the Monty Hall puzzle...
 

On one hand, this is a very famous puzzle that you can easily do a google search for and find the answer. However, that's not really what's important. What's important is that this is a really interesting result and even if you did do a google search, you will most certainly learn something (unless you've learned this before, of course). So, here we go...

You have three doors and only one of them has a prize behind it. So, what can you do? You pick one randomly. After you pick one, Monty Hall decides to be a sport and opens a door for you so now there are only two closed doors. Of course, he's not going to open the door with a prize behind it, nor is he going to open the door you picked.
So, the question now is "should you change your guess or stick with what you picked?"

Of course, the most important question is why? Can you give a good intuitive reason for it?

I learned this one in statistics class...

One of the hardest problems to convince someone of the right answer for.

Yes. Initially you had a 33% chance of picking correctly, and a 66% chance of picking incorrectly. Now you have a 50% chance of picking correctly. You originally had more of a chance of picking wrong.
I don't know any real math behind it, but that is how my mind thinks of it.

Yups, I've seen this one before. There is an interesting variant out there though. let me see if i can find it...

I just got it. if you draw a chart it makes it alot easier. 66% getting it wrong the intuitively, 33% getting it right. If you get it wrong the first time and switch, you win. If you get it right and switch you lose. Thus, there's a 66% chance of winning if you switch and 33% chance of winning if you don't switch. You definately should swtich

I think that the missing "variant" that you are talking about is that each of the doors is an independent probability. So, doing math and saying that you were 66% and are now 50% likely to have the wrong door is incorrect.

Each door is independent. In statistics class, you'll see that there are different formulas for dependent and independent probabilities. In this case, Monty Hall opening the first door for you does not statistically change the percentage of chance that you picked the correct door in the first place.

For instance, if you flip a coin 9 times and get "heads" nine times in a roll, what is the probability of getting "heads" on the tenth time? 50%. The 9 flips beforehand statistically had nothing to do with the 10th flip. It's an independent probability.

That's how I remember it in class, anyway.

ummm what the hell are you taling about? yes both doors are independent. but if you chose to switch... if you chose the right one the first try, you lose and if you chose the wrong one the first try you win. so it doesn't matter... I don't understand why you brought in the fact that the doors are independent.. that's not part of the puzzle at all

The fact that there was a third door that was eliminated shouldn't matter in terms of what you choose the second time around. You have two doors, both have a 50% chance of being the right door. Thus, you're stuck not knowing which to choose from.

Have you been reading the posts???? it's not a 50/50.... it's a 2/3 chance of getting it if you chose to switch... the fact that you have 33% of getting it right the first try and 66% of getting it wrong. If you think about the whole picture.. if you chose the switch, you will get the prize if your first guess was wrong and if you chose to stay, you will only get it right if you guess right the first time. SINCE there is a 33% chance of getting it right, relatively, there only 1/3 of the time you chose not to switch and get it right. on the other hand, if you chose to switch, IF YOU CHOSE THE WRONG DOOR your first try, after switching, you will get it right.

In conclusion: Switching you answer in the end gives you a 2/3 chance of getting the prize. and i am sure of it. message me if you have any question...

You know what? maybe my explanation isn't clear... so let me put it another way.. I googled it so:
http://www.utia.cas.cz/vomlel/mh-puzzle.html
At first you have 33.33% chance of choosing the right door and there is 66.67% chance of the prize being somewhere else. You know that Monty is going to open an empty door so when he does, this should not change a thing about your belief of your door being the right one. You still have 33.33% chance of having selected the right door. Thus there must be 66.67% chance of the prize being somewhere else (behind the last door).

Sorry, I misread the original question. I figured the math out and it makes sense to me now.

I understand what is being said; but wow, this is confusing.

I see it like this...
If you start out with 3 doors, choose one, and Monty reveals one of the wrong doors. That, to me, is the same as you walking up there with only 2 doors to choose from - one wrong and one correct.

There were initially 3 doors to choose from; 33% of them contain the correct door, 66% contain the wrong one.

One door is eliminated.

There is now 2 doors (consider the other one gone - it doesn't matter anymore); 50% of the doors are correct, and 50% are wrong.


The way I see it -- Using the logic stated above, if before the show, there was 5 doors, and Monty decided to remove two of the wrong ones before the show - would that effect your odds? ;)

Think of this:

Say there were a billion doors. You pick one at random. Monty Hall then eliminates 999,999,998 of them, leaving you with two options.

By your logic, your chance of success is 50-50. But seriously, what are the chances that with a billion doors you guessed correctly from the beginning?

The same as if I would have chosen the other remaining door in the begining.

Artsemis, your logic is flawed in the sense that eliminating one door doesn't change the probability of your choosing in the beginnning. monty hall's elimination of one door does not affect the probability of choosing a right/wrong door with your instinct. Put it this way, if instead, after you choose a door in the beginning and monty hall present to you 5 more doors, does that make your original choice less probable? i know that's not a good exampl but you need to understand that your original decision's probability is not at all effected by the elimination later on. feel free to point out any more question though

OK, then let's try an experiment.

