Monty Hall Problem

[lenny]

thanks ThatSoundAgain yours appeared as i was typing...

if you get to the point in the game where Monty has opened a non-prize door, each remaining door will have a 1/2 chance of being the winner.

...so you conclude "my chances of winning" have changed, but that there is no value in switching. i had thought there was, but in fact my argument only supported the claim that the "chances of winning" had changed. in this we agree.

 

[lenny]

lenny, I hadn't thought of your option c. Goes to show that strict definitions are necessary here - my table above operates under the rule that Monty will always choose another door than the player.

ETA: And now I see you wrote about that above. Sorry.

no worries. i was not sure what the rules were!

agreed: strict definitions are required for clear analysis.

 

[ThatSoundAgain]

thanks ThatSoundAgain yours appeared as i was typing...

...so you conclude "my chances of winning" have changed, but that there is no value in switching. i had thought there was, but in fact my argument only supported the claim that the "chances of winning" had changed. in this we agree.

Yes, I agree completely. At that point in time, the player will have dodged the "instant lose" bullets of scenarios 3 and 5 in the table.

 

[Rolfe]

One more possibility, and I apologise if it's been mentioned already: what if Monty needs someone to win the prize every nth time, to maintain interest and viewer levels? On these occasions he steers the game towards you winning, changing from the default strategy of steering the game towards you losing. How often would he do this, i.e. what are your chances of being favoured?

Is there any way this puzzle can be resolved without more information?

That was roughly what I was driving at in my last post. How often is he going to try to steer you to rather than away from the prize? It depends on a lot of things, and audience reaction is one of them. If it's time for someone to win, then he's more likely to steer you in the right direction.

On the other hand you can get too far into second-guessing at that point, and I'm sure the guy was a master of suppressio veri, suggestio falsi.

You can't possibly solve the puzzle for all possible variants, because once Monty is behaving inconsistently it's officially insoluble.

Rolfe.

 

[Rolfe]

....
your terms here appear unclear. please just point me to the relevant post if this was already clarified or if you are interested in a nonstandard definition of the problem (like opening the chosen door).

if monty opens one of the other two doors at random, there are two options:

a) the door he opens is the prize door: your chances of winning are now zero and you know it
b) the door he opens is a goat: you can increase your chances of winning by switching (the standard "two doors versus one door" argument holds regardless of Monty's ignorance).

if he opens the door you have, you know you have won, or you know you should switch; even in this nonstandard verison you gain information and can improve your "chance of winning".

so even opening a door at random introduces more information and thus alters your "chances of winning".
(or: which of the 700 other posts should i have read?)

I think someone has already dealt with this.

I was referring to the point in the game which is dealt with in the standard presentation of the puzzle. You have chosen a door and Monty has opened one of the other two doors to reveal a goat.

If Monty deliberately opened a goat door at that point, your chances of winning are doubled by switching, because your original door remains at the 1/3rd chance it had at the beginning while the remaining door now carries 2/3rds probability of being the car.

If Monty opened either unchosen door at random, indeed there was a chance of an instant lose, another 1/3rd probability. However, once you know that didn't happen, the chances of the car being behind the remaining door are still only 1/3rd, same as the chance of it being behind your original door, so there's no benefot to switching.

This has been demonstrated approximately a gadzillion times since the start of this thread.

Rolfe.

 

[Herzblut]

You're right: I agree that what makes a difference is Monty's behaviour. We have to be careful when we're saying this, though: what matters is not just his behaviour this time, it's his behaviour pattern. To say that he opens a door revealing a goat is not enough information: we need to know if this always happens, or if it only happens when there is a car behind the contestant's door, or if it happens at random, or if some other conditions apply.

My viewpoint is still slightly different and I don't like talking about what "always" happens. Actually, in a game situation nothing else counts but the evidence probability P(E) in this one game. Formally, if

A= car is behind player's door ("Assumption")
E= another door with goat is opened ("Evidence")

Then, in any case, the probability of A, given E simply is

depending only on P(E). Where it's assumed that the inverse conditional probability P(E|A) is 100%. Then

- the standard case is P(E)=100%,
- the random case is P(E)=2/3,
- the bastard case is P(E)=1/3.

 

[lenny]

However, once you know that didn't happen, the chances of the car being behind the remaining door are still only 1/3rd, same as the chance of it being behind your original door,

once you know it is not behind one of the unchosen doors, you have additional information. the probability of the prize being behind the open door is now known to be zero, so the chance of it being behind the remaining doors can not still be 1/3 each; there are only two doors left!

one can argue that the probability of the prize being behind "the remaining door" and the probability of "your original door" are equal, but in that case it makes no sense to argue:

the chances of the car being behind the remaining door are still only 1/3rd

no?

 

[boooeee]

My viewpoint is still slightly different and I don't like talking about what "always" happens. Actually, in a game situation nothing else counts but the evidence probability P(E) in this one game. Formally, if

A= car is behind player's door ("Assumption")
E= another door with goat is opened ("Evidence")

Then, in any case, the probability of A, given E simply is

[qimg]http://www.randi.org/latexrender/latex.php?$$%20P%28A%7CE%29=%20%5Cfrac%7B%5Cfrac%7B1%7D%7B3%7D%7D%7BP%28E%29%7D%20$$[/qimg]

depending only on P(E). Where it's assumed that the inverse conditional probability P(E|A) is 100%. Then

- the standard case is P(E)=100%,
- the random case is P(E)=2/3,
- the bastard case is P(E)=1/3.

Nicely put.

 

[Rolfe]

once you know it is not behind one of the unchosen doors, you have additional information. the probability of the prize being behind the open door is now known to be zero, so the chance of it being behind the remaining doors can not still be 1/3 each; there are only two doors left!

one can argue that the probability of the prize being behind "the remaining door" and the probability of "your original door" are equal, but in that case it makes no sense to argue:

no?

Yes, you're right, I mis-spoke.

Originally, 1/3rd chance of picking correctly first time, 1/3rd chance that it was behind the door that Monty opened, and 1/3rd chance that it's behing the third door.

But after Monty has opened that door (at random), 1/3rd chance on the original door and 1/3rd chance on the remaining door now becomes 50/50 on the two unopened doors.

Sorry.

Rolfe.