Lemieux66
Lemieux66
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March 24th, 2014 at 1:43:47 PM permalink
http://en.m.wikipedia.org/wiki/Monty_Hall_problem

Anyone believe in this? I refuse to. The entire situation changes when you are shown a free door.
10 eyes for an eye. 10 teeth for a tooth. 10 bucks for a buck?! Hit the bad guys where it hurts the most: the face and the wallet.
geoff
geoff
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March 24th, 2014 at 1:50:09 PM permalink
What's not to believe? It's been proving both mathematically and via simulations repeatedly.
AxiomOfChoice
AxiomOfChoice
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March 24th, 2014 at 1:51:47 PM permalink
Quote: Lemieux66

http://en.m.wikipedia.org/wiki/Monty_Hall_problem

Anyone believe in this? I refuse to. The entire situation changes when you are shown a free door.



What, exactly, don't you believe?

In the show let's make a deal, the host would ALWAYS open a losing door that the player did not pick (there is always at least one such door available). Therefore it is always correct to switch.

If the host is opening an unpicked door at random then it makes no difference whether or not you switch. But Monty Hall was not picking at random -- he always picked a losing door.
1BB
1BB
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March 24th, 2014 at 1:56:04 PM permalink
Quote: Lemieux66

http://en.m.wikipedia.org/wiki/Monty_Hall_problem

Anyone believe in this? I refuse to. The entire situation changes when you are shown a free door.



There are some good threads on this if you want to search the site. You should find them interesting. One of the threads was started by mkl54321 who was an excellent and very smart poster here.
Many people, especially ignorant people, want to punish you for speaking the truth. - Mahatma Ghandi
MathExtremist
MathExtremist
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March 24th, 2014 at 2:01:30 PM permalink
Agreed, it's not clear what you're disbelieving. But the premise of the show -- and the assumptions required in the proper formulation of the problem -- was that not only did Monte know where the goats were but that he would always show you one. In other words, Monte was not picking randomly so knowledge of his picking rules can inform your choice.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
Lemieux66
Lemieux66
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March 24th, 2014 at 2:01:43 PM permalink
Didn't know that. I googled about this and a guy posted this on another forum. It makes sense to me now, although the enormity of 100 numbers feels different than 3 numbers.


Quote:
Originally Posted by Grumman
That is not inherently true. It could also mean the host is trying to trick you into moving away from the correct answer. Like I said, it's not about probabilities, it's about knowing how the host responds to information he can see that you can't, to work out what he knows.
Rules of the following game:
1) I think of a number between 1 and 100.
2) I ask you to guess a number between 1 and 100.
3) I tell you that I'm going to rule out 98 numbers, one of which is the number you chose. If you chose correctly in the first place, then I'll choose another number randomly not to rule out. If you chose incorrectly in the first place, then I won't rule out the correct number.
4) I give you the opportunity to switch to the other number, or stick with yours.
5) I reveal whether your final choice is the number I thought of in step 1.

Let's try it out; I'll fill in your guess more-or-less randomly in step 2.

1) I've got my number, but I'm not telling you what it is.
2) You guess that it's 47.
3) I tell you that it's not 1-46, 48-72, or 74-100. It's either 47 or 73.
4) You can now switch your guess to 73, or you can stick with 47.
5) If you stuck with 47, you'd be wrong. If you switched to 73, you'd be right.

Step 3 is important. I'm never going to rule out the correct number, any more than Monty is ever going to open the door with the prize and say, "Oops, too bad!" I'm ALWAYS going to rule out an incorrect choice. That move on my part is what skews the odds so heavily in favor of switching for you.

If you're still not convinced, play the game I described above with a friend. If you follow the rules correctly, if you switch every time, you'll win 99 times out of 100. If you stick with your original number every time, you'll win 1 time out of 100.
10 eyes for an eye. 10 teeth for a tooth. 10 bucks for a buck?! Hit the bad guys where it hurts the most: the face and the wallet.
MangoJ
MangoJ
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March 24th, 2014 at 2:06:15 PM permalink
Believe in what exactly ?
- That the Monty Hall problem exists in real live ? (well, at least some people have been contestants).
- That the Monty Hall problem is likely to produce some confusion ?
- That the Monty Hall is a solveable paradox ?
Lemieux66
Lemieux66
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March 24th, 2014 at 2:12:15 PM permalink
Quote: MangoJ

Believe in what exactly ?
- That the Monty Hall problem exists in real live ? (well, at least some people have been contestants).
- That the Monty Hall problem is likely to produce some confusion ?
- That the Monty Hall is a solveable paradox ?



You pick one of three doors. One is shown not to be correct. So now one is correct and now one is not. If not shown a new door, obviously switch. If shown a new door, information now changes. So to switch seems strange to me.
10 eyes for an eye. 10 teeth for a tooth. 10 bucks for a buck?! Hit the bad guys where it hurts the most: the face and the wallet.
AxiomOfChoice
AxiomOfChoice
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March 24th, 2014 at 2:23:27 PM permalink
Quote: Lemieux66

You pick one of three doors. One is shown not to be correct. So now one is correct and now one is not. If not shown a new door, obviously switch. If shown a new door, information now changes. So to switch seems strange to me.



I'm not sure what you mean by "a new door".

There is a 1/3 probability that the right door is the one you picked, and a 2/3 probability that the right door is one of the other two.

At least one of the other two is guaranteed to be wrong, so the host (intentionally, every time) shows you a wrong door from the other two. Since he is always able to do this, you can't get any extra information about your door. It's STILL true that there is a 1/3 chance that your door is right, and a 2/3 chance that one of the other two doors is right. The difference is, now, if it's one of the other two, you know which of the two it is.

If you stay, you get the door that you picked (1/3 chance of being right). If you switch, you get BOTH of the other doors (2/3 chance of containing the right one). Obviously having two doors is better than having one.
geoff
geoff
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March 24th, 2014 at 2:30:26 PM permalink
You can also think of it by percentages. The door you choose has a 1/3 chance of being the car. Which means that the doors not chosen have a combined 2/3 chance of hiding the car. Someone eliminates one of the 2 doors not chosen and this is where the paradox/ counter intuitive thinking comes in. Even though there is no only 2 doors remaining the first door you chose still only has a 1/3 chance of hiding the car while the one you haven't picked has a 2/3 chance. This is because revealing the goat doesn't actually change anything.

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