Notably different from the Monty Haul problem where additional information is given to the player of the game prior to making the decision to switch.
> Pigeons repeatedly exposed to the problem show that they rapidly learn to always switch, unlike humans.
Or to put it another way how many people would stick with their chosen door if instead of opening a door and showing a goat they were simply asked "would you like to stick with your current choice of door, or switch and take the best result from the other doors?"
That is, effectively, what the choice boils down to. But that being said, since the problem is stated as only getting to pick one "other" door, the "extra information" received is knowing which of the other 2 doors has the best result.
2/3rds of the time, your initial choice will be a goat. In those cases the host deliberately avoids opening the remaining door with a car - thereby telling you where it is.
If you chose the car door initially, then indeed you get no new information. But that only happens 1/3rd of the time.
I am curious if that changes your opinion.
It's true that we can add other options (pick both doors) of equal likelihood, but it's because of our increased information that the reduced option of picking just one door isn't actually worse off.
I have a deck of a billion cards, one entitles you to the car, the others have a picture of a goat. You pick one at random but do not get to see it.
Of the remaining 999,999,999 cards, I discard 999,999,998 goats.
Do you want your initial card, or the remaining card in my deck?
Or to frame it another way, do you want the car iff you picked it first or iff you did not pick it first?
The way they put it, the strategy becomes obvious if you instead reformulate the problem like so: there are three doors with three different prizes. You get to pick one door, but are then given a choice: stick with your original choice, or choose to get the best prize from the other two doors.
The solution (switch!) is more obvious in the second formulation. I have no idea if it's easier to convince someone else of the reasoning in the first formulation, or to convince them that the second formulation is equivalent to the original.
you make your guess Me: Ok, I will reveal that the number is either $YOUR_GUESS or 452,871. You may keep your original guess or switch to 452,871.
Of course, you would switch, because the chances you guessed correctly originally were 1 in a million.
Personally, I'm not sure that it would be easier to convince someone that the second formulation is equivalent to the original problem, but next time I find a dis-believer I will give it a shot!
This problem is completely different, since the number of degrees of freedom remains fixed throughout the decision process. As someone else pointed out, it is trivially resolvable by realizing that no matter how many times you switch envelopes, you're really only making one choice, and doing so with 50-50 odds of picking the more profitable envelope. The offer to switch amounts to meaningless magician's patter, whether the offer is made once or a million times.