Monty Hall Problem (mathematics)

[_Bat_]

Anyone heard of this before? It's a mathematical probability puzzle which is well known for confusing people when they are told the solution because it is completely counterintuitive.

 

The problem is as follows:

 

"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. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you "Do you want to switch to Door Number 2?" Is it to your advantage to change your choice?

 

Note that the player may initially choose any of the three doors (not just Door 1), that the host opens a different door revealing a goat (not necessarily Door 3), and that he gives the player a second choice between the two remaining unopened doors. It is assumed that the player is trying to win the car."

 

The answer is in fact yes, you double the probability of you getting the correct door (from 1/3 to 2/3) if you switch your choice. I, like most others, initially refused to accept this and it took me a few minutes before I could see why the chances are greatly increased if you swap your choice.

 

I think this is rather amusing by just how many people refuse to accept the reality of the situation.

[Jake]

That is indeed confusing...

 

Surely once one door is open, there are 2 remaining doors, which means its a 50/50 chance of picking the right door. Surely now, whether I pick door 1 or door 2, its going to be a 50/50 chance regardless of which of the 2 doors I pick.

 

Changing my choice from door 1 to door 2 isnt going to change the 50/50 chance surely. 50/50 is better than the original 1 in 3, but at this stage, one door is already open so the remaining choice is 50/50 regardless of which door I pick out of the two, so surely there isnt an advantage at that point in changing my choice??

[OutToLunch]

I was with Jake (and 10,000 other Parade readers apparently) - Monty Hall problem - Wikipedia, the free encyclopedia - but after reading that, my head is about to explode.

[Diello]

Remember my father teaching me that many years ago (a Mathematic man!) and being amazed.

[tom_newton]

The way I figure:

If you stick, the chance is one in three - ie the chance of you picking the car under normal situations.

If you change, and you *hadn't* picked the car in the first place, you would *definitely* get the car, because you and monty would have selected both goats (monty cant pick your goat door, so he eliminates the last one). So now the chance to win becomes the chance of you *not* picking the car in the first place (which of course is not changed by the fact you were trying to) so is 2/3.

 

Having said that, you might want a goat. You cant milk a car.

[SYNACK]

They showed this concept in an episode of Numb3rs cant remember the episode name but overall quite an interesting show.

[Jake]

See thats what I cant understand...

 

Now that we know for certain that Door 3 is a goat, then surely its no longer 1 in 3 if you stick, surely its now 1 in 2, as Door 3 has effectively been completely eliminated from the equation.

 

And so its a 1 in 2 chance regardless of whether you change or not. (The third door having now been opened and having nothing to do with anything any more)

 

In my head its like in football, if Arsenal , Man U and Chelsea are all on the same points then its a 1 in 3 if I say Arsenal will win the title, but if Chelsea then lose and fall behind and cant catch up(like opening door 3), they are now eliminated from the title race and Arsenal are now a 1 in 2 chance of winning the title (the other being Man U).

 

I guess in my head I just cant see why 3s have anything to do with it when there are only 2 choices (Door 1 and 2, door 3 having now been rendered irrelevant as we know for a fact its a goat).

 

Then again I am sure half these things are designed to make some stuffy bearded virgin in some basement feel superior to mere mortals anyway

[eduabncs]

I'ts actally quite simple maths & logic (I am c**p at maths, but quite good with logic ) .

 

In the 1st instance you have a 1 in 3 chance that the door you select is the one with the car behind it. This is actually quite low odds in the context if you think about it!

 

Now that Monty has opened one door and it has a goat behind it you odds go to 2 in 3 if you switch & not 1 in 2 as most people assume, so by changing your door choice you actually increase the likelyhood of picking the door with the car behind.

 

If look at the odds of each choice, from the original door choice right through to either winning or losing the car your overall odds are 1 in 6 for each outcome. So again by changing doors you improve the odds.

 

The Wikipedia link sums it all up quite well even if it does go on a bit!

 

Yer heads still hurt..!

Edited July 15, 2008 by eduabncs

[maniac]

If you've ever watched the beginning of the film 21, this problem is the one described in the lecture that features near the beginning of the film, which got me intrigued enough to research it on the net. lol

 

It's a very well know and well discussed problem, my father was even given this problem as part of his 'A' level maths course in the 60's, although it wasn't in the context of a game show.

 

Mike.

[Jake]

Nope, I still dont get it

 

My mind must just be no good at it.

 

The moment the goat is shown behind door 3, my head tells me that it is no longer a question of 1 in 3, or 2 in 3, or 3 in 3. It simply tells me that the car is behind 1 of 2 doors. (the discounted door with the goat simply not being a factor anymore)...so my head just tells me that the car is either behind door 1 , or door 2. So its a 50/50 choice. I cant include door 3 as a probability/possibility because I now know for a fact that it has no car.

 

I guess I just will never get it. (though it does still sound like something beardy people use to show their "cleverness" off, the mathematical equivalent of pointing out a double negative in someones sentence )

[Jake]

Ah, hang on...I think I see why I am saying 50/50

 

I think its down to this quote from that wiki page

 

"The player's probability of winning by switching increases to 2/3 in the original problem because in the two cases above where the host would reveal the car, he is forced to reveal the remaining goat instead"

 

So, if I am reading that correctly, its ONLY 2/3 IF the host would reveal the car and is forced to reveal the remaining goat.

