No, I don't guarantee that everyone who keeps the same door will be 
wrong 2/3rds of the time.  I know that if we carry this out to a large 
number, such as 100, that it is pretty much guaranteed that they will 
be wrong about 2/3rds of the time.  That's what probability is.

The game will be won if the contestant switches doors 2/3rds of the 
time is calculated by the following:

(1/3)(1/3) + (2/3)(2/3) = 5/9

The contestant will win 5 out of 9 times.

Bob's computer simulation shows that switching the door will win twice 
as often as not switching the door.  That's 2/3 and 1/3 probability.  A 
quick look a Bob's simulation shows, by calculation as I did earlier, 
that the overall outcome of 1/2 switch and 1/2 not switching should be 
1/2.  His experiment confirms exactly what we were looking for.

I've already stated over and over again.  If you believe that a person 
will win 1/2 of the time by not switching, then play my game.  I'll 
even up the stakes if you want to make more money off of me.

I could give you arguments all day, but I'm not.  It's simple, will you 
play my game?  Will you challenge my answer?  Unless there is some 
flawed mechanism of how I set up the game, prove to me that you're 
right and earn your $50.  You could buy 5 cubes with that money.

Tyson Mao
MSC #631
California Institute of Technology

On Dec 28, 2005, at 3:56 PM, d_j_salvia wrote:

> You seem to guarantee that everyone will be wrong 2/3rds of the time.
> There is no guarantee.
>
> I'm curious, what percentage of the time will the game be won if the
> computer contestant switches it's choice 2/3s of the time?



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