monty hall problem Lets say if instead of monty showing you the gate , one of the goat ACCIDENTALLY comes out of one of the doors, which is different from the door you have chosen. Will switching still help? I think now it won't. Am i right?
vivek gupta
I don't think so. The reason is that goat is still coming out. I mean after u selected a door, the goat is definitely coming out of one of the doors which is definitely not the one in which money is hidden.
****
Yeah basically it depends on whether monty hall knows or not.
Mike Lee
No, switching will not increase your chances of winning. The goat emerging from a non-chosen box eliminates that box from the calcuations. At that point, the game changes from a one-in-three choice to a one-in-two choice. So, if you compare your odds before the goat exits and after it exits, your odds improve. But the odds on the non-goat, non-chosen box also increases. They both increase to 50%
J. Hancock
The Monty Hall problem lives!
Adrian Gilby
Lets see it this way. In the original case when monty knew what was behind which door, he made use of his knowledge which led to the concentration of probability onto the the single unchosen door. There, you knew that monty will not choose the gate which you have chosen irrespective of the fact whether you have a goat or money behind your door but he could well have chosen the other closed door .Hence there came asymmetry between the two left over doors because one was in the possible choices of Monty and the other was not.
vivek gupta
I'm not sure I see your point.
Adrian Gilby
I agree with you till the point that initially all doors have the probability of 1/3.
vivek gupta
The trouble with the scenario you describe is that suddenly door A has an extra 33% chance of hiding the prize, even though you, the contestant, didn't do anything. That's impossible.
Paul Brinkley
This is very much possible that a contestant makes a choice - and later on the probability of that choice increases even if the contestant does not do anything.
vivek gupta
I think some of the above posters are correct, it *does* depend on Monty knowing which one has the car, because if he knows, he is giving extra information. If it is just random, the chance of each remaining door having the prize is equal (in the case of 3 original doors, 50%)
Mike McNertney
Okay, you've all convinced me. I agree that it matters which doors Monty is allowed to open / the goats are allowed to wander out of.
Adrian Gilby
Arrrrrgh...
Paul Brinkley
Regarding Adrian's post:
Paul Brinkley
In the variant of the question I mentioned in the beginning of the discssion, Monty was totally eliminated from the scene. The contestant just chose one of the doors and then out of one of the gates , the goat came out by itself(thats what I meant when I say accidentally). It just happened that that door was not the door the contestant chose.
vivek gupta
Paul: "I find it mildly puzzling that people would be concerned whether Monty is revealing a door at random. First of all, he would never choose to reveal the door the contestant picked, and then ask whether he wants to switch. That would make no sense. Secondly, if Monty picks an unchosen door at random, and shows a car, it makes no sense to ask the contestant whether he wants to switch. Of course he will! Finally, if the random door hides a goat, we're right back to the original problem."
Adrian Gilby
Sorry, one other thing. If Monty was instead told to open one of the doors with a goat, but with _no consideration_ of which door the contestant had picked, then it would be analagous to Vivek's original question.
Adrian Gilby
Hey, I made a page where you can see the statistical outcomes of playing this game:
levik
The scenario where one card is picked out of 100 and 98 unpicked non-aces happen to flip over, and the scenario where one door is picked out of three and a goat happens to trot out of an unpicked door, are both just variations of the scenario I described with the coin. Another example would be, what's the chance of rolling a 6 on a six-sided die, if a 6 happens to come up on top?
Paul Brinkley
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