One of the most misunderstood and screwed-up areas in business has to be probabilities. People constantly attribute to skill that which can be explained by luck (CEOs getting overcompensated because their stock has risen in a twelve-month period), and they refuse to apply probabilities when making investment decisions (rifles vs. shotguns in venture investing). John Kay make this point nicely in a column today on the subject:
The Monty Hall problem is named after the host of a 1970s quiz show, Let’s Make a Deal. The successful contestant chooses from three closed boxes. One contains the keys to a car and the other two a picture of a goat. The choice made, Monty opens one of the other doors to reveal – a goat. He taunts the guest to change the decision. Should the guest switch to the other closed box?
When the solution was published in an American magazine, thousands of readers Рincluding professors of statistics Рalleged an error. Paul Erd̦s, the great mathematician, reputedly died still musing on the Monty Hall problem. But the answer is, indeed, yes: you should change.
This is not the only case where intuition does not correspond to the mathematics of probability. One person in a 1,000 suffers from a rare disease. A friend has just tested positive for this illness and the test gives a correct diagnosis in 99 per cent of cases. How likely is it that your friend has the disease? Not at all likely. In random groups of 1,000 people an average of 10 would display false positives and only one would be correctly diagnosed with the disease. But most people, including most doctors, think otherwise. “The human mind,” said science writer Stephen Jay Gould, “did not evolve to deal with probabilities.”
Last month, the General Medical Council struck off Professor Sir Roy Meadow, the paediatrician, from the medical register. He had given misleading evidence in the criminal prosecution of Sally Clarke, whose two infants died in their cots. When Mrs Clarke was charged with their murder, Sir Roy told the jury that the chances of two successive cot deaths in the one family was “one in 73m”.
But although the disciplinary committee heard evidence from distinguished statisticians, it does not appear that they understood the application of probability theory to such cases any better than Sir Roy. The committee found that he had underestimated the incidence of cot deaths, and that he had not taken account of genetic and environmental factors that mean a household that experiences one cot death is more likely than average to suffer another. But even if you recognise these effects, his key conclusion remains valid. It is unlikely that such an accident would have happened at all. It is very unlikely indeed that such an accident could have happened twice in the same family.
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The Monty Hall problem slightly blew my mind the first time I heard it (although now the solution seems retrospectively kind of obvious, as they always do). The probability of winning the car if you don’t switch is 1:3. The probability if you do switch is 2:3.
Here’s the most intuitive way I’ve found to explain why switching makes sense. Imagine that, instead of three doors, there are one million. Now imagine that Monty opens up 999,998 of them: all except for the one you originally chose, and one other remaining closed door. Would you switch now? Obviously, yes. The odds of your original guess being correct are miniscule (one in a million, to be exact), and the odds of the other door being the right one are correspondingly high.
There are other ways to think about this, but that’s the one I like best.
[obDisclosure: I'm an MBA, so my qualification to work with numbers is highly suspect.]
Three good books on the subject of probabilities,
1) the mathematician reads the newspaper,
2) innumeracy, and
3) fooled by randomness
Sorry but neither the answer to the monty hall problem or the sickness problem is correct. (Many high level mathmeticians have wrestled with this problem and it amazes me how easily people can dimiss the musings of great minds when they themselves have never shown any prediliction towards mental greatness.)
The fallacy occurs by trying to include time in the equation in an haphazard manner. Similar to division, in which most division works just find until you find yourself dividing by zero, the itroduction of temporal factors can really skew the logic.
For example in the monty hall question, before you know what is in one of the boxes, your chances are 1 of “3″. After you know the contents of one of the boxes it would be insane to say you have a two in three chance since there are only 2 chances left. Yes compared to having three choices 1 of 2 is better than 1 of 3, BUT THAT IS NOT THE QUESTION! The question is not whether you can go back in time and knowing which of the boxes to ignore, improve your odds, MONTY IS ASKING YOU A COMPLETELY DIFFERENT QUESTION, NOW. Even though you made your choice by choosing from three possibilities, there are only two now. The reason you chances appear to improve is because the domain of the equation was CHANGED from two to three. The only way the Monty Hall conundrum is true is if after showing you the goat box, he allows that box to remain in play. Now out of three choices only one of two can be correct thereby improving your chances “out of three”, NOT “OUT OF TWO”.
