing here. Sorry to interrupt – as someone who has taught the Monty Hall problem as part of a probability class, I thought I’d mention what I think of as the key point.

Ghizlane,

This is your mistake:

When the host eliminates a door, he gives the player some information – namely, that the eliminated door wasn’t the winning door. By switching, the player takes advantage of this information.

If you’re still puzzling over it, take another look at the table that quantsuff posted above. It amounts to a full and completely rigorous mathematical proof that the player should switch. If you can explain why you’re not convinced by quantsuff’s table, then it would be easier to see where your confusion lies.

November 16, 2009 at 1:33 pm |

qusilobashevski

I just did.

Am I the only one reading here what others post with enough scrutiny?

If you prefer me to spell it:

– You consider the problem as involving the choice from 3 doors from the beginning to the end.

– I do consider the problem as changing; it as involving two doors only at the second tour; which means in tables:

1 2 L W

1 2 W L

It’s a 50% chance winning

Now regarding the “guess”: it is either right or wrong, but it does not concern the reasoning on it self: it is not a direct result neither even an interpretation… It is just a guess of the “why?” not logically relevant to the reasoning. So you can not count it as a mistake to invalid the reasoning.

November 16, 2009 at 4:43 pm |

Ghizlane,

OK, let’s try again. I promise I have read your post thoroughly! Let me try a different explanation.

Let’s think about a different game for a second. There are three doors, 1, 2 and 3, with a car behind one of them, as before.

a. The player chooses a door. Let’s say it’s door 1.

b. The host asks the player if they would like to switch.

c. If they switch, the host gives them door 2 AND door 3.

d. If they don’t switch, the player keeps just door 1.

e. Finally, the player gets to open the door or doors that they have chosen, and keep the prize if they have it.

Now, it’s obvious in this new game that you should switch. But if you think for a minute, you should see that this game is actually THE SAME as the original game.

In both games, if the car is behind door one then you will get it if you don’t switch, but if the car is behind doors 2 or 3 then you will get it if you do switch.

Does that help?

November 16, 2009 at 6:49 pm |

quasilobashevski

I already understood the reasoning and the maths behind that way of seeing the problem the first time quantsuff gave me the links. You are just say in it in your own words; still, I don’t think it’s THE SAME GAME. Why? your point c holds the answer: when the player has to decide to switch or not, one door is already opened so HE KNOWS already that only two remain: one losing and one winning. So the problem changes it’s no longer a problem involving the choice of three doors but only two.

I am thinking we have an issue of continuity or discontinuity: is it THE SAME problem or NOT?

I think physicists are still trying to figure out a unifying theory: if such theory exists (hopefully not)!

Here is a joke I made up about it:

Two particles past very close to one another and get to “talk” to each other for a quantum bit: one is named Einstein and the other Heisenberg.

The second particle says to the first: «you are so predictable!» negatively surprised, the second one replies: «How dare you…?» and before it can finish its words, the second one disappears.

I hope you will find it funny!

November 16, 2009 at 8:41 pm |

Quasilobashevksi

Thank you for your support for what I am attempting to do. I hope you find the future submissions useful.

Ghizlane

I am sorry if you feel we haven’t addressed your concerns. For what it’s worth, I think quasilobashevski has done so–and more eloquently than me. I’m grateful for anyone reading my work, of course, and am bound to say I feel your contributions are reinforcing my point. But as the Monty Hall Paradox was really a side issue, I

~~wonder if you have~~would really appreciate any thoughts on the substance of the articles so far?Thanks once again for your interest.

November 16, 2009 at 9:52 pm |

quantsuff,

I’m enjoying your posts very much. I’ll stop cluttering your comments thread with my attempts to explain “allopathic” mathematics to Ghizlane forthwith. But before I do, I’d like to propose a challenge.

Ghizlane,

Since you understand the mathematics so well, I propose a deal. We will play 1200 rounds of the Monty Hall game. You will be the host, I will be the player. The prize each time (provided by you) will be $1000 (US). Afterwards, I will give you $700,000.

If your reasoning is correct, then by the end, you can expect a tidy profit of $100,000. So I imagine you’ll jump at my generous offer?

November 16, 2009 at 10:51 pm |

quantsuff

I am really sorry my posts are so far around what it seams to be a side problem. I’ll stop right now!

quasilobashevski

I am not a gambler, never was, never would be! Playing maths on money: why does it always has to be around money at the end of everything… even a discussion over logic?

I never pretended I understood maths so well either 🙂

See (I mean read) you around!