Here’s a great example/explanation of the Monty Hall math problem:
This is the same as Margin Gardner’s Three Prisoners problem, and is based on the earlier Bertrand’s box paradox (also known as the ”three-card swindle”).
It’s a good reminder that our intuition isn’t always reliable.
(via)
Our brains are optimized for seeking survival, not truth. I’m amazed sometimes we even have the circuitry in our heads to construct math.
I thought Monty Hall was a type of D&D adventure where the DM keeps his players interested by giving them excessive amounts of loot?
I thought Monty Hall was a type of D&D adventure where the DM keeps his players interested by giving them excessive amounts of loot?
Monty Hall was a game show host (I forget the name of the program) where a person won whatever prize was behind one of three doors, that they picked. “I want what’s behind door number three!”
Same thing, different context. “Monty Haul” Campaigns were criticized by some as being antithetical to role-playing in that they encouraged less character development and more “let’s see what’s behind door number three, kill it, and take its stuff”.
I remember when I first saw this in “Ask Marilyn” back around 1990. I had heated arguements with both my dad, as well as people at work who held their positions with religious ferver. Even when you explain the math (which can be hard to do clearly and convincingly in this case), it does not convince someone who is emotionally attached to their position. I was finally able to “convert” one of my coworkers after writing a computer simulation that ran through thousands of iterations and displayed the total results — which proved that if you did switch, your odds were 2/3 of winning the car.
I love that Nope thought I was being serious. :)
I love that Nope thought I was being serious. :)
My humor detector sometimes gets false negatives over these here internetz.
Low bandwidth, you see.
Damn, I have to take up watching Numb3rs again. I started to watch it, but I don’t remember why I never continued. Every time I see a clip though, I remember that “hey, this looks like a really cool show, I should watch this”.
Then again, for complex decision-making, our ‘intuition’ – our subconscious mind – is often the best tool we have to make the right decision. Dutch psychology professor Ap Dijksterhuis wrote a fascinating book about the power of our subconscious mind.
@Adamus
Absolutely just think about what we do when we catch a ball and then how much longer it takes before we are taught how to describe the trajectory of that ball.
@Johnny Fargo
I, too, had to write a computer program to convince a friend.
Of course, their response was something to the tune of “how can the computer be sure.”
math makes my head hurt.
stupid logical equations with their seemingly easy solutions that turn out to be hard to understand.
Math is a tool of the devil.
@ Ty
So will you join me in my sincere efforts to ban once and for all, the dangerous weapons of maths instruction?
@Teleprompter
I wouldn’t joke to much relativity was denounced in Russia mainly on political grounds that it wasn’t Marxist Science.
Monty Hall– Let’s Make a Deal (that was the show)
I’m not much of a gambler. I’d take my prize and go home. Knowing my luck, there’d be a goat eating a can behind door #3.
I prefer to take the extreme example, I think it is presented on the Wikipedia page. Multiply the number of goats out to 1000 instead of 2. Pick a door. The host then closes all but two doors, including the one you picked. If your intuition still tells you that the chance you picked the right door is now only 50/50 you need to get out and read some information theory or probability theory or SOMETHING very quickly. It should be immediately clear that the host is revealing information.
“I’m gonne reveal to you one of the cards that you did not choose”
This implies a random choice, which goes against the problem. The structure is that a “goat” must be shown. That being true, consider the choices again
G C G
If a goat must be shown, then by choosing a goat, both goats are eliminated and switching means winning – 2/3.
If it’s a fluke whether or not a goat is chosen, the base assumption is incorrect because in one of the three choices you lose no matter what you do and the probability goes back to 1/2.
My favourite explanation of this problem is that you have a choice between the door you first picked, or the better of both of the other doors. By changing, you’re getting to choose two doors (including the one that’s been revealed). I’m not convinced that the explanation given in that clip is particularly convincing.
It annoys me when he reveals the car at the end and everyone sighs and smiles as if they are now convinced. It’s as though it’s a magic trick and everyone is cooing in amazement at some sleight of hand. The point is that it should be just as convincing if he turned over a goat card.
This problem always annoys me because whenever it comes up, it usually results in a heated debate with both sides being really condescending to each other. In truth, neither side is wrong, they are just arguing two different things.
If you play the experiment over hundreds or thousands of trials, and always switch, the probability of winning is 2/3.
If you consider one particular case, and examine it from the point in time that you’ve been offered one or two remaining doors, it’s 1/2.
