Monty Hall Problem has struck again

[midcan5]

I have always been fascinated by game theory and rational choice theories. How is it we make decisions particularly ones that involve choice or doubt. Thought this worth sharing.

Monty Hall Problem has struck again

"Here’s how Monty’s deal works, in the math problem, anyway. (On the real show it was a bit messier.) He shows you three closed doors, with a car behind one and a goat behind each of the others. If you open the one with the car, you win it. You start by picking a door, but before it’s opened Monty will always open another door to reveal a goat. Then he’ll let you open either remaining door.

Suppose you start by picking Door 1, and Monty opens Door 3 to reveal a goat. Now what should you do? Stick with Door 1 or switch to Door 2?"


http://www.nytimes.com/2008/04/08/science/08tier.html

http://www.nytimes.com/2008/04/08/science/08monty.html

 

[Gunny]

I have always been fascinated by game theory and rational choice theories. How is it we make decisions particularly ones that involve choice or doubt. Thought this worth sharing.

Monty Hall Problem has struck again

"Here’s how Monty’s deal works, in the math problem, anyway. (On the real show it was a bit messier.) He shows you three closed doors, with a car behind one and a goat behind each of the others. If you open the one with the car, you win it. You start by picking a door, but before it’s opened Monty will always open another door to reveal a goat. Then he’ll let you open either remaining door.

Suppose you start by picking Door 1, and Monty opens Door 3 to reveal a goat. Now what should you do? Stick with Door 1 or switch to Door 2?"


http://www.nytimes.com/2008/04/08/science/08tier.html

http://www.nytimes.com/2008/04/08/science/08monty.html

I'd stick with the one I originally chose. You can second-guess all you want, but all it ever comes up as is a guess. Second-guessing, in my opinion and based on my personal experience, usually leads to a wrong choice with the original choice being correct.

There really is no rational choice in such a game. It's random choice as no pattern is established. At least as far as I recall from the show, there was no real pattern estblished to which door the booby prize was behind.

The odds on the show were different though. IIRC, there were two good prizes and one boooby prize, not one good one and two bad ones. It's just that one of the good ones would be like a car while the other would be more like consolation furniture.

 

[RetiredGySgt]

Yup a total guess, no reason at all to change your choice.

 

[midcan5]

interesting as you're both wrong.

 

[Gunny]

interesting as you're both wrong.

I was kind enough to answer your question. Would be nice if you could explain why it is you believe I am wrong.

If their was something to the show that made the choice not random, I don't recall. Not that I have even SEEN it in 40 years.

Oh, and I cannot be wrong insofar as my opinion concerning second-guessing and my personal experience goes. If you hump a rifle for a living, that can get you dead REAL quick.

 

[jillian]

I have always been fascinated by game theory and rational choice theories. How is it we make decisions particularly ones that involve choice or doubt. Thought this worth sharing.

Monty Hall Problem has struck again

"Here’s how Monty’s deal works, in the math problem, anyway. (On the real show it was a bit messier.) He shows you three closed doors, with a car behind one and a goat behind each of the others. If you open the one with the car, you win it. You start by picking a door, but before it’s opened Monty will always open another door to reveal a goat. Then he’ll let you open either remaining door.

Suppose you start by picking Door 1, and Monty opens Door 3 to reveal a goat. Now what should you do? Stick with Door 1 or switch to Door 2?"


http://www.nytimes.com/2008/04/08/science/08tier.html

http://www.nytimes.com/2008/04/08/science/08monty.html

I changed my choice each time. That netted me about a 70% or so win rate.

 

[Foxfyre]

I changed my choice each time. That netted me about a 70% or so win rate.

I changed my choice each time which netted me about 33% or win rate. I don't really understand that if a goat is revealed behind a door you didn't pick how there isn't a 50%-50% chance that you picked the right door. Seems like if there is a truly random distribution, over time you would wind up right approximately 50% of the time.

Maybe you guys who are math whizzes can show me the logic of my reasoning here.

 

[jillian]

I changed my choice each time which netted me about 33% or win rate. I don't really understand that if a goat is revealed behind a door you didn't pick how there isn't a 50%-50% chance that you picked the right door. Seems like if there is a truly random distribution, over time you would wind up right approximately 50% of the time.

Maybe you guys who are math whizzes can show me the logic of my reasoning here.

