algorithmic modeling for Rhino
Yup, you should always switch.
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That's better!
@Vicente, ah indeed, there's no point to the jitter without different seeds.
@David, I prefer to explain it thus:
First imagine we do not switch, ever. The chances we picked the right door were 1/3, and the chances we picked the wrong one were 2/3. We don't change our minds, Monty doesn't open any doors either, so this is fairly straight-forward. We win 33% of the time and lose 66% of the time.
Now Monty opens one of the doors without a prize and we always switch to the third door. What happens if we picked right the first time? Well, we lose and that sucks. But what happens if we picked wrong the first time? In that case we cannot help but pick the right door this time.
So switching after a non-winning door opens is akin to flipping your win/lose chances.
I've had a few conversations with people about this, and if they remain skeptical, I try to illustrate it by asking if it was ten doors instead of three, and Monty opened eight, would they change their mind then. That usually helps all but the most recalcitrant.
I think there's a small mistake but shouldn't affect the result. The seed in the jitter component doesn't change, so for every pair of available doors to open it will always choose the same one.
Yes it is, but how much it is true depends on how many doors are opened. If Monty always opens up only 1 door which does not contain the prize then your odds will still be better if you switch, but very rapidly diminishing returns. If however Monty opens ALL doors except one, then the odds increase overwhelmingly. In the 3-door case both of these options look the same.
Very curious. This is also true with more than three possible choices?
I think I have to change my vote in the upcoming elections. xD
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