Why is that nonsensical? The game you are being offered is essentially this: “Pick a card. Then you get to guess whether that card is the winning card, or one of the 4,999,999 losing cards. If you guess correctly, you win!”. Only, instead of calling it that, they throw in this flash and sizzle about revealing 4,999,998 other losing cards, and call the “guess my pick is the winning card” option “Stay” and call the “guess my pick is the losing card” option “Switch”. But it’s the same game. If you were to play “Pick a card and then guess whether it’s the winner or one of the 4,999,999 losers”, which option would you go with?
Put another way: remember, sometimes Monty Hall has a choice between two doors to open. Let’s suppose he decides based on a coin flip. And let’s suppose he flips the coin ahead of time just in case he’ll need it. Then we have 18 possibilities:
I pick Car Monty's coin Monty opens
------ --- ------------ -----------
A A Heads B
A A Tails C
A B Heads C
A B Tails C
A C Heads B
A C Tails B
B A Heads C
B A Tails C
B B Heads A
B B Tails C
B C Heads A
B C Tails A
C A Heads B
C A Tails B
C B Heads A
C B Tails A
C C Heads A
C C Tails B
If we assume each door is equally probable to be picked by me, each door is equally probable to have the car, each side is equally probable to come up on the coin, and each of these three pieces of data is probabilistically independent of the other two, then these 18 possibilities are each equiprobable. Remember, it’s not always true that where there are n possibilities, each has equal probability 1/n; that just happens to be the case for this setup using the assumptions I outlined.
Alright, so what? So, now, imagine I’ve reached the stage of the game where, say, I know that I picked door A, and I know that Monty opened door B, but I don’t know anything else. Then just three of the original possibilities remains in play:
I pick Car Monty's coin Monty opens
------ --- ------------ -----------
A A Heads B
A C Heads B
A C Tails B
And out of these three live equiprobable possibilities? One is a win for staying, and two are wins for switching. Accordingly, a 1/3 probability of winning on staying and a 2/3 probability of winning on switching.
As it must be, since staying is equivalent to betting “My original door was right” and switching is equivalent to betting “My original door was wrong”, and you’ve gained no new information at any time about whether your original door was right or wrong, and you originally had only a 1/3 probability of your original door being right. The whole song-and-dance about opening doors and so on is distracting you from the actual bet you are being asked to take sides on.
Wow, lots of Monty Hall posts. I’m going to suggest that Superhal does some experiments, to see what happens with this problem. You’ll need a die and a coin to do this experiment.
Pretend that you’re Monty Hall:
[ol]
[li]Choose which door obscures the car (door 1, 2 or 3)[/li][li]Roll the die to let the contestant choose which door.[/li][ul]
[li]1 or 4, contestant chooses door 1.[/li][li]2 or 5, contestant chooses door 2.[/li][li]3 or 6, contestant chooses door 3.[/li][/ul]
[li]If the contestant did not choose the correct door, open the single remaining door. If the contestant did choose the door with the car, flip the coin to decide which door to open:[/li][ul]
[li]Heads: open the door on the left (which will either be 1 or 2).[/li][li]Tails: open the door on the right (which will either be 2 or 3).[/li][/ul]
[li]After being asked, the contestant decides not to switch.[/li][li]Record the number of times the contestant wins the car, and the number of times the contestant does not win the car.[/li][/ol]
Play many iterations of the experiment, say 50 or 100. Afterwards, you’ll notice that the number of losses is about twice as large as the number of wins. This is because staying wins with only 1/3 probability.
If this isn’t convincing, I suggest that we open another thread to discuss Monty Hall.
On the contrary, if it were not possible that Monty followed that script, then Ellis Dee’s problem statement should have said so (especially since Monty really did sometimes not offer a switch). You might as well ask “What color is the car that Monty hid”, and argue that it’s red, because if it were possible for it to be black, the problem should have said so. When you’re not given enough information, the correct response is to say “I don’t know”.
A man with no eyes
Spies apples on a tree
He neither takes them
Nor leaves them
Now how can that be?
a one eyed man took one of two apples
How is a one-eyed man a man with no eyes?
As I see it, Ellis Dee stated the problem by way of example, and explicitly marked all the portions that might change within particular realizations of the game. Giving an explicit listing of all the things that can change is the same as giving a listing of all the things that cannot change.
But, I suppose this might be the difference between a “mathematical puzzle” and a “lateral thinking puzzle,” in the latter, you assume that the problem was stated in an intentionally deceivingly way. Like this.
Regarding your parenthetical remark, “especially since Monty really did sometimes not offer a switch,” it is my understanding that Monty Hall never actually used the problem as we know it today on Let’s Make A Deal. See below.
He has exactly one eye. Not several eyes. Although every sane person would interpret the first line as you did, the author wants it to mean “a man who does not have multiple eyes.”
Ah, I see.
Work better if you say “a man without eyes”. You have to spin pretty hard to make “with no X” mean “with exactly one X”, but it is literally true that he does not have “eyes”.
Incidentally, this is not true:
Chronos is right; Monty’s behavior does matter, insofar as it affects what information you learn from which door he picks. And learning new information could change the conditional probabilities of your first guess being right (insofar as they change the relevant events to condition on).
Suppose Monty’s behavior is to only ever open a goat-door if the player’s door has the car, and to open the car-door otherwise. Then you sure as hell don’t want to switch upon Monty opening a goat-door, because, under this pattern of behavior, Monty opening a goat-door is definitive proof that staying wins the car.
