Monty hall problem

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I suppose you all know this famous problem. It is pretty clear to me why switching doors is beneficial, but I'm however unable to counter this argument from my friend:

What is the difference between having picked a door and then the host revealing a goat, compared to not having picked one and then the host revealing the goat. Clearly there is difference but can someone elaborate in understandable terms because I tend to get very confused.

Boorglar

The difference is that when you pick the door, the host must open ANOTHER door. Therefore if you had the goat in your door (which is 2/3 likely) then the door left closed will have the car 2/3 likely. This gives you information.

If you hadn't picked a door, then the host could choose any of the doors with goats at random, thus not giving any other information about the closed doors.

D H

To hammer this point home, imagine if there were 1,000 doors, one of which hides a car and each of the the remaining 999 doors hides a goat. You pick randomly a door. The probability you picked the right door is a paltry 1/1000. Monty then opens 998 doors, each of which shows a goat. Do you switch? Of course you do. The probability that the car is behind the unchosen, unopened door is 999/1000.

Another way to look at it: Monty has just given you information, a whole lot of information in this case.