you're on a game show, and you're given the choice of three doors:
behind one door is a million dollars and the others are empty. You pick Door 1 and the host opens Door 3 to reveal an empty room. You're given the option to switch doors.
Assuming the host knows which door has money, should you switch doors?
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No
http://en.wikipedia.org/wiki/Monty_Hall_problem
Yes.
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yes. when you make your original decision, you have 1/3 probability of choosing correctly (payoff = 1000000/3). after he eliminates a door not chosen, the probability of staying with your original choice doesn;t change, while the joint probability of the other two doors is 2/3. with one of those choices eliminated, there is only one other choice to make and your payoff function becomes 2000000/3. haven't you ever seen 21
nice, Jay Gatsby
:(
What if you know that the game show host is familiar with the Monty Hall problem?
Haven't you ever seen The Princess Bride?
It depends on whether the revealed door can have the prize behind it or not.
Regardless, I think everyone can agree on the fact that there is a 2/3 chance you pick the wrong door.
Let's try a thought experiment with 1 million doors. Monty reveals that there's goats behind 999,998. Now all that's left is your door and this other one that's randomly left unopened. Do you switch now? Bear in mind, 99.9999% chance you picked the wrong door.
When you first picked the door: 1/3rd chance of winning
If you switch: your chance of winning is 2/3 --if you pick the right door (1 out of 3 times), and one empty door is revealed, switching means you lose 100% of the time --If you pick the wrong door (2 out of 3 times), and one empty door is revealed, then switching means you win 100% of the time
1/3x0 + 2/3x1=2/3
slightly counterintuitive at first
Don't believe me and Solidarity, just fill out the probability diagram and cover all of the possible outcomes, bearing in mind that revealing the prize door before you have a chance to switch is impossible.
Dude, I think he just misread my post. I have a tendancy to finish thoughts midsentence and not give a lot of context.
Regardless, I've gotten into a debate with two quants- one with a Math PhD- and a number of other really smart folks on this. The 2/3 number is pretty controversial. Perhaps when- err, if- you get a job in industry, you'll actually be able to tell what actually makes someone a retard or not.
The 2/3 probability is not controversial at all if the car's location is random. The odds that you pick the right door if you switch is 2/3.
If the car's location is arbitrary then you can't really say that it's 2/3. But I think the problem is usually stated as the car being behind a random door (i.e. uniform). In which case, since the car's location is random, you have a 1/3 chance of picking it originally. If you don't pick the car (2/3), then the host must reveal the only other goat door with certainty, so you have 2/3 odds of getting the car when you switch.
EDIT: The wiki article confirms:" Although not explicitly stated in this version, solutions are often based on the additional assumptions that the car is initially equally likely to be behind each door". With this assumption, the solution is clearly correct.
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