Thanks to David Killoren for directing me to this excerpt from Bloggingheads. Science writers John Horgan and George Johnson spend a few minutes disucssing the Monty Hall problem. Johnson recently reviewed Leonard Mlodinow's book The Drunkard's Walk: How Randomness Rules Our Lives, which contains an explanation of te problem. After Johnson gets frustrated trying to explain the problem to Horgan, the following exchange takes place:
Johnson: I thought I understood this when I wrote the review. It's already eluded me again.Horgan: This is like explaining the two-slit experiment in quantum mechanics which has messed me up a number of times.
Johnson: This is even harder, I think.
Goodness! I notice that in his review Johnson describes Mlodinow's discussion as the clearest explanation of the problem he has yet seen. But, of course, my book on the subject has not yet been published.
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I'm not sure. But I do think it may be easier to spell than quantum mechanis. j/k
D'oh! Sorry about that. The error has been corrected.
The NYT got it wrong.
If "a women has 2 children and one is a girl" _nothing_ has been said about dependance so the odds of the other child being a girl remain 50%..
But the with the 2 goats and a car,
If the arbiter reveals a goat from your 2 unchosen doors, he has seen thte goat and he possible car, and chosen a goat,, ie NOT independant. so the 'goatyness' of the remaining unchosen door has been reduced.... so change to that door
NB we are told that there are 2 goats and a car NOT that the ods of a car are 33% - ie the objects are not independant
But even the NYTimes can get it wrong .
If I tell you I have two children, and my oldest child is a girl; what are the chances my second child is a girl? The chances are 50/50; my chldren are either GG or GB. If I do not specify the order, then the possibilities are GG, GB, and BG. Marilyn vos Savent had a lot of fun with this one. This is one of those times where less information lowers your ability to predict the outcome.
I think the problem with "a women has 2 children and one is a girl" is semantic rather than mathematical. Does it mean "the first one to come through the door is a girl", or does it mean "[at least] one is a girl"? i.e. is the child who is a girl specified in some way?
I really don't see what is so hard about the Monty Hall problem. As long as one understands that strategies are the same as long as they are declared before hand there's no issue.
Well i've tried to explain the MHP to several friends and family members, some with B Sc's and more and none of them understood it.
I have to agree with Johnson. Once you get your head around the basic idea of wave/particle duality, the two slit experiment isn't that hard to cope with. Monty Hall, on the other hand, remains confounding to non-mathematicians even when we know why the answer is what it is. Perhaps it's just closer to home when it comes to intuition. Photons operate on such a different scale to our ordinary world that on some level we can accept weirdness, whereas we deal with probability (badly) every day. There's lots of stuff in QM that [i]is[/i] insanely hard to conceptualise, of course.
I don't know about that. I'm a grad student in physics and I definitely find QM harder then the monty-hall problem, even just explaining double-slit experiment. Or to put it another way, it took me much longer to understand, although I was an undergrad when I learned about the double-slit experiment and a grad student when I first heard of the MH problem, maybe that had something to do with it.
I think the Monty Hall problem is a little harder than the double slit experiment. Not just because I have a soft corner for double slits, but because I found it easy to understand the first time around (and found it, well, amazing).
The Monty Hall problem is counter-intuitive, and it's harder to explain to friends. With double slit you have the wave analogy.
My most successful strategy in explaining the MHP is to extend the problem to more than 3 doors. Imagine a hundred doors. Contestant selects 1. Then MH opens 98 doors showing all goats.
If that doesn't work, try a million doors. Even the most obstinant, entrenched opponent will switch doors then.
http://en.wikipedia.org/wiki/Monty_Hall_problem has a good discussion that finally made sense to me. I approached the problem as a probability calculation after the door is opened and 50:50 seems right. But it isn't in the context of looking at all the possibilities of 3 doors.
I've found the quickest way to explain the probability of a single number from the throw of two dice is to list all 36 possible outcomes and show the number of times say four occurs (1:3,2:2,3:1). Counting three and dividing by 36 makes quick sense to many people. No messy formulas.
The Monty Hall problem is nearly always described wrong, which makes it hard to understand. Two crucial points are often left out - the game leader must ALWAYS open one of the remaining doors (or the rules must be such that he must open one of the remaining doors), and the game leader MUST know where the prize is. Leaving out any of those have huge impact on the result. If for example the game leader does not know were the prize is, then the probability with a three door scenario will be 50% for either of the remaining doors, if the prize is not behind the opened door. If, on the other hand, the game leader knows were the prize is, and are free to have any strategy when it comes to opening or not opening doors, noone knows what the probabilities will be.
Neither of them are difficult to understand if someone explains them.
I still don't understand why people can't understand the Monty Hall problem. I can understand not grasping probability and specifically Bayes' Theorem, but seriously, how hard is it to understand when it's explained as what happens if you switch vs. not switching. If you choose door 1, if the goat's in 1, if you switch you lose. If in 2, you switch, you win. If in 3, you switch, you win. 2 out of 3 times, if you switch, you win. It's not that difficult!