2009-03-18, 09:44 PM
TehMatt Wrote:So:
Choose A + Stay = Win
Choose A + Switch = Lose
Choose B + Stay = Lose
Choose B + Switch = Win
Choose C + Stay = Lose
Choose C + Switch = Win
C is struken since it was eliminated by the host, revealing it contained a. This makes the probabilty 50% chance of winning if you stay and 50% chance of winning if you switch.
No.
Choose A + Eliminate B or C [does not matter] + Stay = Win
Choose A + Eliminate B or C [does not matter] + Switch = Lose
Choose B + Must eliminate C + Stay = Lose
Choose B + Must eliminate C + Switch = Win
Choose C + Must eliminate B + Stay = Lose
Choose C + Must eliminate B + Switch = Win
For choosing A, there are not two difference possibilities if he picks B or C. This makes it 1/3 if you stay and 2/3 if you switch.
If you really want more proof, wikipedia it or something. This problem is very very famous and used in tv shoes, movies, etc... It is proved that the above is true.


. This makes the probabilty 50% chance of winning if you stay and 50% chance of winning if you switch.