In mathematics there are definite rules for the substitution of terms. The terms in this “problem” are switched willy nilly to help perpetuate the mystery. If you simplify the problem by removing one box you change the problem: x times 3 = % is not congruent to x times 2 = %. They are two different equations. You could graph them and they would give you two different lines. To connect them as if one is a simplified version of the other is ridiculous.
Generally, intuition gave us everything from PI to relativity. don’t be so quick to throw in away no matter how intelligent the person you are talking to is supposed to be.
That example with the rare disease is badly put. Even if 1/1000 get a rare disease it doesn’t mean you’d apply the test randomly to 1000 asymptomatic people.
Actually, Adam is completely correct. Bayes’ Theorem addresses all this stuff.
If you take Adam’s logic a bit further, you can make some money at a bar with friends and strangers.
“Here, let’s play a game. I have picked a number and written it down on a piece of paper. It’s in my pocket. For $1, if you guess my number, I will give you $10 bucks. The numbers are between 1 and 10.”
Say, you have chosen 2, and he picks 8.
Now eliminate half the numbers. Eliminate 1,4,5,6,10.
Ask the person the person if they are willing to call it quits if you give them $2.
Nope, he’s getting excited that he might win.
So eliminate, 3 and 7. Only 2, 8, and 9 remain.
Offer him $4 to quit. Most won’t. Even though they believe their chances are 1/3 with their number (when in reality their odds are only 10%), they still won’t take the $4.
Now you reveal your “2″ and take the dollar.
As far as rare disease is concerned, I think it is fairly stated. For example, let’s say you go have some routine tests done with a blood sample. As part of that test, you are screened for Disease X. Further suppose that “Disease X” exists in 1 out of 1000 people in the general population. The test is 99% accurate. They just do the one test and nothing else. What are the odds you have the disease?
Answer: Unlikely.
Obviously, in real life they follow up with other tests to confirm and repeat the test and whatever else they do.
An example might be West Nile Disease. You have volunteered to donate blood, and now the Red Cross as part of its screening process checks for West Nile.
hrm no I still think the rare disease is badly stated. It says “A friend has just tested positive for this illness and the test gives a correct diagnosis in 99 per cent of cases. How likely is it that your friend has the disease? Not at all likely.”
The friend ALREADY TESTED positive. I totally agree that if he hadn’t already tested positive it would be unlikely.
haha oh yeah in a screeing situation u’d be right
I disagree that the disease case was badly stated — it is straight forward:
If 1 in 1,000 has the disease, and 10 in a 1,000 tests issue a false positive, (assuming he was randomly tested, asymptomatic, etc.) your friend has only a 1 in 11 chance of having the disease. (10 false positives plus 1 true positive).
Actually, now that I have thought about it, Adam’s example (1st comment with many doors instead of 3) proves absolutely nothing. Whether there are a 3 doors or a trillion, what makes it irrlelevant is that — pre switching option — Monty ain’t opening the one with the car!
So you always end up with two. The basic problem comes down to this — either your door has the car or not. Its always going to be a choice between the one you picked (as of that moment, still unknown) and the remaining door (also unknown).
The initial starting # of doors has precisely zero relevancy.
I’ve long found the theoretical solution to Monty Hall problem unsastisfying. Perhaps I’ll post why soon . . .
OK, I’ve thought about it, and I am reversing my self.
Adam is correct. If Monty starts with a billion doors — you pick one — and then he then flips over 999,999,998 doors, none of which show a car. All that remains is the last door, plus your original pick.
Of course you switch (duh!)
The odds are 1 in B against your initial selection being correct, so it makes sense to swap, and only 1 in 2 against the other door.
Now why is the same solution so unsastifying when its only 3 doors?
Barry, this is a mindbender, isn’t it?
You’re right. How many doors he opens says absolutely nothing about the original pick. The original pick remains closed to the very end.
Thus, its odds always remain throughout the game at 1:number of doors available.
And as you have pointed out, you want to switch.
But once you soak on it for a while, it all becomes clear. I like Adam’s method too.
