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Dresdan · 12-08-2001, 09:19 PM

quote:
The results are counter-intuitive, I know. That's what makes it a PUZZLE! It does make sense, though, when you look at it right: Let's say you pick Door A. Now, there's a 33% chance the prize is behind Door A, and a 66% chance it's behind (Door B OR Door C). These numbers will NOT change. When the gameshow host opens Door C to reveal a broken Rubik's Cube, the chance of Door B OR Door C being the winner is STILL 66%.....except that Door C has been eliminated, leaving that 66% sitting behind Door B.

The way I see it is that there is more than one math problem here.
In the first problem you have three doors meaning that you have a 33.3..% chance of choosing correctly. If you fail to do so you are wrong. So the new the problem is that you have two doors to choose from which means that you have a 50% chance of choosing correctly.

12-08-2001, 09:19 PM	#139
Dresdan Zhentarim Guard Join Date: November 18, 2001 Location: Bay City, MI. , U.S.A. Age: 55 Posts: 331	quote: The results are counter-intuitive, I know. That's what makes it a PUZZLE! It does make sense, though, when you look at it right: Let's say you pick Door A. Now, there's a 33% chance the prize is behind Door A, and a 66% chance it's behind (Door B OR Door C). These numbers will NOT change. When the gameshow host opens Door C to reveal a broken Rubik's Cube, the chance of Door B OR Door C being the winner is STILL 66%.....except that Door C has been eliminated, leaving that 66% sitting behind Door B. The way I see it is that there is more than one math problem here. In the first problem you have three doors meaning that you have a 33.3..% chance of choosing correctly. If you fail to do so you are wrong. So the new the problem is that you have two doors to choose from which means that you have a 50% chance of choosing correctly.