The answer to the question “should you switch?” is, if you’re running 1000 trials, yes, you should switch every time. If you are running one trial, it doesn’t matter, it’s 50/50.
It’s designed to be tricky, so it bothers me that people delight in berating people who are tricked by it.
@3D
“It’s designed to be tricky, so it bothers me that people delight in berating people who are tricked by it.”
And I think you can count yourself as one of those people.
When it comes to maths, at least, there is in fact a right and a wrong answer. So when you say they are both arguing from different positions and, therefore, both correct, you are wrong, I’m afraid.
By saying if you’re doing 1000 trials, you should switch every time, but if you’re only doing one it doesn’t matter, you indicating that you don’t really understand probability. What difference does it make how many trials you’re doing?
For anyone still having trouble with this, here’s the way I explain it when there is disagreement:
When you make the initial choice, the chance of you being wrong is 2/3. By switching, you maintain that probability of being right because you’re not going to choose the door you know is wrong. So if your initial guess was wrong, you switch and you’re right. If your initial guess was right, you switch and you’re wrong.
So the question you should be asking is “what’s the chance of you being wrong initially?” The answer is, inescapably, 2/3.
In other words, no matter how many trials you’re doing, you should always switch.
Fleegman wrote:
((@3D… “It’s designed to be tricky, so it bothers me that people delight in berating people who are tricked by it.”))
“And I think you can count yourself as one of those people.”
No, what I meant to say was that it’s designed to elicit one correct layman’s answer by wording the question in such a way that there is a different, also correct, theoretical statistician’s answer. It’s a parlor trick.
But thanks for making my point about being condescending.
“When it comes to maths, at least, there is in fact a right and a wrong answer. So when you say they are both arguing from different positions and, therefore, both correct, you are wrong, I’m afraid.”
True, there is only one correct answer to a question. However, the two people who are answering “1/2″ and “2/3″ are answering two different questions, and both are giving correct answers. Your answer is correct, but so is the 1/2 answer.
What is the probability that you were wrong to start the trial? The correct answer is 2/3. What is the probability that if you agree to switch every time before running the trial, that you will win? The correct answer is 2/3. What is the probability that you are wrong RIGHT NOW, given that your host has shown you the fact that one of the bad doors is out of play? That answer is 1/2.
It’s a completely contrived situation designed to elicit an intuitive, correct answer which is also a wrong answer to another question which can be worded the same way. It’s designed in doubletalk so that people can call each other nasty names in bars and on college campuses, and eventually, later, on the internet.
“For anyone still having trouble with this, here’s the way I explain it when there is disagreement: When you make the initial choice, the chance of you being wrong is 2/3. By switching, you maintain that probability of being right because you’re not going to choose the door you know is wrong. So if your initial guess was wrong, you switch and you’re right. If your initial guess was right, you switch and you’re wrong.”
Except that using the new information revealed by the host, there is *currently* a 50/50 chance that your initial guess was wrong or right.
“So the question you should be asking is “what’s the chance of you being wrong initially?” The answer is, inescapably, 2/3.”
That’s true, except that the probability at the time of the original choice is different than what it is at the time when there are 2 doors left and you know one of them is the winner.
Someone tried to make a point by increasing the number of doors to 1,000 and removing 998 of them. Here’s a better example: we can change the number of doors to 2, you pick your door, and a host shows you a goat behind the other one. Is your chance of winning a car the same 50/50 it was before he showed you a door? No, it’s changed to 100%.
In that case, your theoretical chances of winning the car are still 50/50, because half the time your initial choice was wrong. But that is completely impractical.
Similarly, in the original example, when he opens the third door, your chances of winning *that trial* went up from 1/3 to 1/2.
What’s true for a series of trials does not necessarily apply to one particular instant during a single trial after new info has been revealed. This is the trick of the question and it’s designed that way on purpose, to exploit the occasional gap between statistical theory and practical usage.
This problem used to confuse the hell out of me and I only got it now. I didn’t watch the video, but read over the Wikipedia article. I’ve done that many times before, but it never clicked.
The best way I can explain it now is that the only time switching would be bad for you is if you picked the car on the initial door choice. Since the odds of picking the car initially is one out of three, two out of three times you should switch.
What’s getting to me is sort of what 3D is getting at:
Contestant A picks a door and Monty shows a goat behind one of the other doors. Contestant A walks away and contestant B now must picks out of the two remaining doors. For contestant B, the odds are 1/2 since he doesn’t know anything about the initial choice. What’s really confusing is if contestant A comes back and tells B what door he picked initially. From A’s point of view B should pick the door opposite of A, but form B’s perspective, A is just pointing at a random door and doesn’t help him at all.
