RE: The role of probability in solving the Monty Hall problem
March 13, 2016 at 8:46 am
(This post was last modified: March 13, 2016 at 8:49 am by Excited Penguin.)
Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.
I mean you'll choose to switch to the wrong door once. Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.
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It is misleading to think that it's truly 2/3 to switch to the car. And that's because there is nothing to say that you will select those three doors the same amount of times while playing the same game repeatedly, for instance door X 3 times, door Y 3 times, and door Z 3 times as well, which would have to hold true in order for the probability you're claiming to be truly accurate. You with me?
(March 13, 2016 at 8:45 am)Mr.wizard Wrote:(March 13, 2016 at 8:41 am)Excited Penguin Wrote: Rob, you just don't get it, do you. It's not about probability in the least. We already know that mathematically your answer is correct, but the fact is that we're saying you shouldn't use probability at all in that instance. Why? Well, because there's nothing to guarantee that you won't end up infinitely choosing to switch to the wrong door.
When you add up the number of options however, that problem dissolves and your probability theory proves to be effective. For instance when there are a million doors to choose from. But not when there are only a few of them.
--- THe problem is still there, but it's much more of a leap to say that you won't get the right answer as the number of doors increases.
What do you mean infinitely choosing the wrong door?
I mean you'll choose to switch to the wrong door once. Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.Then again. Then again. Then again.
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It is misleading to think that it's truly 2/3 to switch to the car. And that's because there is nothing to say that you will select those three doors the same amount of times while playing the same game repeatedly, for instance door X 3 times, door Y 3 times, and door Z 3 times as well, which would have to hold true in order for the probability you're claiming to be truly accurate. You with me?