| | THE MONTY HALL MATH PROBLEM...
As explained by my physical chemistry professor (it came up today while discussing the Boltzman Distrubution...he was explaining another type of probability, so I asked about this one)
Aside: this conundrum came up during Spring Break...on the show "Let's Make a Deal" the contestant has a choice of three doors, and after he or she picks, one of the other three doors is opened by the host, who knows which door hides the prize. You are supposed to guess whether it's statistically favorable to change doors or to stick with the original door. You are then told that your chance of winning if you switch is 2/3...and here is why:
At the beginning, you have a 1/3 chance of winning, no matter which door you pick. You have a 2/3 chance of not winning at this same time, as in, a 2/3 chance that it is behind one of the other doors. The thing is, when one of these doors is eliminated, the chance that it WAS behind one of those doors is unchanged. So, by switching, you will have a better chance of winning.
To give a similar, but more obvious example: if I pick a number between 1 and 100, and tell you to guess it, you have a 1% chance of picking the right number. If I then "eliminate" 98 numbers and tell you that the number is yours or the one I have left; you would feel like you have a better chance with the one that I kept, right? Yes--there would be a 99% chance that this was the right number.
And there you have it!
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| | Posted 3/28/2008 3:08 PM - 107 Views - 4 eProps - 3 comments
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I adopted a cute lil' pikachu fetus
from Fetusmart! Hooray fetus!
