A better card-based analogy to the Monty Hall problem is achieved by envisaging a scenario in which the dealer has three cards, one card is red on both sides, one card is white on both sides, and the remaining card is red on one side and white on the other.
The dealer puts the three cards into a bag, shakes the bag, lays it on the table and slides one card out.
The visible face of the card is red. So there are only two possibilities, it is either the red/red card or the red white card. So it’s an even chance, right ? OR IS IT ???
Forgive me if this has been posted before, this is my first post , and I wish everybody here a very Happy New Year.
Yes, it’s an even chance. The essential aspect of the Monty Hall problem is that Monty knows where the prize is, and since the dealer doesn’t see anything the player doesn’t, that feature is missing here.
Well, yes it does, actually, it has a very close analogy.
Both problems have a scenario in which there is an apparently even money chance which turns out on closer analysis to have a 33.3333% - 66.6666% split.
Consider an alternative version of the Monty Hall game where Monty offers you a switch with probability 2/3 if you pick the correct door, and probability 3/4 if you pick an incorrect door. In that case, your probability of winning by switching is 9/13. Would you argue that this is less similar to the original Monty Hall problem than your card problem is?
The setup is as usual: there are three doors, one with a car and two with goats, and you pick one. Monty knows where the car is, and he gets to decide whether you get a chance to switch or not. He’s going to give you the option randomly, with a probability that depends on whether you’ve picked the door concealing a prize. If you pick the door concealing the car, there’s a two-thirds probability that you get a chance to switch. On the other hand, if you pick a door concealing a goat, there’s a three-fourths probability that you get the same chance.
Here’s what I don’t get: how did all those Ph.D. mathematicians get this wrong? I absolutely understand why the average person would get tripped up, but not only did a bunch of superbrainiacs get it wrong when initially thinking about it, they wrote in to Parade Magazine insisting she was wrong! That blows my mind.
It’s nice to just append this to a thread that’s once again discussed the matter already.
As for why, here’s a bit from the NY Times article to explain a bit of it:
The basic answer is that the problem is ill-formed. It’s very easy, and probably quite common, for people to come to the wrong conclusion about an ill-formed problem. This is because they make the wrong assumptions, and usually solve the wrong problem properly.
Most problems, and especially those not couched in careful or technical terms, require at least some inference and assumptions to be made about the system. Someone with an advanced degree may have a lot of understanding of how to solve not just that type of problem, but a group of similar problems. Sometimes the assumptions are made almost without even realizing that they have been made; this is often the source of confusion when the person who asked and the answerer are working from a different set of assumptions.
What is a problem it is that even when the faulty assumption is revealed as such, many people who ought to know better will stick to their original thinking. That’s bad but sadly true for people at any level of skill. Possibly even worse for those who regard their knowledge as superior, such as the professor quoted in the piece as recanting only after a long time fighting against it.
This problem in particular is one that seems simple enough not to require certain assumptions, which is why some simply disregarded what was given and created a faulty problem.
A good example of a common problem that requires deep, but very natural, assumptions to be well-formed is the 0.999… = 1? question. The answer relies on the structure of basic arithmetic as most know it; which is to say it can have another answer but not without throwing out a lot that might be considered necessary.
A good example a totally ill-formed problem that seems quite simple is one that also appeared on Mythbusters, which I won’t say anything more about than that it had to do with airplanes and treadmills.
I dunno, I’m with Marilyn on this one. The people arguing with her didn’t appear to raise the (slight) ambiguity in the wording; they were just caught up in their erroneous idea that it was a 50/50 proposition so switching or not switching was equally advantageous.
Yeah, they didn’t point out this objection until they were shown to be wrong. They’re trying to save face but they were wrong.
It’s conceivable that one could interpret the problem to be that Monty Hall only sometimes offers to open a door, but if that were the basis of their reasoning, they sure as hell would have mentioned it the first time. It opens up about a zillion alternate scenarios.
I think the question is misleading, and those guys were mislead. But the question doesn’t contain a lie or a fallacy, so these guys just got fooled, answered hastily, and don’t want to admit they fell for it. I didn’t think the solution sounded right at first, so to prove it I set out to write a program to test it. I didn’t have to write the program though, simply working out the details of the problem to simulate it revealed the answer. And my knowledge of statistics is minimal.
Let me be sure I understand the problem. There are 3 doors. The contestant picks one. If the door is a winner the game doesn’t end right?
If the game doesn’t end when the contestant picks a winning door on their first pick, I can’t see how the contestant’s chances of winning are any greater than 33%. Any of the doors they wind up with has a 33% chance of winning.
Monty offering a switch does not mean the contestant’s pick isn’t a winner. It’s inconclusive either way, unless the show had some consistent formula or policy for it.
You name a door to start with. No winner is determined yet. Monty opens one of the doors you didn’t name, and it must be a door that doesn’t have the prize behind it. Now you get to pick between the door you named, and the unopened door that you didn’t name.
Since the door you named has a 33% chance of winning, then there must be 66% chance that the prize is behind one of the other doors. Monty shows you what’s behind one of the unopened doors that does not have the prize behind it. So there must be a 66% chance that the prize is behind the other unopened door. Now it’s time for you to pick a door. Do you go with the 33% chance, or the 66% chance?
There’s always the possibility in real life that Monty would make you stick with the first door you named and not let you choose one of the others because he knows you have won and doesn’t want to give you the chance to make a wrong decision. But that’s not part of the logic problem.
It was clear when I read the MvS column that she clearly misstated it. If you leave ambiguities, multiple correct answers apply. People fill in the holes in their own way. You don’t deserve to make fun of people for the error you made. Especially if you write a book explaining how Wile’s proof of FLT is wrong since it’s a discrete Math problem and he used continuous Math techniques. Did she ever issue an apology for that? (Or the many other errors she’s made?)
Increasing the number of doors makes it clearer for me. Try it this way:
There are 100 doors but only one has a prize behind it. You pick door #1, because you have no imagination (actually, it just makes the problem easier to state).
Now, Monty opens up all the other doors, except for door #73, revealing that there are no prizes behind the open doors. Now, do you switch to the one door that Monty decided not to open, or stick with your original door?