PARADE!

ceasar

@flabdablet said:
I don't believe it's complicated at all.
<long complicated probability discussion>

;)
Amusingly, the real mistake is thinking that you need to calculate any probability after you've chosen the door. You don't. It stops being a probability problem the minute you make your choice, because at that point the outcome is fixed: at that point you've either picked a goat or a car, so the probability of switching wins is already either 0 or 1. Everything that happens after you pick a door is - as far as probabilityis concerned - irrelevant.
And that's why the problem isn't complicated: you only need to calculate one probability.

You could try to calculate the probabilities beforehand, but that assumes that you knew the entire rules of the game before picking a door. The rules as originally stated in this thread do not state that YOU knew Monty was going to open another door, even if you assume that MONTY was always going to open another door (which, as I previously brought up, is also not stated in the rules and has the potential to massively change everything, but let's assume for this post that he was aways going to open another door.) At the very least, the first time this game was played, presumably the person did not know that Monty was going to open another door. That person would have had to calculate more than one probability - first, the probability that picking a particular door would win (which is rather a trivial 1/3 for all doors, of course, but it had to be calculated!) and then the Monty Hall problem after the goat was revealed at and became apparent that it WAS the Monty Hall problem (it wouldn't have had that name yet, but that doesn't matter.)

See, on a game like this, it's often dangerous to make assumptions. You may think you know the rules, but game shows love changing the rules. They pretty much have to honor what they've explicity told you (for example, they can't have 3 goats and no car if they tell you there's a car) but they could easily add some other twist.

And a common twist (seen in the actual game show) is for Monty to offer you some sum of money to take a particular action. This offer could potentially be made at any time during the process. At this point it would be helpful to know the current state of probabilities based on all available information at whatever point this offer is made. Is Monty ignorant, does he non-randomly pick the goat door, is he more likely to offer money in particular circumstances, is Monty going to offer you even MORE money if you refuse his first offer, etc.

Now, maybe it turns out that the game plays out exactly as you expect. But it would be silly to shut your brain off and assume everything will go exactly as planned with no additional conditions.

Of course, if you add the possibility of Monty offering you money, you can say that's not the Monty Hall problem anymore. True - but it's something that actually happens on his game show, so it would be kinda silly to ignore that possibility. Anyway, the point is: you can't say "you only need to calculate the probabilities once" unless you know that you know ALL of the rules.

PalmerEldritch

@flabdablet said:

@PalmerEldritch said:
If you like phrase the problem as follows:
"Suppose you're on a game show, and you're given the choice of 3 doors. Behind one door is a car; behind the others, goats. You pick a door and the host, who knows what's behind all the doors, opens one of the two doors which you didn't pick and reveals a goat. He then says to you, "Do you want to stay with your 1st pick or switch to the door I didn't open?" Is it to your advantage to switch your choice?"
You now have: 'The Door you Picked', 'The Door Monty Opened', and 'The Door Monty didn't Open' . In order to evaluate whether it's advantageous to switch doors you need to calculate the conditional probability of both 'The Door you Picked' and ''The Door Monty didn't Open''. To do this you use Bayes Theory which isn't dependent on any specific door numbering.
I'd like to see you work that calculation step by step, if you don't mind.

Here you go, it's a standard derivation.

We have 3 doors: 'The Door You Picked' , 'The Door Monty Picked', and 'The Door Monty Didn't Pick'

Lets rename these as doors x,y, and z.

Also let X,Y, and Z denote the events that the car is behind doors x,y, and z respectively.

Let O denote the event that Monty opens door y.

We want to calculate:

Pr(X|O) The probability the car is behind door x, given Monty opens door y
Pr(Z|O) The probability the car is behind door z, given Monty opens door y

Bayes Theory tells us that:

Pr(X|O) = Pr(O|X).Pr(X)/Pr(O) and
Pr(Z|O) = Pr(O|Z).Pr(Z)/Pr(O)

and since events X,Y, and Z are mutually exclusive the Law of Total Probability tells us that:

Pr(O) = Pr(O|X).Pr(X) + Pr(O|Y).Pr(Y) + Pr(O|Z).Pr(Z)

Calculating the conditional probabilities for event O, we have

Pr(O|Z) = 1 (if the car is behind door z, Monty is forced to open door y)
Pr(O|Y) = 0 (if the car is behind door y, Monty can't open it)
Pr(O|X) = ? (if the car is behind door x, Monty has a choice of doors to open. Let's say this probability value = q. If it's superfluous to requirements it'll drop out of the equation anyway)

Since we know the prior probabilities for events X,Y, and Z are:

Pr(X) = Pr(Y) = Pr(Z) = 1/3, it follows that
Pr(O) = (q*1/3) + (0*1/3) + (1*1/3) = q/3 + 1/3 = (q+1)/3

Substituting these values into our Bayes equations we get:

Pr(X|O) = q*1/3/[(q+1)/3] = q/(q+1), and
Pr (Z|O) = 1/3/[(q+1)/3] = 1/(q+1)

So the conditional probabilities that the car is behind door x or door z depend on the value of q, the value assigned to the probability Monty opens door y when he has a choice of 2 doors to open.

If q = ½, that is Monty opens a door at random then

Pr(X|O) = 1/3 and Pr(Z|O) = 2/3. These probabilities mirror the proportion of times we'd expect to win over multiple trials by a) always staying, and b) always switching.

If q = 1, that is Monty always opens door y then

Pr(X|O) = Pr(Z|O) = ½. In other words there is no advantage to be gained by switching doors.