Another good one (and I'm sure most of you have heard it before -- even MythBusters did a thing on it):
You're on a game show and there are three doors, of which you can pick one. One door has a prize behind it, the other two doors have nothing behind it (or a big pile of crap, whatever you prefer). So you pick a door (but don't open it), and the host (who knows what's behind the doors) picks another door and opens it, and behind it is a big pile of crap. The host then gives you the option to choose another door and open that one instead.
Do you have a better chance of getting the prize if you switch doors, or are your chances the same either way?
Actually, no: believe it or not, switching IS beneficial and does increase your chances. Originally your chances were 1 in 3; after the host reveals one door of crap your chances are still 1 in 3, unless you switch, in which case your chances are apparently 2/3.
After hearing this was the case (on Mythbusters), I found it easy to understand why your chances are higher if you switch by just increasing the number of doors in the problem (which I see is also listed on Wikipedia as a "solution").
So instead of 3 doors say there are 100 doors instead. You pick one. At this point, your chances of picking the door with the prize are 1/100. Okay, so now the host opens 98 (instead of 1) of the other doors that he knows has a pile of crap behind them, and the host gives you the option to either keep your choice, or switch.
Obviously, your chances are way higher if you switch and choose between one of the two remaining doors you didn't originally choose, compared to keeping your original choice (when your chances of picking the right door were only 1/100).
If I make a whole bunch of decisions that don't affect my final decision of choosing between two doors, How does changing the final decision get me the prize?
What if I've decided(beforehand) that I'll change my decision at the very last moment. Am I sure to pick the crap door first?
Apparently, regardless of the number of doors, one will always pick a crap door first.
With the whole Monty Hall paradox, rather than just the odds, they are also taking into account there's the Good Monty Hall and the Bad Monty Hall, the former who is urging you to switch to get the prize, and the later who is just trying to trick you because he knows you have the prize. Of course this is all for effect since it's a game show and there are ratings to consider. What no one in the discussion knows is that the producers know ahead of time who will win and who will not, and the prizes behind the doors are on a rotating circle, like a giant lazy susan, so if you were chosen to win and you picked door B which has a goat, while Monty is blabbering away, a stage hand rotates the floor behind the door so the prize now sits behind door B. Then Monty gives you a chance to switch to Door A, which now has a goat. If you do switch, since it's been decided you're going to win, the stagehand rotates the floor again so the prize moves to door A. If you stand firm with door B, they don't move the floor and you win.
Figure rotating floors and TV ratings into your statistics.
Comments
You're on a game show and there are three doors, of which you can pick one. One door has a prize behind it, the other two doors have nothing behind it (or a big pile of crap, whatever you prefer). So you pick a door (but don't open it), and the host (who knows what's behind the doors) picks another door and opens it, and behind it is a big pile of crap. The host then gives you the option to choose another door and open that one instead.
Do you have a better chance of getting the prize if you switch doors, or are your chances the same either way?
After hearing this was the case (on Mythbusters), I found it easy to understand why your chances are higher if you switch by just increasing the number of doors in the problem (which I see is also listed on Wikipedia as a "solution").
So instead of 3 doors say there are 100 doors instead. You pick one. At this point, your chances of picking the door with the prize are 1/100. Okay, so now the host opens 98 (instead of 1) of the other doors that he knows has a pile of crap behind them, and the host gives you the option to either keep your choice, or switch.
Obviously, your chances are way higher if you switch and choose between one of the two remaining doors you didn't originally choose, compared to keeping your original choice (when your chances of picking the right door were only 1/100).
See also: http://en.wikipedia.org/wiki/Monty_Hall_problem
If I make a whole bunch of decisions that don't affect my final decision of choosing between two doors,
How does changing the final decision get me the prize?
What if I've decided(beforehand) that I'll change my decision at the very last moment.
Am I sure to pick the crap door first?
With the whole Monty Hall paradox, rather than just the odds, they are also taking into account there's the Good Monty Hall and the Bad Monty Hall, the former who is urging you to switch to get the prize, and the later who is just trying to trick you because he knows you have the prize. Of course this is all for effect since it's a game show and there are ratings to consider. What no one in the discussion knows is that the producers know ahead of time who will win and who will not, and the prizes behind the doors are on a rotating circle, like a giant lazy susan, so if you were chosen to win and you picked door B which has a goat, while Monty is blabbering away, a stage hand rotates the floor behind the door so the prize now sits behind door B. Then Monty gives you a chance to switch to Door A, which now has a goat. If you do switch, since it's been decided you're going to win, the stagehand rotates the floor again so the prize moves to door A. If you stand firm with door B, they don't move the floor and you win.
Figure rotating floors and TV ratings into your statistics.