Am I mixing something up here?
@Zhatt
From B’s perspective it’s not a random door if it’s explained how it was chosen and he can use this information to his advantage. It all comes down to conditional probabilities (which are brains are rather crap at) and the fact that the host doesn’t, unless you choose the car, have a free choice in which card to pick — this is the information that A or B can use i.e. their choice is not random. Think of it this way if the host gets to remove a goat first then the odds do change and swapping makes no difference as the choice now is random.
Hum. So as soon as A tells B what door A picked, it’s almost like Monty opening the goat-door in that it’s giving you information about the situation. But the thing is, that situation is A’s situation. B’s situation is different. B doesn’t know if the door that A picked has the car. It either doesn’t (loose) or it does (win) no mater what door A points at.
I have a feeling you’re right, I just want to figure out why.
I don’t quite understand your idea of removing the goat. Do you mean after Monty reveals it? It shouldn’t matter if you sent the goat to the moon, that third option is gone either way.
Here, let me try something else…
A new contestant -we’ll call him “C”- gets to pick between two doors. One has a car, one a goat. That’s a 1/2 chance of wining, right? After C picks a door, Monty rolls a third door onto the stage and opens it, reviling a goat. Should C switch? As I under stand it, C’s predicament isn’t any different than B’s. Now what if A comes along and points out the door he picked last time? Assuming the same setup, does C get a 2/3 chance? What if C picked the other door that A picked?
I guess C would get the same 2/3 chance that A had if A intervenes no matter how many doors Monty brings in and opens.
@Zhatt
Ok think of it another way – imagine that the card A picked is always called position 1 and the other card is called position 2. Because of the host’s choice the card in position 1 has a probability of 2/3 of being the car. In this situation if B picks randomly between the two positions you wouldn’t consider it a 50/50 chance that he picks the car i.e. the cards aren’t actually in a random order. When described in this way it seems a lot more obvious what the probability is.
The idea of removing the goat is that if before you even start the host removes a goat and *then* randomly places the cards it really is a 50/50 chance.
For you last point with new contestant C the conditional probabilities have all changed so switching will make no difference. An important thing to remember is what we are really talking about is the probability of a card being a car or a goat and the order in which things are done does make a difference. It’s much better to think of purely in terms of the probability of a card being a goat or a car and not the probability of a contestant picking a particular card.
@3D: “But thanks for making my point about being condescending.”
Oops. I guess irony doesn’t come across that well in a comment thread; my bad.
Anyway, onwards:
@3D: “What is the probability that you were wrong to start the trial? The correct answer is 2/3. What is the probability that if you agree to switch every time before running the trial, that you will win? The correct answer is 2/3. What is the probability that you are wrong RIGHT NOW, given that your host has shown you the fact that one of the bad doors is out of play? That answer is 1/2.”
No. The probability of you being wrong RIGHT NOW, given that your host has shown you the fact that one of the bad doors is out of play, is 2/3. Not 1/2. Are we talking about the same problem? I’m assuming we are, but videos are barred at work, here, so I can’t watch it. Here is the problem I’m talking about, simplified:
1) Three doors, one prize.
2) You choose one door
3) Host shows you a different door that isn’t the prize.
4) Host asks if you want to switch to the remaining door.
5) What are the odds of winning if you switch?
Is that the same problem?
Assuming it is, your statement above about the 1/2 probability of being correct lies at point 4, if I’m not mistaken. And you say that the chance of being correct AT THAT POINT is 1/2? Well, in that scenario, the chance of being off the prize 2/3. The fact that there are only two choices left, including your original choice has no bearing on the probability of you initially choosing incorrectly.
@3D: “That’s true, except that the probability at the time of the original choice is different than what it is at the time when there are 2 doors left and you know one of them is the winner.”
Yes, I know one of the remaining doors is a winner, but I also know which door I originally picked. And I know that the probability of having chosen the wrong door is 2/3. Always. Your first pick defines the probability of being wrong. And that probability doesn’t change regardless of any new information. So even at that point, I should still switch doors.
If it’s a different problem we’re discussing than the one I outlined above, well, egg on my stupid face.
I like numbers, but they jacked that scene from the movie 21. Even down to the Car and the Goat as the prize.
No, the puzzle, goats and car included, is much older than 21. It’s been around since at least 1975.
Heh, not only is it older than 21… it’s 37!