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The most intuitive and simple explanation that worked for me is: * if on the 1st try you choose the correct box (33% chance), then the one you can switch to will be wrong * if on the 1st try you choose the wrong box (66% chance), then the one you can switch to will be correct one therefore your goal is to pick the wrong box on the 1st try and then switch, and you have 66% chance to do it
For me the hangup was always the hidden rule: host won’t open a door with a car. That is unstated and remains unstated even in modern discussions of the problem (see Pinker’s “Rationality”). Once explicitly states the outcome becomes intuitive.
However, if the rule is not explicitly stated, how can the player know that the rule exists? Perhaps "Monty" is evil and will not always open a door, "evil Monty" will only open a door when he knows you've chosen correctly.
IOW, without that rule explicitly stated, the answer "Switch" is simply incorrect. Without that rule, the answer is "I don't have enough information to know."
The Host knows which box
has the prize.
You choose box A, leaving B and C.
The Host opens box B and,
revealing it empty, gives you
information about box C
that you don't have about your box
since the Host cannot choose your box
and must choose an empty box
from an information
perspective, box C has
better odds.Earlier quoted context omitted.
This doesn't do anything for me. (I understand the Monty Hall problem, I just don't see how changing the number of doors makes a difference to anyone's intuition.)
It's because it makes the initial choice so increasingly unlikely (increasing with the number of doors) to be correct that when the doors are taken away and you're left with only two, one of which must be right, it means that the other door is incredibly likely to be the right one.
But you still need to conivnce people that the one remaining unopened door is more likely than the door you originally selected. They were both unlikely to begin with, ramping up the number of doors doesn't explaing why one of them should be preferred.
If, on the other hand, the presenter didn't know which box is the right one, and just opens one of the other two boxes at random (with a 1/3 chance of opening the box with the prize), then, if the opened box turns out to be empty, the chance of the remaining box being the right one drops to 50%.
The difference between these two scenarios becomes obvious if we expand it to 100 boxes: You pick a box, 1% chance of being right. Of the other 99, at least 98 are empty. The presenter knows which, opens the 98 empty boxes, and now there's a 99% chance of the remaining box being the right one.
Other scenario: The presenter opens 98 boxes, not knowing which are empty, so he has a 98% chance of opening the box with the prize. On the unlikely chance that all are empty, there prize must have been in either the box you picked, or the remaining box, but we still have no information about which it is, so there's a 50% that you're holding the right box.
Of course if you don't know whether the presenter knows, and you don't know if he was lucky that the 98 boxes he opened were all empty, or that he knew, then the situation becomes quite a bit more complicated, but the chance of switching being the best option is going to be larger than 50%.
How much? Let's say there's an a priori 50% chance that he knows or doesn't know. If he doesn't know, then opening 98 empty boxes is pretty unlikely, so it's pretty likely that he knows. It's probably possible to calculate those odds, but I'm not going to try that now.
And if the presenter's behaviour changes depending on whether you picked the right box or not, for example he uses his knowledge to actively tempt you away from the right box, then all bets are off. Or maybe him opening another box is proof that you've got the right one. Or maybe that's what he wants you to think...
There's another version of the Monte Hall problem that highlights why this is such a counterintuitive problem. Imagine that after you pick your box, Monte Hall invites an audience member up on stage and instructs them to choose one of the remaining two doors to open. This audience member doesn't know anything at all and just randomly picks one of the two doors. When their door is opened we see that it's empty. You're…
Correct me if I'm wrong, but in your particular example (spectator opens an empty door and I am asked if I want to switch), nothing changes in regards to the original Monty Hall problem. If a spectator opens a random remaining door, one of two things can happen: - a car is revealed, I lose immediately (there is no option to switch anymore) - no car is revealed, which means I again have 2/3 chances when switching, not…
Consider instead a Monte Hall scenario with 100 boxes (vice just 3), maybe we call this the "deal or no deal" variant of the problem...
The user picks 1 box and has a 1/100 chance of selecting the box with the prize. Now, the host opens 98 boxes that they know do not have keys in them, leaving two unopened boxes (the one the user picked and the one the host left unopened).
Now, pick your box from the remaining selection and claim your prize. You bet your sweet bippy I know which box I'm picking.
Information is gained by observing the winnowing of the field of options.
The thing that is often de-emphasised in the presentation of the problem, in order to make it seem more mysteriously paradoxical, is that the presenter knows where the car is and this knowledge is always used perfectly. If the question always ended with "remember: Monty knows where the car is and will use this information", it would be more obvious. Imagine a universe with many simultaneous Monty Hall clones playing…
> If the question always ended with "remember: Monty knows where the car is and will use this information", it would be more obvious. And perhaps also, “remember: Monty will always open a door, and the contestant knows it”. Makes me wonder if there were similar shows where the host can choose not to open a door.
If you choose the door with the car, the host will open another door to tempt you, so then you don't switch, unlike in traditional Monty Hall. And if he didn't open a door, well, that means you choose a goat, so you should switch for a 50% chance. Except the host knows you're thinking this. And you know he knows. And he knows you know he knows. And...
Perhaps, iterated infinitely, this eventually resolves into a limiting expected value for switching and not, but I have no idea how to compute such a scenario.