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MR2Guy Regular user Nashville 179 Posts 
Guys
It's probably been discussed before, but this one busted my chops: Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors? The answer, is yes. If you stick with your original choice, your odds are 1/3, if you switch, your odds are 2/3. Jason
Question every rule.
There are no absolutes. 

Scott Cram Inner circle 2678 Posts 
Don't feel bad. This busted Marilyn Vos Savant's chops, as well. Since this puzzle has a viruslike ability to generate argument and discussion everywhere it goes, I might as well put forth a futile effort to prevent it by offering the following links to check it out for yourself:
Cut The Knot.org: Monty Hall Google Search: Monty Hall Puzzle 

MR2Guy Regular user Nashville 179 Posts 
Scott
Thanks, I meant to post some links to generate some, uh, discussion. It took me a couple of weeks to wrap my head around the solution, it made sense, but I couldn't make it click in my head. Here's a magician friendly example to prove the theory: Imagine that you know where one card is in the deck, and you spread them out on the table, and ask a spectator to pick, say, the 9 of clubs. They touch a card. There is also a second spectator watching this. At this point, their odds are 1 in 52. Let's say they picked any card other than the 9 of clubs. You proceed to turn over 50 cards, face up, leaving 2 face down on the table, the original selection, and the 9 of clubs. Ask the second spectator which card he thinks the 9 of clubs is. To me, it seems obvious that his odds are much better if he picks the card that the 1st spectator did not choose. Hope this helps. Jason
Question every rule.
There are no absolutes. 

Slim King Eternal Order Orlando 17862 Posts 
The virus has gotten me! You are telling me that my odds remain the same even though one negative choice has been removed. I don't see it that way. Are you saying that picking the second door IMPROVES my odds? What if there were two contestants choosing doors? I choose number one and he chooses number two. Would our odds increase if we switched our choices???? How could we BOTH improve our odds?..... Doesn't gell with me...Not Yet.
Dave
THE MAN THE SKEPTICS REFUSE TO TEST FOR ONE MILLION DOLLARS.. The Worlds Foremost Authority on Houdini's Life after Death.....


therntier Special user 681 Posts 
Quote:
On 20050629 23:06, Slim King wrote: The game changes if there are two spectators. The big reason for this is you are taking one choice away from Monty. When I teach this example, there are a few ways I teach it. One is to imagine that there are 100 doors, of which you select door 46. Monty knows where the correct door is. You know that there is little chance that you got it correct. Monty starts showing you doors one at a time. He shows you door 1, 2, 3, 4, 5, skips 6, then continues to show the rest of the doors. Do you think that there is a reason he skipped 6? Most people should agree that the door has some significance. About 10% of my students like that explaination. It is my favourite solution. The other way I teach it is to ask them which of the three doors they want, for example they select door 1. Then I ask if they would like what's behind door 1, or the combined contents of what's behind door 2 and 3. Obviously they would like two doors instead of just 1. They know that at least one of the doors are empty, but it still doubles their chances of getting the prize. Since they know that they will be getting at least one empty door, I shouldn't be able to change their decision by showing them that one door is empty. I like this solution a little bit because it emphasizes the fact that the porblem only works because Monty Hall knows where prize is. That is, he always shows you a non winning door. If he doesn't know where the prize is, there is absolutely know advantage in swithching. Trevor 

darylstudios New user 5 Posts 
A mathematical proof is as such:
Suppose we have, as in the original, 3 doors (A,B,C). Let us also assume that the car is in door C. The gameshow host will always open a door that does not contain the car but instead contains a goat. Scenario 1: The contestant does not change his mind. (i) He picks A, B is opened, he stays with his choice, and LOSES. (ii) He picks B, A is opened, he stays with his choice, and LOSES. (iii) He picks C, A or B is opened, he stays with his choice, and WINS. That is a 1/3 chance of winning if he does not change your mind. Scenario 2: The contestant changes his mind. (i) He picks A, B is opened, he changes his mind to C, and WINS. (ii) He picks B, A is opened, he changes his mind to C, and WINS. (iii) He picks C, A or B is opened, he changes his mind to A or B, and LOSES. This is a 2/3 chance of winning if he changes his mind. The first time I saw this puzzle, I too could not believe it. But give it some thought, and soon you will understand. Hope this helps! 

drkptrs1975 Elite user North Eastern PA 452 Posts 
Interesting concept.


Mishel New user Israel 100 Posts 
Trevor,
I was only aware of the first solution you mentioned. the second solution you gave is realy great! I think both explenation should convince almost anyone.
Mishel.
Don't let the same dog bite you twice. 

idris New user St. Louis, MO 38 Posts 
The problem with the discussion I've seen for this problem is that you are assuming that Monty will always give you a chance to change your mind.
As I recall, in real life that wasn't the way it worked. Sometimes he offered a choice and sometimes not. Since he knows where the winning door is, why would you assume that he would let you change to it once you've picked a loosing pick. It is more likely that he will only offer a choice if you've already picked the winner. I vaguely recall hearing an interview with him (or maybe it was a reference to a biography he did or had done) where he said the choices were ususally to allow the contestant to blow the prize. I can't swear to this though. Only if the choice is always give does the logic for changing apply. Jerry
Jerry


Scott Cram Inner circle 2678 Posts 
Jerry,
The puzzle is simply called "The Monty Hall Dilemma" because that's the easiest point of reference to relate it to. It's just a name that can get people to understand the nature of the puzzle, and never claims to be accurate to every episode of Let's Make A Deal. There's also a puzzle known as "The Prisoner's Dilemma", despite the fact that, as far as I know, no prisoner in any penal system has ever been put through this challenge. 