I just want to point out that John Kay got the Monty Hall problem wrong in his second letter a week after the initial one. He was responding the hundreds of emails from confused people he received. Or so he assumed. He says in the second paragraph:”He might have spoiled the game by opening a box that contained the car keys, but he didn’t. Monty might have been blind drunk and could as easily have opened your box by mistake – but he didn’t. It does not matter what Monty knows, or is thinking. All you need know is that the box Monty opens is empty.” This can’t be further from the truth. The crux of the matter is exactly that the opening of a door by Monty of a door is not random. If it were, you would be indifferent about the remaining two doors. Turns out he is confused, and too arrogant to think that he might be wrong, even after speaking og the history of the problem so eloquently.
Wow, go camping for a few days and look what happens.
Barry, I think you’re right that my explanation fails as a “proof” — it’s meant to be an aid to intuition. As you mention, the logic seems unsatisfying when there are only three doors in total, but far more compelling when there are a million. And I think I know the reason why:
Reformulate the problem so that, rather than opening any doors, Monty leaves all the doors closed and says, “You can either stay with your original pick, or you can instead flip your bet: if it turns out the car *isn’t* under the door you originally picked, then you win the car.” In this case, a naive person might argue that no new information has been revealed, and so flipping the bet confers no advantage. Obviously, this logic is false: 1/3 of the time the original bet will be correct, and 2/3 of the time it will be wrong. Flipping the bet is always your better choice.
Now consider that, effectively, this is what Monty is saying when he opens the door. Monty knows where the car is, and he is not going to reveal it to you. (As another poster pointed out, John Kay goofed on this part of the explanation.) So either you guessed right originally (1:3 odds) and you should stick with your original bet, or you didn’t guess right originally (2:3 odds) and Monty has just revealed to you which of the remaining doors is the correct one.
That’s why the million-door formulation is an aid to intuition. Monty is allowing you to flip your bet — that is, to declare that your original guess was wrong — and it’s far more obvious that you should bet against yourself when the odds are 1 in a million than when they are 1 in 3, even if you aren’t quite aware that that’s what you’re doing. (Note that if there are only two doors to start with, flipping the bet in fact doesn’t help — or hurt.)
And, yes, I think this is pretty much straight up Bayesian math, although here my memory gets fuzzy.
A more interesting question is why Paul Erdös died musing this problem. Obviously there’s more to it than I can hope to fathom!
I would like to chime in with some REAL statistics here…
The Monty Hall Dilema
Definitions : The price is randomly placed behind door A, B, or C.
Let’s assume you choose door A. Everyone agrees that you’re chances of picking correctly is 1:3. When Monty reveals an empty door you are given new information, which means it is an entirely new problem.
Now let’s assume that door C is revealed to contain nothing. Everyone should agree the prize must be behind door A or door B. Whether you change your choice from A to B or keep your original selection the odds of you winning are 50% because there are only TWO possible outcomes. There is no reason for door B to be more likely to contain the prize than door A and vice versa.
To further illustrate this idea, imagine 2 different people (say, Josh and Matt) are playing the same game without knowing it. Josh chooses door A and Matt chooses door B. Monty reveals door C contains nothing. You are trying to argue that Josh should change to door B and Matt to door A…totally absurd.
The confusion comes because your initial guess has a 33% chance of being right, or in other words, there is a better chance of you being wrong. Therefore, if you are most likely to be wrong then it would make sense to switch. That argument sounds good, but it just isn’t right. It falls apart because it is taking information from the initial scenario and applying to the second, a big NO-NO in statistics.
Let’s take picture…
(I’m frenchspeaker but I will do my Best to be understood:)
Imagine a Road: By walking on this road it means there are 3 doors and you , as player, will choose a door. After, monty will choose another door.
Then road is separated in two ways.
First way will be taken if monty open door and it’s empty. Monty and player will take “way of empty door”.
Second way will be taken if monty open door and car is behind. Monty and player will take “way of door which have a car”.
Now define some protocols.
A protocols is just a list of steps to follow.
Protocol “random” says: You will walk till the road is separated in two ways. So player will choose at random a door and then monty will choose randomly one of the other doors.
Protocol “empty” says: You will walk till to be in way of empty doors. So player will choose a door at random and then monty will look behind two other doors and open a door that is empty.
Protocol “car” says: You will walk till to be in way which have a car. So it’s the same than previous one but monty open door which have a car.If he can obviously
So we have three protocols
By following protocol “empty” you can be sure that monty will open empty door and so you can calculate chances of winning before starting game. Because by définition a protocol must be followed, will be followed.
