Is anyone familiar with The Monty Hall Problem?

[GimpyIntellect]

Monty Hall from Let's Make a Deal fame.

He offers a contestant three doors, with hidden prizes behind each one.

Behind one of the doors is a brand-new car.

Behind the other two doors is a goat.

Obviously the object is to randomly pick the car.

Monty knows in advance which door is hiding the car.

Hypothetically, let's say the contestant chooses door #2. Monty then reveals to the contestant that hiding behind door #1 is a goat. He then offers the contestant the ability to either keep door #2, or switch their choice to door #3.

If you were the contestant, would you switch doors?

 

[rocket_surgery]

Monty Hall from Let's Make a Deal fame.

He offers a contestant three doors, with hidden prizes behind each one.

Behind one of the doors is a brand-new car.

Behind the other two doors is a goat.

Obviously the object is to randomly pick the car.

Monty knows in advance which door is hiding the car.

Hypothetically, let's say the contestant chooses door #2. Monty then reveals to the contestant that hiding behind door #1 is a goat. He then offers the contestant the ability to either keep door #2, or switch their choice to door #3.

If you were the contestant, would you switch doors?

This his always been an interesting problem for me even though I don't claim to know what is really the correct answer if there even is one.
I seem to remember some group figured it out in a manner they claimed was correct but even that was disputed.
There is also the possibility that Hall will steer a person towards a certain door without realizing it simply because he knows ahead of time which one is right.

 

[Magicscreen2]

Switch.

 

[LeinadVirgo]

Always switch. 50% is better that 33%

 

[VaticanAssassin]

The biggest problem Monty was how to watch the hour or two that is funny without having to watch the 973 hours that aren't.

Doesn't matter if you switch or not.

 

[Colonel Hogan]

Initially the contestant had a 33.33333333....% chance of selecting the door concealing the car.

Upon revelation that one of the unselected doors concealed the goat (something that would have OBVIOUSLY been true no matter which door the contestant selected), we now know that the contestant has a 50% chance of getting the car based on the two doors remaining. Again, that "new, improved" odds of success has nothing to do with the contestant's actions, but only with the elimination of a single unsatisfactory result.

Assuming Monty ALWAYS offers the option to switch doors after the initial selection, Monty's OFFER is irrelevant to the remaining mathematical odds for a favorable outcome. There is no reason to switch, nor is there a rational argument to stand pat. The odds remain at 50/50 no matter what the contestant decides.

Having said all that, it seems I have heard from the past that there IS a rational reason to switch. But I'm damned if I understand it if that's true.

 

[GimpyIntellect]

The correct answer is to switch.

https://www.youtube.com/watch?v=mhlc7peGlGg

 

[KingOrfeo]

This is the correct answer.

 

[Hard_Rom]

Lets bring down our next contestant... Bob Bitchin!

 

[Sean]

Doesn't matter if you switch or not.

Don't you claim to be an actuary or something?

 

[Wrong Element]

Initially the contestant had a 33.33333333....% chance of selecting the door concealing the car.

Upon revelation that one of the unselected doors concealed the goat (something that would have OBVIOUSLY been true no matter which door the contestant selected), we now know that the contestant has a 50% chance of getting the car based on the two doors remaining. Again, that "new, improved" odds of success has nothing to do with the contestant's actions, but only with the elimination of a single unsatisfactory result.

Assuming Monty ALWAYS offers the option to switch doors after the initial selection, Monty's OFFER is irrelevant to the remaining mathematical odds for a favorable outcome. There is no reason to switch, nor is there a rational argument to stand pat. The odds remain at 50/50 no matter what the contestant decides.

Having said all that, it seems I have heard from the past that there IS a rational reason to switch. But I'm damned if I understand it if that's true.

The "correct" answer, apparently, is switch ... but I see it the way you do. Either the car is behind the door you chose, or it's behind the other door.

 

[Sean]

Initially the contestant had a 33.33333333....% chance of selecting the door concealing the car.

Upon revelation that one of the unselected doors concealed the goat (something that would have OBVIOUSLY been true no matter which door the contestant selected), we now know that the contestant has a 50% chance of getting the car based on the two doors remaining. Again, that "new, improved" odds of success has nothing to do with the contestant's actions, but only with the elimination of a single unsatisfactory result.

Assuming Monty ALWAYS offers the option to switch doors after the initial selection, Monty's OFFER is irrelevant to the remaining mathematical odds for a favorable outcome. There is no reason to switch, nor is there a rational argument to stand pat. The odds remain at 50/50 no matter what the contestant decides.

Wrong.

 

[Saint Peter]

I would rather have a goat than a shitty car.

 

[Colonel Hogan]

The "correct" answer, apparently, is switch ... but I see it the way you do. Either the car is behind the door you chose, or it's behind the other door.

Wrong.

I get it now, but it truly IS counterintuitive. The key is you cannot "divorce" the probabilities of your second chance from your first selection and the information you receive by revealing the location of one goat.

IF you switch, you WILL win the car 66% of the time.

That show is still on, right? Maybe I should work up a goofy costume and get in the audience. Hmmmm.

 

[Saint Peter]

I will give you a hundred dollars if you have a goat in your purse.

 

[Parminderjeet]

I will give you a hundred dollars if you have a goat in your purse.

Now you're just asking for hashtag to show up.

 

[Colonel Hogan]

I will give you a hundred dollars if you have a goat in your purse.

Uhhh.....okay, okay, I got this.....so then I switch to the garage after your girlfriend walks out of the bedroom, right?

 

[quietlylooking]

This is the correct answer.

The simple explanation here is very good. I think I may finally understand the correct answer, but I will need to try to explain it to my husband to see if I've really got it.

I love xkcd.