I'm not a math whiz by any stretch of the imagination. But I kind of like statistics (as opposed to, say, geometry or algebra). The way it supposedly works is that on your first choice, you have a 1 in 3 chance of being correct. Once one choice is eliminated, the odds are reduced to a 1 in 2 chance of you being correct, so you go from a 33% chance of being correct if you don't change your choice to a 50% chance. So without changing your choice, you would generally be right 33% of the time. I'm probably not explaining it very well.... but that's the best I can figure

 

[Gunny]

I'm not a math whiz by any stretch of the imagination. But I kind of like statistics (as opposed to, say, geometry or algebra). The way it supposedly works is that on your first choice, you have a 1 in 3 chance of being correct. Once one choice is eliminated, the odds are reduced to a 1 in 2 chance of you being correct, so you go from a 33% chance of being correct if you don't change your choice to a 50% chance. So without changing your choice, you would generally be right 33% of the time. I'm probably not explaining it very well.... but that's the best I can figure

I don't get that reasoning. If you have a 33% chance of being right with 3 choices, and one choice is eliminated, leaving two choices and one of them is yours, it stands to reason you then have a 50% chance of being right whether you change your choice or not. You still have chosen between what is now two options.

Nor do I see where mathematics and/or science come into play unless there is a pattern established as to which door the goats are usually behind/which door the prize is behind.

Perhaps Midcan will come along and enlighten us.

 

[Larkinn]

Its not that hard to explain, or understand. Intuitively it makes no fucking sense, it just happens to be right thats all.

Ok. Imagine you are picking between 3 doors. Now there is a 33% chance of picking the right door. So, you choose one and there is a 33% chance of it being there and a 66% chance of it being in the other 2. But, one of the other 2 is then eliminated and shown there is nothing in it. Remember that *nothing has changed from your original probability of your guess*. It makes no sense that showing that another door is wrong somehow increases the probability of your original guess being right.

Take, for example, the game Deal or No Deal. Lets assume there are 50 different suitcases (no idea how many are actually there). If you pick a random suitcase hoping that the $1 mil will be in there, you have a 2% chance of getting the right one. Now, assume you pick #1 as the right suitcase. Remember there is a 2% chance of it having 1 mil. Now, the person who runs the game shows you that the $1mil is NOT in boxes 3-50. So, it can only be in box 1 or box 2. The action of opening up boxes known to not contain the $1 mil cannot raise the probability of you picking the 1 mil when you first picked it.

Its not intuitive and takes a while to grasp, but its definitely the case that you should switch if Monty asks you if you want to switch. Probability theory is sort of a bitch.

 

[Foxfyre]

I don't get that reasoning. If you have a 33% chance of being right with 3 choices, and one choice is eliminated, leaving two choices and one of them is yours, it stands to reason you then have a 50% chance of being right whether you change your choice or not. You still have chosen between what is now two options.

Nor do I see where mathematics and/or science come into play unless there is a pattern established as to which door the goats are usually behind/which door the prize is behind.

Perhaps Midcan will come along and enlighten us.

When the choice is three doors, you have a 1 in 3 chance of picking the right door or a 33.3% chance of being right. When the choice is then narrowed to two doors, it seems logical that you now have a 1 in 2 chance of picking the right door or a 50% chance of being right.

But then I guess I'm jaded because I am always aware that, until the door you picked is opened, the game managers can switch around the prizes any way they want to. So they actually have the power of deciding whether you will win or lose.

 

[RetiredGySgt]

I disagree. Your choice whether you change it or not will still have the same chance the other door would have.

3 doors, you pick one. One is revealed. Whether you change your pick or not there is still the exact same chance as originally that the win is behind other of the other doors.

Now if Monty only asks you if you want to change when HE knows your wrong then ya that changes the odds and is a clue.

Given NO change and no way to figure out which is which and no pattern based on the wrong one opened, the odds do not change. You have the exact same chance of being right as you did to begin with. Why? Because by NOT choosing a new door, you have in fact CHOSEN a door. It is still random which door the winner is behind. You gain no advantage by selecting a different choice. It will still be a matter of luck.

Now if one door where opened and the remaining two were randomly shuffled..... odds would change.

 

[RetiredGySgt]

There is another way to look at this. Since it is a GIVEN one door will be revealed, in effect you always had a 50 percent chance of being right.

 

[midcan5]

I don't get that reasoning. If you have a 33% chance of being right with 3 choices, and one choice is eliminated, leaving two choices and one of them is yours, it stands to reason you then have a 50% chance of being right whether you change your choice or not. You still have chosen between what is now two options.

Nor do I see where mathematics and/or science come into play unless there is a pattern established as to which door the goats are usually behind/which door the prize is behind.

Perhaps Midcan will come along and enlighten us.

Jullian and Larkinn explained the math behind it. You select one and Monty shows you a second one. Your first choice was one in three, now knowing one for sure, changing it gives better results. 1/3 = 33% chance - knowing one 1/2 = 50% when Monty shows you one goat your chance changes - what you picked the first time was just least (statistically) likely to be right one, so changing it works statistically.