But Protocol “random” allows also to walk on way of empty doors. We agree on the fact that following protocol “random” is not the “safest”, because it can happens to be in way which have a car. But it can happens…
So arriving in way of empty door by using protocol “random” will occur less than using protocol “empty”. And so if you test it with a software: results will be less than 2/3 because you simulate more than one game.
But assuming we analyse one case: (where empty door was open) we are already in way of empty door. chance are that moment are 2/3 (by switching) doesn’t matter the path you followed.
For example we are all aware that if monty open door with car by following protocol “random”: Chances of winning for player are 0 (in case of player loose if monty chooses door with car).
Nobody will say after opening door with car that chances of winning are chances calculated for protocol “random”. And You are right not to do because in this case chances of winning are chances calculated for protocol “car”. Because protocol “random” becomes useless as soon as door is open…And give the floor to protocol “car”.
So to make a summary
Protocol “empty” and protocol “car” are two specific cases of “protocol “random” . Chances of winning for protocol “random” are just a combination between chances of winning of protocol “car” and “empty”
Is it almost clear?
Canneel, there’s no way that’s a clear explanation. Once you start using random, ill-fitted terminology like roads and protocols and your explanation is longer and more convoluted than the original, how could it possibly be clearer? As for Josh’s counter-argument, he’s ignoring the fundamental premise of the whole problem. The reason it is better to switch is that you’re given more information, which changes the probabilities. If two people were playing the game and they both chose the doors with goats, which door would Monty open? He wouldn’t be able open the unpicked door because it has the car, so your game would never work. The point of the third door being opened when you’re playing the game with only one contestant is that more information has been injected into the situation. With this new information, it’s better that you switch.
Dear, I don’t really understand what you are trying to demonstrate.
Question is “does monty intent influence chances of winning by switching as soon as door with a goat is opened?” So if monty knows or not, is there an impact on chances of winning when door which has a goat is opened?
Answer is no. Game with two players has nothing to do with that because as you said: in some games monty can only choose door with car.
When you choose a door at random, your chances of winning are 1/3. If not it means you did not choose at random.
So chances you did not choose the right door are still 2/3 even if monty open a door (at random) wich has a goat. You cannot assume that if monty open a door which has a goat it is because you were lucky. This assumption is in contradiction with first one that says you choosed at random.
So once a door with goat is opened by monty; chances that monty open a door which has a goat are 1/1 even if he did know where was the car because he did it. But probability you choosed door with car are still 1/3 and probability that door with car is one you did not choosed are 2/3.
So elements to solve this problem
probability that first choice is door with car is 1/3
probability that door which has a goat opened by monty is the door with car is 0/1
Probability that doors you did not choosed contains a car is 2/3.
Swap. If you don’t believe the theoretical reasoning then experiment. Get three cards with GOAT, GOAT, CAR written on them and also get a friend who will be Monty Hall. Your friend must know which card is the CAR – this is the important factor in the problem and is also necessary to simulate the situation in the game show as was. Play the game 99 times and see what happens if you don’t swap. Now play the game 99 times and see what happens if you do swap. Play the game more than 99 times if you wish – play it as many times as is necessary to convince you that swapping is the right option.
If you play the game 99 times I predict that if you go for the swap option you will win approximatley 66 times. If you play the game 99 times and don’t swap I predict that you will win approximatley 33 times.
My theory bit:
A third is less likely than two thirds. Two thirds of the time you will have chosen the wrong door and so you should swap – remember Monty Hall knows where the car is. One third of the time you have chosen correctly and shouldn’t swap. You don’t know in what situation you are in but chances are you have chosen poorly and should swap.
Bob,
I think that every body agreed on the fact you have to swap.
Problem we just discussed about is: should we swap when Monty does not know where is the car but still open door or card with goat.
Because as you said monty has to know if he must open door with goat each time. But what happen with probability when monty open a door at random and reveal a goat.
I can’t believe that such intelligent people just don’t get it! This is not about the odds of winning the car, it is about the odds of winning the car IF you changed your mind. The host removes a door so that if you change you mind your result will always be negated. Even if the second choice was made at random it can still be shown that 2 out of 3 times the succesful result was due to the contestant’s original choice being changed. Try playing the game with 100 doors. Make a choice then take 98 doors with goats away. Your left with 2 doors, a goat and a car. Do it with paper and you will very quickly realise that the solution is very obvious.