Probabilistic Distribution

[Roberto]

but your chance is now 50% not 33% sticking to your original door - so it makes no difference if you change or not?

as youve decreased your risk with the original choice of door wouldnt it make sense to stick with your winner?

which ever way you look at it, after the 1st door has been opened, you now have 2 choices so a 50% chance regardless if you stick or twist.

is this not correct?

No; it's certainly not! The odds are 2/1 in favour of changing your mind!

 

[Rudy]

Have a look over here: http://en.wikipedia.org/wiki/Monty_Hall_problem

The solution is simple:

1. You start with an even prob. distribution for the car: Door1: 1/3 Door2: 1/3 Door3: 1/3

2. You choose eg Door1. You have a 1/3 prob that it contains the car. Door2+Door3 have 2/3 prob. to contain the car.

3. Monty opens eg Door3 and reveils a goat. But still Door2+Door3 have 2/3 prob to contain the car ! Opening one of the doors does not alter the original prob. The new prob. distribution is:
Door1: 1/3 Door2: 2/3 Door3: 0

4. So now you better switch to Door2: it has a 2/3 prob that it contains the car !