 

[Ravi]

If you play the game and then click through for the explanation, it almost makes sense. Almost.

 

[Larkinn]

I disagree. Your choice whether you change it or not will still have the same chance the other door would have.

Err...feel free to disagree but this isn't really a matter of opinion.

3 doors, you pick one. One is revealed. Whether you change your pick or not there is still the exact same chance as originally that the win is behind other of the other doors.

Not quite. Separate the two odds into two different groups. There is a 33% chance that the one you picked is right. There is a 66% that the one you picked was wrong. What makes it wrong? Well that its in one of the other two. Well if you take away the other box that has nothing in it there is STILL a 33% that you were right and a 66% chance that you were wrong. The 66% chance that you were wrong is now entirely in the one reminaing box that you haven't picked. Therefore it is 66% to be right and the probability is NOT 50/50.

Given NO change and no way to figure out which is which and no pattern based on the wrong one opened, the odds do not change.

Right, they don't change. They STAY at 33/33/33. Only now its 33/66 since one of the boxes goes away.

You have the exact same chance of being right as you did to begin with. Why? Because by NOT choosing a new door, you have in fact CHOSEN a door. It is still random which door the winner is behind. You gain no advantage by selecting a different choice. It will still be a matter of luck.

You have the exact same chance of being right...IF you don't choose to switch. If you do switch it now doubles, since it is now 33+33%

 

[Larkinn]

There is another way to look at this. Since it is a GIVEN one door will be revealed, in effect you always had a 50 percent chance of being right.

Umm thats a completely wrong way of looking at it.

Take 100 boxes. I'll put a dollar in one of the boxes. Pick which box you think the dollar is in. Now I will remove 98 of the boxes that I know don't have the dollar in it. You think there is a 50% of the dollar being in your box? Thats absurd.

 

[RetiredGySgt]

Umm thats a completely wrong way of looking at it.

Take 100 boxes. I'll put a dollar in one of the boxes. Pick which box you think the dollar is in. Now I will remove 98 of the boxes that I know don't have the dollar in it. You think there is a 50% of the dollar being in your box? Thats absurd.

No I think I have just as much chance of it being in the box as there was originally. Changing my mind now doesn't somehow make my guess any better then it was to begin with. It is still either in the box I chose or not, just like it was to begin with. The odds have not gone up from the original choice. Changing my mind does not give me a better chance of being right.

 

[Larkinn]

No I think I have just as much chance of it being in the box as there was originally.

Only if you are talking about the box you originally picked.

Changing my mind now doesn't somehow make my guess any better then it was to begin with.

Sure it does, because now you are basically getting 2 for the price of 1. So there is a 66% probability that you are right if you switch.

It is still either in the box I chose or not, just like it was to begin with. The odds have not gone up from the original choice. Changing my mind does not give me a better chance of being right.

RGS...you are wrong. As I said before, its not a matter of opinion.

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

Read that if you are actually interested in understanding it.

By the way...this should be a lesson to all of you here who like to claim that "common sense" is always right and that simple=true and that all that derned learning just makes people wrong more often.

 

[Gunny]

Its not that hard to explain, or understand. Intuitively it makes no fucking sense, it just happens to be right thats all.

Ok. Imagine you are picking between 3 doors. Now there is a 33% chance of picking the right door. So, you choose one and there is a 33% chance of it being there and a 66% chance of it being in the other 2. But, one of the other 2 is then eliminated and shown there is nothing in it. Remember that *nothing has changed from your original probability of your guess*. It makes no sense that showing that another door is wrong somehow increases the probability of your original guess being right.

Take, for example, the game Deal or No Deal. Lets assume there are 50 different suitcases (no idea how many are actually there). If you pick a random suitcase hoping that the $1 mil will be in there, you have a 2% chance of getting the right one. Now, assume you pick #1 as the right suitcase. Remember there is a 2% chance of it having 1 mil. Now, the person who runs the game shows you that the $1mil is NOT in boxes 3-50. So, it can only be in box 1 or box 2. The action of opening up boxes known to not contain the $1 mil cannot raise the probability of you picking the 1 mil when you first picked it.

Its not intuitive and takes a while to grasp, but its definitely the case that you should switch if Monty asks you if you want to switch. Probability theory is sort of a bitch.

Nah ... I don't get it. If I understand what you are saying correctly, you are saying that decreasing the choices does not change the odds. From my POV, if you decrease the choices from a 1 out of 3 chance to a 1 out of two chance, the odds HAVE to correspondingly change from 33 to 50%. It doesn't seem logical that the odds would remain at 33% with only two choices remaining.

Not saying you're wrong. Just saying I'm not seeing the logic to the theory.