In response to nothing...

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aneuzil21

Active Member
#1
#1
Regarding the famous Monty Hall Problem, I saw a thread somewhere offering much confusion on the equation. For the famous, pick one of three door's to reveal a car or a goats problem...here's how it works:

Out of three doors you choose door # 1. You have a 1/3 chance (a.k.a. 33%, a.k.a. 33/100) of being correct. The ever-knowing host then shows you one of the two remaining doors that has a goat behind it. We now can eliminate that door, that 1/3 chance from the equation leaving us with 2/3 (a.k.a. 66%, a.k.a. 66/100). If we're given the option to switch doors, we have a 50% out of 66% of being correct. 50/66 equates to about 76%. So, switching gives us a 76% chance of winning versus our original 33% chance.

Take three cards (An Ace and two 2's) and simulate. If you choose a card, and then switch you will get it right about 75% of the time. Do it 10 times, it will work about 7.
 
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callipygian

Well-Known Member
#2
#2
aneuzil21 said:
Regarding the famous Monty Hall Problem, I saw a thread somewhere offering much confusion on the equation. For the famous, pick one of three door's to reveal a car or a goats problem...here's how it works:

Out of three doors you choose door # 1. You have a 1/3 chance (a.k.a. 33%, a.k.a. 33/100) of being correct. The ever-knowing host then shows you one of the two remaining doors that has a goat behind it. We now can eliminate that door, that 1/3 chance from the equation leaving us with 2/3 (a.k.a. 66%, a.k.a. 66/100). If we're given the option to switch doors, we have a 50% out of 66% of being correct. 50/66 equates to about 76%. So, switching gives us a 76% chance of winning versus our original 33% chance.

Take three cards (An Ace and two 2's) and simulate. If you choose a card, and then switch you will get it right about 75% of the time. Do it 10 times, it will work about 7.
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I can't believe you started a thread just to post an incorrect solution. :rolleyes:

By switching, you have a 66% chance of winning, not 50/66.
 
A

aneuzil21

Active Member
#4
#4
No, it's 50 out of 66. You have a 50% chance of choosing the correct card...out of the 2 remaining cards, which represent the remaining 66% or 2/3.
 
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callipygian

Well-Known Member
#6
#6
EasyRhino said:
You know, that's a good idea, because I still don't intuitively believe in the mathematical answer to that question.
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The key part you're probably not understanding intuitively is that this is not a strict probability problem; this is a conditional probability problem.

Let's say that, after a door is opened, an outside observer is brought into the room with no knowledge of what has happened. This observer sees two closed doors, one open, and has no idea which door was picked or which door was not. This observer only has a 50% chance of guessing correctly!

The original door picker has additional information - the fact that, if he picked the wrong door from the beginning, Monty was forced to open the 2nd wrong door (and consequently reveal the location of the goat). As a consequence, what is being bet on at the end of the problem is not which door has the goat - the bet is whether the original door picked was correct or not.

And to be complete, mathematically, this is how you work it out:

Case 1: Picker picks door #1. Car is in door #1. => Monty opens either door #2 or door #3, picker switches, picker LOSES.

Case 2: Picker picks door #1. Car is in door #2. => Monty is forced to open door #3, picker switches, picker WINS.

Case 3: Picker picks door #1. Car is in door #3. => Monty is forced to open door #2, picker switches, picker WINS.

Edit: Thought the goat was the prize! Apparently, car > goat. lol
 
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vingtetun

Active Member
#8
#8
Another key

Another key is that the host has to know where the goat is and thus not pick it when he shows you a door. If the host doesn't know where the goat is and he randomly opens a door after you pick a door then there is no value in switching. It's just 1/3 he shows the goat, 1/3 you're right initially and 1/3 it's in the third door.

It's a very neat problem though. I like it.
 
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Canceler

Well-Known Member
#10
#10
This was done to death in the original thread, but I can’t help myself… Arrrgh!

callipygian said:
the bet is whether the original door picked was correct or not.
Click to expand...
Exactly! IF you pick a goat first, you are a lock to win, because you will be shown where the OTHER goat is, then you switch, and get the car.

Now, what are the chances of picking a goat first? 2 out of 3, or 67%.
 
Pelerus

Pelerus

Well-Known Member
#11
#11
The best explanation of this situation that I have heard is a follows:

-There are three doors, behind one of which is a car, and behind the other two goats.

-The contestant picks door 1. There is a 1/3 chance that the car is behind this door. Call this door unit A.
[Unit A = door 1], {Unit A = 1/3}

-Consequently, there is a 2/3 chance that the car is behind doors 2 and 3, taken together as a unit. Call this unit B.
[Unit B = door 2 & door 3], {Unit B = 2/3}

-The host eliminates one of the two doors, between doors 2 and 3, which does not conceal the car. In this case he eliminates door 2. Unit B now contains only door 3.
[Unit B = door 3], {Unit B = 2/3}

-There is still a 2/3 chance that the car is behind unit B. Eliminating door 2 did nothing to change that, but only served to redefine unit B as door 3 only.

-For unit B to decrease in probability upon the removal of one of its components (door 2), would mean that the removed component still has a fraction of chance that it takes with it. But door 2 has been revealed to have a zero chance, and thus its removal from unit B does nothing to alter unit B's total chance: 2/3. Unit B is simply redefined as door 3.

-->Thus, there is a 2/3 chance that the car is behind door 3.
 
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