Monty opening either of the doors you didn't pick, at random, irrespective of whether or not he reveals the car.
If that's his game, switching won't make any difference.
Rolfe.
? I think that depends on what is seen as the outcome.
I'm going to regret this. Disagree. Your point would be valid if Monty was choosing his door entirely at random. It is not valid if Monty is barred from choosing the same door you chose.
Rolfe.
The problem as set out by Vos Savant, does prevent Monty from opening the contestant’s door. Variants can make a difference, but are a side-track. I am referring only to the "original" puzzle as expressed by Vos Savant, and the nonsense about
knowing, avoiding and
probability.
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"Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?"
—Whitaker/vos Savant 1990
And these explanations:
“Here is the way Monty explains it. When the contestant made his first choice his probability of being right was 1/3. When Monty opened the second door the contestant would think his chance of being right had gone up to 1/2. It hadn't, though, it was still 1/3; and since the only other place the auto could be was behind the third door, the probability of it's being there was 2/3."
“This is the way Marilyn first explained it: "Suppose there were 100 doors and Monty opened 98 of them. You'd switch pretty fast then, wouldn't you?"
“You are on a game show with three doors. A car is behind one; goats are behind the others. You pick door No. 1. Suddenly, a worried look flashes across the host’s usually smiling face. He forgot which door hides the car! So he says a little prayer and opens No. 3. Much to his relief, a goat is revealed. He asks, “Do you want door No. 2?” Is it to your advantage to switch?”
—W.R. Neuman, Ann Arbor, Mich.
Nope. If the host is clueless, it makes no difference whether you stay or switch. If he knows, switch."
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The puzzle is a rather bad joke. If there is any probability, it is imputed, or inferred by the reader. All the other stuff: Monty, game, winning, swapping is narration, and obfuscation by dint of word.
1) If probability is invoked, then it never changes from the initial state.
2) If there is any probability, it is not conditional.
3) It’s a game if you want it to be.
4) The contestant can never learn to advantage by watching previous iterations, unless the advantage to any given door is huge.
5) My idea of the garages and recruiting Jeeves is narration.
6) The 100 door example is irrelevant.
7) The statement “if Monty knows, swap, but if he doesn’t know, don’t” is
narrative Voodoo.
I have re-written this post few times, because the more I think about it, the more trivial the puzzle becomes. I started with the earlier garages idea, where Dave wants to find the car in the most
efficient manner, and where Monty/Jeeves is replaced by a deterministic Boolean logic device operating the doors. A sort of game, where Dave and Monty are co-operative. That “works”, but is still over-egged.
I started with the idea that Dave knows nothing at all, so what could he do?
Start simple and just look.
1) Opening just one door is not going to find the car, so that’s out.
2) Opening all 3 doors will work, but is the worst case, so that’s out.
3) That leaves two doors or steps, as the only possibility for improvement. (So “swapping”)
Dave knows nothing, so perhaps there are 3 cars, and therefore 2^3=8 possible states. My next line of thought was to say that the puzzle limited
that set ( perhaps by Karnaugh mapping or Veitch diagram
http://en.wikipedia.org/wiki/Karnaugh_map), to the point where Monty’s simple Boolean engine could do the rest.
Like this:
But, the third table tells all. It’s a “walking one”, which may be applied to a range of puzzles built upon it. For example:
Dave is given the task of writing down the colour of a car he finds behind one of three doors. He knows nothing other than that.
He approaches Door A and opens it, and finds a black car is there, so writes that down, and then leaves. The next day, he tries what worked the last time, but this time the car is not there, but for some unknown reason, Door C has opened. He looks, but the car isn’t there, so he tries Door B, and the black car is there. He notes that, and leaves.
On the third day, he again tries Door A, but this time, Door B opens, but the car is not there, but he finds it behind C.
On the fourth day, it’s back at Door A.
That is because the car is on a train which advances one door at a time, and then loops back to Door A. The doors are closed when he is not there, and they are not independent, but related. As the train moves, it sets a mechanical device that opens the relevant door, should it be triggered by Dave opening any door.
Winning = knowing the color of the car
Swapping = walking a bit further to the relevant door on each occasion
Even distribution = the train’s motion.
Door logic = Monty.
Contestant = trigger on the door.
Of course, Dave can learn the pattern, and cut down on the amount of walking, if the rule can’t be inferred from the question.
Monty is just the interconnection of the doors, and his stage crew, the train, and where they just happen to put the car.
The logic of Monty’s doors means that the contestant is faced with one door, where the car may be there in 1/3 of the trials, but two other doors that are
effectively one, and were the car is 2/3 of the trials.
Instead of you choosing, and
then being asked to swap, what would you say if Monty first asked “would you like to choose the door were I put the car 1/3 of the time, or the door where I put it 2/3 of the time?”
It’s not “I have 1/3 of chance with a door, and if I swap, then 2/3 with the remaining door” but “with that
system of 3 doors, I have 2/3 of a chance”.
If Dave were to watch the show, he could learn nothing of value. If the car were evenly
distributed, and he swaps, he is guaranteed a 2/3 success rate. To beat that, a car would have to be behind any one door more than 2/3 of the time. One way or the other,
some process has to be there, if the probability is to be other than 1/3.
The 100 door claim is superfluous because 3 doors is the
only number that allows one input state to be mapped to two output states. More or fewer doors do not allow for that.
It’s so dumb. If there is no Monty, and you “choose” a door, and then tell yourself that there is “2/3 chance with the other two”, that is a truism of no value, because you still have two doors to select from, so the chance remains at 1/3.
If Monty is there, and he shows an open door, then you may say that the chance with the other two is`1/2. Many say that, and it’s correct, but Monty does
not open the door you have chosen, so the car will certainly be behind the other door. (if it's not behind your first choice)
Variants play around that inanity.