 

Whereas the way I was reading that original quote, was that the host has not revealed a car initially, but has ONLY revealed a goat. Which if I am reading that wiki quote correctly would mean that the odds would indeed be 50/50 because the 2/3 probability isnt triggered unless the host reveals a car and has to open an alternative door showing a goat instead.

 

Which is indeed listed on that wiki page in the little table where it says

 

"The host does not know what lies behind the doors, and opens one at random without revealing the car (Granberg and Brown, 1995:712).

=

Switching wins the car half of the time."

 

Which of course is the 50/50 I have come to.

Edited July 15, 2008 by Jake

[EduTech]

Ok i dont get it :|

 

Never been any good at maths.. never mind puzzles

 

[eduabncs]

This diagram from the wikipedia page explains it best.

 

http://upload.wikimedia.org/wikipedia/commons/thumb/d/de/Monty_tree_door1.svg/555px-

[SeanVin]

I agree with Jake. It is similar to the coin tossing misconception. The probability of throwing two heads in a row is 1 in 4. Now suppose the first coin is a head, people assume the chances of getting the second head is still 1 in 4, however there can only be two possible outcomes for the second throw, therefore the chances of getting two head in a row is now 1 in 2. The odds have changes because you know the outcome of the first throw.

 

When Monty opens a door you will know the position of one of the goats, at this point there are only two possible outcomes, the odds have changed and you now have a 1 in 2 chance of getting the car, it doesn’t matter if you change your mind.

Edited July 16, 2008 by SeanVin

[SeanVin]

Here’s another maths problem that is counter intuitive. Suppose you drive one lap of a 1 mile race track at an average speed 60mph. How fast would you have to drive on your second lap to complete both laps at an average speed of 120 mph?

[tom_newton]

There's definitely no 50:50 odds - because whilst you are down to one car, one goat, the chance of the door you are switching to being a goat is reduced, because in two out of 3 situations (ie those wher eyou haven't picked the car in the 1st place - doesn't matter that you don't know) Monty (with knowledge) has explicitly avoided that door because its a car.

 

So, two thirds of the time, you will pick Monty's avoided car. One third of the time, you will already have the car, and you'll pick monty's discarded random goat.

 

I think this problem is skewed by the fact that the first 1-in-3 random choice, people believe they might have the car, and switching is intuitively disadvantageous.

[Jake]

So in effect, its only a 2/3 IF the host knows where the car is. If the host doesnt know where the car is, then its 50/50 as per the wikipedia table

[JJonas]

Having said that, you might want a goat. You cant milk a car.

 

with the rate milk is going up the goat will probably have better resale value

[ConTheITGuy]

Here’s another maths problem that is counter intuitive. Suppose you drive one lap of a 1 mile race track at an average speed 60mph. How fast would you have to drive on your second lap to complete both laps at an average speed of 120 mph?

 

180?

 

And, in the original problem, the host does know where the car is - he will always show one of the remaining goats. I first came across this in "The Curious Incident of the Dog int he Nighttime", quite a nice novel.

 

[_Bat_]

The best way to put it (in my opinion) is, pretend that there are 100 doors. You choose one, then monty hall gets rid of all the other doors except the one you chose and one other, one of which has to be the door with the car behind it. You can see there that it would be in your best interest to swap your choice because the door you chose has much less chance of being the correct door then the one that monty hall leaves you with.

 

That is effectively what happens, even though there are only three doors

[SeanVin]

That is a very good expination Bat looks like I was completely wrong. Having had a little think it would appear you are actually betting on what Monty is left with after your first choice, there is a 2/3 chance of him being left with a goat and a car in which case he must leave the car.

 

Andy the answer to the race track problem is not 180mph!

[somabc]

Trick question my friend. Can't be done. It's all down to time, and you will have none left after the first lap.

 

edit: I was wrong it's not 240mph

 

LaTeX Code: \\frac {\\bar v}{2L} = \\frac {2 v_1}{2L} = t_1

Edited July 16, 2008 by somabc

[SeanVin]

Trick question my friend. Can't be done. It's all down to time, and you will have none left after the first lap.

 

Your absolutely right, it would take you a minute to do two laps at an average speed of 120Mph but it already took you a minute to do the first lap. Most people come up with 180mph reasoning that 180+60 is 240 and dividing it by the two laps giving 120mph. What they have done is taken the average of two averages.

 

In my RAF days we used to produce monthly statistics for the number of engines damaged by foreign objects (FOD, a significant problem for jet engines). At the end of the year the management produced their annual FOD report with an exceptionally low number of incidences per 1000/flying hrs. It looked all wrong to us, we had calculated a much higher figure. It took us a while to work out what was going on, but as you can probably guess, they had averaged out our monthly figures that were themselves averages. We pointed this out and gave them the correct data. We were of course totally ignored, it wouldn’t have reflected well on them if they had published the higher figure.