 

[Conager]

Doesn't matter if you switch or not.

Exactly. You may have had a 33% chance to start, but the remaining odds are 50% if you switch, and 50% if you stand pat.

 

[Conager]

I get it now, but it truly IS counterintuitive. The key is you cannot "divorce" the probabilities of your second chance from your first selection and the information you receive by revealing the location of one goat.

IF you switch, you WILL win the car 66% of the time.

That show is still on, right? Maybe I should work up a goofy costume and get in the audience. Hmmmm.

No, Sean is wrong, unless Monty only offers the switch under certain conditions. If he offers the switch after he reveals the sacrificial goat every time, he had either one or two goats to choose from. One, and only one if you initially chose a goat, and either of the non-chosen doors if you chose the car. Nothing about the reveal changes anything about the remaining odds of whether you got lucky when the ods were 1 out of 3 or not. 50% you were lucky, 50% ypu were not.

Once the reveal has happened, choosing again is an absolutely independant event. Stand pat, or switch, and each choice is equally valid.

 

[Colonel Hogan]

No, Sean is wrong, unless Monty only offers the switch under certain conditions. If he offers the switch after he reveals the sacrificial goat every time, he had either one or two goats to choose from. One, and only one if you initially chose a goat, and either of the non-chosen doors if you chose the car. Nothing about the reveal changes anything about the remaining odds of whether you got lucky when the ods were 1 out of 3 or not. 50% you were lucky, 50% ypu were not.

Once the reveal has happened, choosing again is an absolutely independant event. Stand pat, or switch, and each choice is equally valid.

The phrase "the remaining odds" is where your and my initial logic fails. If you are choosing between two unknown and equally probably results, the odds ARE always 50/50. But you were initially choosing between THREE unknown results. And the odds associated with that first choice DO have a probability impact on what you SHOULD do, given a second chance.

Also, the second choice is NOT an "independent event." It would be if there was a chance the prizes behind the doors were switched AGAIN, but they weren't. Since they remain in the same position as when you first selected, THAT is part of the material consideration that affects the odds related to your second choice.

You absolutely DO increase your chances of winning from 50% to 66%.

 

[Conager]

The phrase "the remaining odds" is where your and my initial logic fails. If you are choosing between two unknown and equally probably results, the odds ARE always 50/50. But you were initially choosing between THREE unknown results. And the odds associated with that first choice DO have a probability impact on what you SHOULD do, given a second chance.

Also, the second choice is NOT an "independent event." It would be if there was a chance the prizes behind the doors were switched AGAIN, but they weren't. Since they remain in the same position as when you first selected, THAT is part of the material consideration that affects the odds related to your second choice.

You absolutely DO increase your chances of winning from 50% to 66%.

Nope.

Because the offer does not change the remaining reality that you were either right, or you were not.

You are right that contestants from a standing start will win 66% of the time, but that applies equally to those who did switch, versus those that did not switch.

There is no hold-over voodoo left on your door because initially it was longer odds. As of right now, there are only two possibilities. You were lucky or you weren't. The only thing the reveal told you was what you already knew was going to happen was that you're going to be shown one of the doors that you didn't choose, and did not have a car.

Same problem as saying you flipped a coin twice and it came up heads both times should you pick heads or tails for the third time?

Working the problem from the outset and calculating odds will tell you that those that pick heads should win 66% of the time because the odds of tbe third flip (from the outset) being tails thrice is low. However when you were actually making that third choice the fact that something else happened before is completely irrelevant.

It's a red herring. The coin and the doors do not care what happened before.

 

[astuffedshirt_perv]

Consider a slightly different version of the problem: there are 100 doors. You pick one at random. Monty opens one door, showing no prize. He repeats this 97 more times, all revealing no prize. Now there are only 2 doors, one you picked with a 1% chance of being correct and a second door, which has a 50% chance of being correct. Do you switch? The answer is yes.

Now reduce the number of doors to the minimum--three--and we have the classic Monty Hall problem. You switch. This problem is a wonderful way to get students, math teachers, lay people all wound up for no particular reason, since the show is off the air.

 

[Super-Surprise]

http://sharenxs.com/images/wz/cabc/-b/ai/ac/cb/gdelaurentiis360.gif?1489701547

 

[Colonel Hogan]

Nope.

Because the offer does not change the remaining reality that you were either right, or you were not.

You are right that contestants from a standing start will win 66% of the time, but that applies equally to those who did switch, versus those that did not switch.

There is no hold-over voodoo left on your door because initially it was longer odds. As of right now, there are only two possibilities. You were lucky or you weren't. The only thing the reveal told you was what you already knew was going to happen was that you're going to be shown one of the doors that you didn't choose, and did not have a car.

Same problem as saying you flipped a coin twice and it came up heads both times should you pick heads or tails for the third time?

Working the problem from the outset and calculating odds will tell you that those that pick heads should win 66% of the time because the odds of tbe third flip (from the outset) being tails thrice is low. However when you were actually making that third choice the fact that something else happened before is completely irrelevant.

It's a red herring. The coin and the doors do not care what happened before.

IT IS NOT THE "SAME PROBLEM" as choosing heads or tails based on the result of prior coin flips. Each coin flip produces an INDEPENDENT result. In the 'Hall' problem the coins have ALREADY BEEN FLIPPED AND THE COINS MERELY COVERED UP -- AND YOU KNOW THAT THE RESULT IS TWO TAILS AND ONE HEAD. And you are trying to determine the location of those results -- first with no location information and then with partial location information.

If you don't understand that that is dramatically different probability wise than the premise you are constructing, I can't help you.

In any event, the probability experts say it matters, and since I suck at math, I'm going with them. If you can produce online academic references that persuasively argue otherwise, I'll reconsider.