Mike Powers Inner circle Midwest 2959 Posts 
Actually Marilyn Vos Savant gave the correct solution and then was called on the carpet by some PhD's and other folks that should have known better. Shame on them! She gave another solution that hopefully showed these folks how wrong they were.
When trying to convince a friend I was reduced to playing the game 20 or 30 times so he'd see himself losing two thirds of the time. The premise was that since his theory was that it didn't matter if he switched, he would always stay with his first choice. After a few cycles, instead of showing him one card and then the other, I just started turning both cards over at once. He said "hey you get two chances to win!" I said "now you see why you should always switch!" Mike
Mike Powers
http://www.mallofmagic.com 

Slim King Eternal Order Orlando 17862 Posts 
I am Thick as a Brick....I'll keep chipping away at this. Maybe as Mike did, I should just "DO" it a whole bunch of times. I just got a new half gallon of RUM sooooo...maybe now is a good time!
THE MAN THE SKEPTICS REFUSE TO TEST FOR ONE MILLION DOLLARS.. The Worlds Foremost Authority on Houdini's Life after Death.....


Kevinh5 Regular user 108 Posts 
I still don't get this. Once one door has been eliminated, the door you selected has a 1/2 probability of being the good prize. The other door has a probibility of one out of two also. You don't change your odds by changing doors.


owen.daniel Inner circle England 1048 Posts 
Kevin,
You are right to feel uneasy about why eliminating a door has any relevance. Suppose that the problem was stated as follows. Before you choose a door, the game show host shows that one has a donkey behind it, and then lets you choose freely. Now (assuming its not a donkey you are after) you know that the car is behind one of the two remaining doors, and the probability is then a half. The difference in this case is that you selected your door before any were shown to you, and so you force the hosts actions. So lets consider the options. CASE 1. The door you selected was a donkey. In this case the host has no choice in which door he shows you: he cannot open your door, that would tell you that you are wrong and you'd change your mind. He cannot show you the door with the car behind it (then you'd know which one to choose), so he chooses to show you the one that has the other donkey. In this case you should switch: the game show host has unwittingly identified the correct door for you, and if you switch you are guaranteed yourself a new car. CASE 2. The door you chose had the car behind it. In this case the host has more freedom: he can in fact open either of the remaining doors and show you a donkey. If you change in this case, you will lose. So in one of these two cases you always win by changing, and in the other you always lose by changing. This is an important point: once you have chosen your first door, there is no more probability involved, and the actions of the host are deterministic. So rather surprisingly, the whole problem is entirely dependent on only your first selection! If we adopt the strategy of changing when given the option, then we win if our initial selection was a donkey, and lose if it was the car. But since there are 2 donkeys and only 1 car, we have a probability of 2/3 of winning. 

owen.daniel Inner circle England 1048 Posts 
Now here are a few follow up remarks that I wanted to keep separate.
* From the point of the game show host, he would be better off if he relaxed his rule that he never shows you what's behind the door you chose. In particular, if you chose a donkey, and he opened your door, then you would now have a 50% chance of winning, which is much lower than the guaranteed win that you get if he opened the other donkey door. In this game you do always have a 50% chance of winning if you switch door. * Now consider the following alternative problem which I like to give to people who already know Monty Hall. In this version there are two competitors trying to win the same car. There are still three doors, two of which house donkeys. Now each person is instructed to choose a different door. Once they each have a door the host eliminates one of the contestants, and shows that there was a donkey behind their door. Should the remaining contestant switch or stay? Owen. 

Jonathan Townsend Eternal Order Ossining, NY 27242 Posts 
Quote:
On 20050316 14:35, MR2Guy wrote:... If you stick with your original choice, your odds are 1/3, if you switch, your odds are 2/3. When you make your first choice  it's 1 out of 3. If you make a second choice you have a 1 out of 2 chance.
...to all the coins I've dropped here


Kevinh5 Regular user 108 Posts 
That is exactly my thought, Jonathan.


TomasB Inner circle Sweden 1144 Posts 
Kevin, I have 100 doors with a car behind one. I promise to open 98 without cars after you have made your selection, since I know I can always do that regardless of what you select.
You go for door number 5. I open all doors but 2 and 5 and you see that there is no car behind any of those 98 doors. I give you the choice to switch to door number 2 instead, or stick with your original choice of door 5. What do you do? /Tomas 

Kevinh5 Regular user 108 Posts 
My probability of being correct got higher and higher with each reveal you made. At the end, there are only two doors, one of which is a winner. If I keep my original choice, my odds are 1 out of 2. If I change my choice, my odds are 1 out of 2. What matters is the probability NOW, not the probability back when it was 1/100.
So my answer is that really doesn't matter at that point. I'll keep my initial choice. 

Kevinh5 Regular user 108 Posts 
Ahhh, Tomas8 and Owen.Daniel, I see your point about the host KNOWING which was the winner. In that case, the host HAS revealed the answer, I agree.
My answer is based upon NEITHER the host NOR I knowing. A fair blind choice. 

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