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Richard Carrier no can haz conklushun?

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See his post here.

I've made McCain McFast & McFurious!

You can find pics of all the "Easter eggs" in the video here on my xanga.

My Monty Hall Problem Explanation

Monty Hall problem:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Apparently, it is always to your favor to switch doors and this makes absolutely no sense to most people.

The easiest way for me to understand it is to pretend like two different probabilities are being asserted and then merged. And this seems to work easily across any number of scenarios from 3 doors to millions of doors. Anyone can see that with 5 doors, for instance, YOU have a 1/5th chance of being right. But the game show host (if we take that person on their own) has a 5/5 chance of being right, because they know which door is correct. Duh. [I would say 1/1 but that side tracks us] However, when you merge the two, the game show host has a 5/5 chance of being right, minus your selection, making it a 4/5th chance of being right. Hence, rather than being a 50/50 shot, you are actually choosing between your probability (1/5th chance of being right) and the host's adjusted probability (4/5ths chance of being right). From that basis, no matter how many doors are in the mix (especially as the number of doors climbs) the host's odds are always better.

With 1,000,000 doors, you have a 1 out of a million chance of being right (aka, a big whopping guess), and the host has a million out of a million chance of being right, minus your door, making his chances 999,999,999 out of 1 million. Clearly way better odds than your sucky guess as you would intuitively expect at this level.

With the most minimal classic three doors situation, you have a 1/3rd chance of being right initially and he would have a 3/3 chance of being right initially (if there was an initially), minus your selection again, taking his chances down to 2/3rds. Clearly still better odds, but only slightly.

With 4 doors, your ignorant chances are 1/4th and as a result the all knowing yet impaired host's chances are taken down to 3/4ths.

With 6 doors, your ignorant chances are 1/6th and as a result the all knowing yet impaired host's chances are taken down to 5/6ths.

With 7 doors, your ignorant chances are 1/7th and as a result the all knowing yet impaired host's chances are taken down to 6/7ths...

And so on and so forth by extension...

I think that makes it really easy to understand and apply consistently across the board. Let me know if that does it for you guys.

Ben

Truly inspiring...

I just wonder how much the awe will pan out...