The Monty Hall Problem Explained

http://www.philipbrocoum.com/?page_id=91

The Monty Hall Problem Explained! ...

I would like to add another very simple way of looking at this problem from Edward on this thread: http://groups.google.com/group/leo-editor/browse_thread/thread/be6d8b031f6842...

If you don't switch, the only way you can *win* is if the car is behind your door. If you switch, the only way you can *lose* is if the car is behind your door. There is a 1/3 probability that the car is behind any particular door, so switching *doubles* your odds of winning. This is a plain fact.

Monty knows where the car is. His showing you an empty door in no way alters the odds.

The analogy with picking a card is exact. If you switch, your odds of winning are 51/52. That is, the only way you can lose is if you *correctly picked* the ace of spades originally. The person showing you 50 cards **cannot change this fact**. If you don't switch, your odds of winning are 1/52. If yo switch, the probability *must* be 51/52.

Similarly, Monty can not change the fact that if you switch the only way you can lose is if you initially picked the right door. The probability of that is 1/3, so the probability of winning is 2/3 if you switch.

Edward

Category:

Education

Tags:

• Monty
• Hall
• problem
• puzzle
• riddle
• probability
• mathematics
• three
• doors
• math

[SandBarAndYou]

The whole point of conditional probability is the understanding that knowledge of an event changes the probability distribution of the event. Therefore to use an argument that says "the probability that THESE doors had the prize at the start was X and so it should STILL be X" is not explaining anything. This argument is refusing to allow the distribution to CHANGE with knowledge. And it CAN change with knowledge.

[Stedwick]

@SandBarAndYou Well, the odds do change… with more information, it becomes apparent that switching is better, which wasn't apparent before. Please see my extended description on this video by clicking "show more" above for more details.

[AsbjornOlling]

I am only 14 years old, and I have two uncles (<50yrs) who studies mathematics at the university of Copenhagen. So they think they're pretty damn good at stuff like this.

About a year ago, I discovered the Monty Hall problem, and told them about, thinking they might find it interesting. A year later, I am still trying to prove to them how it works, and they refuse to believe it.

How do I show these old ignorant fools that I am right?

[Stedwick]

@AsbjornOlling I'm not sure, this is a pretty standard problem in mathematics circles, so anyone with a degree in mathematics should be very well familiar with this problem. The reason it's famous is not because mathematicians had trouble with it, but because the general public finds it so fascinating. Maybe your uncles never took probability? Maybe 50 years ago the puzzle wasn't famous so they never learned it? I don't know.

[henryporter101]

Good video,however,when you say"Since he's telling you which door is empty,he might as well be giving you both doors,since you wont mistakenly walk away with the empty one ".No,Monty can only tell you that 'a' door is empty.If he gave information that revealed that two doors are empty,then one would be able to get the car every time,instead of 2 every 3 times.Great video,but I don't think the analogy holds.

[Stedwick]

@henryporter101 He opens a door and it's empty. What more do you want?

[XxMaplerzxX]

DAYUM this was a great explanation. great monotone voice but easy to understand thnx

[henryporter101]

@Johnnyradionic.In original game,Monty can't show the car door-even if selected correctly.You have a choice,either you 'guessed' it,which has 1/3 probability or the door he's concealing has it.By switching,you go from 1/3 guess to an informed 'either/or' which makes it 2/3 probability.In your scenario,since 2 doors are selected,Monty is forced to tell the scholars that one of them 'guessed' correctly,so they may as well stick to their original guess which was 1/ 3 and became 1/2

[SandBarAndYou]

@Johnnyradionic No, at that point, the probability of one contestant's door and the others is the same.

[Johnnyradionic]

What if there were TWO contestants and each of these two contestants were ardent mathematicians, both with an innate understanding of the principle behind the Hall paradox. After each choose a door, and the third unchosen door was opened to reveal a zonk, would each of the two players, if given the chance, leap at the opportunity to accept the other fellow's door? (And in this scenario the 2 scholars are rivals who would not agree to sell the car and split the proceeds) Hmm?

~ Johnny Radionic ™

[CapumOfOuttaSpace]

I thought you were absolutely wrong until you got to the part about 'my hand' vs 'your hand'.. then it clicked and I understand now. Great explanation!

[REDTEAM22003]

12 people need to learn math

[SandBarAndYou]

@Stedwick Let's say the card scenario is changed slightly so that you pick a card initially but the other guy does not get to know which card you picked. Next he flips one card over and it's not the one and fortunately it's also not your card. The chance of your card being the one now becomes 1/51. It WOULD be at 1/52 had he known your initial pick and PROMISED not to flip it over(i.e original scenario). I know why this is true. Do you?

[SandBarAndYou]

Everything said before 4:10 was beautiful. You explained that the reason why you should gain no confidence in your initial pick is because the other guy is SUSPICIOUSLY leaving other cards face down. You are then correctly implying that there is nothing SUSPICIOUS about why he left YOURS face down. He HAS to leave yours alone simply to follow the game rules. Throughout the entire game you have no reason to EVER gain confidence in your pick. After 4:10 you go off the rails.