The Monty Hall problem is driving me nuts

When it just doesn't fit anywhere else.
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PeterG
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The Monty Hall problem is driving me nuts

Post by PeterG »

Read about it here.

The correct explanation makes no sense to me. The math is way over my head, and the attempts to put it in layman's terms are completely unconvincing to me. I'd be strongly inclined to reject the correct solution entirely and refer to it as the "correct" solution...

...except that the problem is easily simulated (I might even try it myself sometime), and by all accounts the simulations consistently produce the outcome predicted by the correct explanation.

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I'm baffled.
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appleman2006
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Re: The Monty Hall problem is driving me nuts

Post by appleman2006 »

I have looked at that puzzle in some length and it makes perfect sense to me. Never have really been able to understand why it does not make sense to some even though I know there are some very smart people that simply do not get it.

To me the simple explanation is that the first time you pick you only had a 33 percent chance of being right, The second time you have a 50/50 chance. But always remember that you picked the first time at lesser odds. So more times than not you should switch your choice. To me it is that simple.
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Hats Off
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Re: The Monty Hall problem is driving me nuts

Post by Hats Off »

There is a saying bs baffles brains. This would seem to be such as occasion. By not switching he also has a a 50:50 chance of being right since one goat door has been eliminated.
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PeterG
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Re: The Monty Hall problem is driving me nuts

Post by PeterG »

appleman2006 wrote:The second time you have a 50/50 chance.
Hats Off wrote:By not switching he also has a a 50:50 chance of being right since one goat door has been eliminated.
That's how it seems to me too, but it's not correct. The whole point is that the second time you still have a 1/3 chance of winning if you don't switch, but a 2/3 chance if you do switch, because there always was a 2/3 chance that the winning door would be one of the doors you didn't choose.
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One of those doors has now been revealed to be a loser. So there's a 2/3 chance that you'll win if you switch to the door you didn't previously choose.
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What I don't get is why you can't just say that there was also a 2/3 chance that the car would be behind either Door 1 or Door 3, also giving Door 1 a 2/3 chance and canceling out the advantage of switching. That's the explanation I would stick with...

...if it weren't for the proof offered by the simulations.
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MattY
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Re: The Monty Hall problem is driving me nuts

Post by MattY »

I was surprised when I first read this problem, but it made sense after I studied it. To me the easiest way to understand it is to suppose that there are not just three doors, but 1,000 doors. Just like in the three-door example, the host knows which door the prize is behind, and opens all other doors but one. How likely do you think it is that you chose the correct door initially? Can you see that in this case you should change to the one door out of the 999 others that the game show host left closed?

The chance that you chose the correct door didn't change - it's still 1/1,000. And the same applies when there are only three doors - the chance of it being correct is still only 1/3, not 1/2.

Or to put it another way, the chance of the door you chose being the right one is initially 1/3. Consider the other two doors collectively -the chance of the car being behind one of the other two doors is 2/3. Once the host opens one of those doors, you still have a choice to switch to those two doors together, and the chance of the prize being behind one of those two is still 2/3 - but it's just that one of them is open. The full 2/3 chance now rests on the unopened door of those two.
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