Suppose Monty’s behavior is to always open door 3 if the player picked door 1, regardless of where the car is. Then, conditioned on the case where the player picks door 1 and Monty throws open door 3 revealing a goat, the odds for the car being in door 1 vs. door 2 are 50-50; the revelation of a goat behind door 3 gave enough information to shift the car door probabilities from 1/3, 1/3, 1/3 to 1/2, 1/2, 0, but the fact that Monty chose door 3 specifically gave no further new information to change the probabilities beyond this, as there’s nothing more revealed by this fact than the already known fact that the player chose door number 1.
Etc., etc. But, yes, Dr. Love is right; the Monty Hall discussion is reaching the point of belonging in another thread.
It’s also worth noting that Wikipedia itself, where you copied your posing of the problem from (the version published by Marilyn von Savant in 1990) immediately follows it up with
Those exceptions are pedantic nitpickery. If Monty reveals the car after I pick, that’s a completely different, wholly unrelated question. The question is a very simple one: I pick a door, Monty then reveals a goat, I’m offered to switch to the only other remaining door. That’s the question, and I have a 1/3 chance to be right with my first guess, period.
No, not necessarily; it depends on what information Monty’s revelation of a goat gives you. Like I said, suppose Monty’s behavior is to always open door 3 if the player picked door 1, no matter what, and it happens to be the case that this time, it reveals a goat. That’s a case where switching would be of no advantage to you; the odds would be 50-50, as explained above.
In the setup that makes the problem tick, Monty doesn’t just happen to open a goat door, he’s guaranteed to always open a goat door, and so the probability that your original guess was correct doesn’t increase upon seeing Monty open a door with a goat in it, because that’s not new information: you’re guaranteed that Monty will open a goat door, so you don’t learn anything about your original guess’s winning chances to see it happen.
However, if it’s just contingently the case that the door Monty opens had a goat in it, even though it was a priori possible that the opening would have revealed a car instead, then that is new information about your original guess: now the probability of your original guess being correct does shift when conditioned upon the new information, from 1/3 to 1/2. For the same reason that, if you pick the number 523989 for a million-ticket raffle, and hear the announcer say “5-2-3-9-8-… Drum roll, please”, your odds have momentarily shot from 1 in a million to 1 in 10. You didn’t previously know the first 5 digits would be 5-2-3-9-8; upon learning that they were (i.e., upon seeing that all the tickets that didn’t start that way were losers), you’ve gained enough information to become much more confident in your original choice.
Again, the question is a very simple one: I pick a door, Monty then reveals a goat, I’m offered to switch to the only other remaining door. How the problem changes if Monty might sometimes reveal the car is irrelevant because that’s a different question. Next thing you know we’ll have to factor in the chance that Monty always has the car moved to the first door chosen before revealing a goat. What happens to the chances then, huh? Who cares! That’s a different question.
But you can’t answer a probability question without specifying a probability distribution. And if you don’t specify within your probability distribution any information about the probability that Monty would reveal a goat [and thus complementarily, the probability that Monty would not reveal a goat], then you can’t compute probabilities conditioned on Monty revealing a goat.
Suppose I gave the setup as follows: You pick door 1. Out of the corner of your eye, you happen to glimpse that there’s a goat behind door number 2, even though no one opened the door. Should you now try asking Monty for permission to switch to door number 3? What’s the probability that you win if you stay?
Or, put another way, suppose you pick door 1, and another contestant picks door 3, and Monty opens door 2 to reveal a goat. Monty asks the two contestants: “Would you like to switch?”. Should you both be itching to switch with each other, since you were both in a situation where you picked a door, and Monty revealed another door to have a goat? What are the probabilities for each door to win? If the probabilities in this problem are different from in your setup, why? Does the mere presence of another contestant change anything? And is the relevant probability distribution completely specified?
You tell me. This latest hypothetical seems to argue that the conventional answer to the conventional question is wrong. All I’ve asserted is that the conventional answer to the conventional question is correct. The other examples given (“what if he reveals the car?”) aren’t the conventional question.
EDIT: Note that in your penultimate hypothetical, after spotting the goat out of the corner of my eye I would indeed like to switch to door #3, since that information of seeing the goat after I picked door #1 acts the same way as Monty revealing a goat. (1/3 if I stay, 2/3 if I switch.) It’s different if I spied the goat before I chose a door, but that’s not how you phrased it.
My position is that the way you’ve phrased the conventional question leaves it incompletely specified; to get the conventional answer, you need to add assumptions such as those that Chronos and I have pointed out, as does the very Wikipedia article you took that specification from. And, yes, these assumptions may be of the form “This particular situation arises as a special case of a number of other possibilities, with the following prior probability distribution upon them: …”
Otherwise, there are any number of probability distributions we could pick to model the situation. There is not enough information to answer the question. The described setup is compatible with a 50-50 answer, for the same reason as the question “You buy a raffle ticket in a 3 person raffle. Then you learn your mother and father bought tickets too, and you learn that your mother’s ticket was a losing ticket. What’s the probability your ticket is a winner? What’s the probability your father’s ticket is a winner?” is compatible with a 50-50 answer. The described setup is compatible with all sorts of things; there’s nothing in it mathematically preventing us from using a probability distribution giving rise to a 89.75% probability of winning on staying. There just isn’t enough information there.
Still waiting for an answer to your own hypothetical.