Ran across this problem the other day, and it is quite an interesting brain teaser. No fair googling, use your own mind!
The Monty Hall Problem
The following problem is taken from a quiz show on TV that really existed (or still exists). It is an interesting topic to discuss in almost any group of people because even the most intelligent often get into trouble, and is (in other languages) referred to as the Goat Problem.
The quiz show candidate has mastered all the questions. Now it’s all or nothing for one last time: He is lead to a room with three doors. Behind one of them there’s an expensive sports car; behind the other two there’s a goat. (Don’t ask me why it’s a goat. That’s just the way it is.)
The candidate chooses one of the doors. But it is not opened; the host (who knows the location of the sports car) opens one of the other doors instead and shows a goat. The rules of the game, which are known to all participants, require the host to do this irrespective of the candidate’s initial choice.
The candidate is now asked if he wants to stick with the door he chose originally or if he prefers to switch to the other remaining closed door. His goal is the sports car, of course!
The question now is:
Is the candidate better off if he sticks with his original choice,
aaah, good ol' Monty Hall.Maybe I have an unfair advantage but hey, I didn't google.
I remember discussing this in my Stats class last semester.Found it very interesting that the contestant had a higher chance of winning if he changed his door choice after Monty revealed one of the goats.I believe he had a 66% chance of winning that way than the intuitive 50%.
Could dig up my notes and prove that but too lazy right now.
chances that the dude's hit the car with initial choice - one in three. chances of having hit the sheep - two in three. given the choice of one wrong choice being eliminated, and given that there was twice as much of a chance to have made a wrong decision initially, switching is going to give the wrong result is only one in three tries.
FF they are all absolutely correct, even though the analysis seems counterintuitive.
The easiest way I have heard that explained is that the chosen door has a one in three chance of being correct, while the non-chosen doors have a two in three chance of being correct. This never changes. When one of the non-chosen doors is revealed, the remaining non-chosen door assumes all of the initial 2/3rds weighting, and is therefore twice as likely to be the correct door. This has been proven by computer models. The key is that the initial 1/3 2/3rds odds remain constant, and that new information revealed does nothing to change those initial odds.
queer ji u dont quite understand..this is my inspired version:
whatever first try is, either he is right first time (1/3) probablity, or he is wrong first time( 2/3)
if he is right, then either of the other two doors is wrong, but that implies a hit of 1/3....if he is wrong, with a prob of 2/3, then the other person has eliminated the other wrong choice for him,
so the original probablity holds! becoz in second case she has shown what the wrong door is, besides the first one assumed to be wrong by 2/3rd possiblity.
its like (1/3) versus (2/3*1/1). the one/one is imp.