I am Monty Hall. We have 1000 doors, but by your logic, you still have a 50-50 chance after the elimination.

I have picked a winning door. It is somewhere between 1-1000. I have written down the answer on a post-it on my monitor.

Pick a door, any door. I will then eliminate all doors but yours and one other. One of those two will be the winning door.

I will then give you the opportunity to switch, if you like.

I give you my pledge now -- switching will give you a 99.9% chance of winning.

Of course, since you think the odds are 50-50, we can go through several trials if you like.

Pick a door, Artsemis!

Monty program......
 

I have devised a test to help us all......

This program simulates the setting with 100 doors to begin with. My results have been that I NEVER loose if I switch doors.

It is just a simple basic program with no error checking so please only enter numbers when you try it etc.

I'm posting the source because I'm not sure if I can attach an EXE in a post or not. Give me a MP and I can e-mail you an EXE to play with if you want. You may wants to mess with the logic a bit or the number of possible doors....

This is for MS Quick Basic, so adapt if needed.

Maybe someone will make a web interface for us.

COLOR 7, 1
CLS
RANDOMIZE TIMER
FOR c = 1 TO 10
door = INT(RND * 100) + 1
PRINT
PRINT
INPUT "Hello I am Monty - please type in a door number between 1 and 100:"; an
PRINT
PRINT "Thank you - you have chosen door #"; an
PRINT
PRINT "I will now elimante 98 doors leaving you with a choice of 2"
PRINT
IF an = door THEN
DO
kept = INT(RND * 99) + 1
LOOP UNTIL kept <> an
ELSE
kept = door
END IF
PRINT "I have removed 98 doors - remaining are "; kept; " and "; an; ""
PRINT
INPUT "Do you wish to switch ? [Y/N]"; yn$
IF yn$ = "Y" OR yn$ = "y" THEN
an1 = kept
kept1 = an
ELSE
an1 = an
kept1 = kept
END IF
PRINT
IF an1 = kept THEN
PRINT "CONGRADS! - you have won !!"
ELSE
PRINT "SORRY - try again :-("
END IF
NEXT c

You, sir, have a way with words.

Not sure exactly what it was you said that worked, but it just "clicked" for me when I read your post :)

Great! :)

I was worried you might just get lucky and guess the right number the first time...

Here's a way to think about it:

You choose one of three doors A B or C
Say the prize is behind C.

If you choose A, door B will be gotten rid of and if you then switch you win.
If you choose B, door A will be gotten rid of and if you then switch you win.
If you choose C, one of the other doors will be gotten rid of and if you switch you lose.

So out of three possibilities, 2 of them make you win.
So switching is the sensible option...

the right door was 78 wasnt it jeebus

Monty Hall eliminates all doors except 78 and 840

Shpoop, would you like to switch? :D You seem pretty confident with door #78....

It makes much more sense when you use the 1000 door sample instead of three. Cause, really, you know that your original door was 1/1000, and the ones they eliminated weren't the right ones, leaving only the right one and your 1/1000 door. It all makes sense now... my brain just tingled.

That's not the point... the exampl is to solve the problem.. the problem/riddle IS'NT suppose to be easy -_- most people, might fall into the logical trap but if you bother to think twice, you can tell just as well if the puzzle were about 3 or 300 doors.

For anybody who doesn't get it yet, here's how I explained it to myself:

There are n doors.

When you pick the first door, you have a 1/n chance of being right.

When the other doors are opened, the prize is no more likely to have moved to behind your door so the chance it is there is still 1/n.

The chance that is is berhind one of the two doors is simply 1.

So the chance it is behind the the other door is 1-(1/n).

since n>2; 1-(1/n)>1/n

So your odds are always improved by swapping.

The intuitive way of seeing it is that when you first choose, there are more wrong doors to choose from, but when matey boy opens a wrong door, there is one less wrong door to choose from.

Thanks. Nice puzzle.

I always loose things.. so the solution is to choose the door I didn't ;)

I like the puzzle though

This is all out the window however if Monty has the prizes on wheels and 1/2 of the time or so he moves the prizes after he has eliminated one door from the 3rd to the door the player picked.

And since this might be the case what to do what to do ????

yup, the answer is you should change your guess. in the first try, you had 33% chance of getting it right. on the second try, if you change, you now have a 50% chance. if you dont change, even though there are only 2 choices left, you still only ahve a 33% chance of being right. you can experiment if you want, get a friend, and do this 100 times and see what happens. then do it 100 times w/out changing after he shows the wrong option.

it sounds stupid, but it's true.

Of course, this is all assuming that Monty knows where the prize is...

http://www.straightdope.com/classics/a3_189.html

hah great explanation lordjeebus! It's all crystal clear in my mind now :)

i mentioned this to my girlfriend last night and she flipped out at me because i was being pigheaded and not open to the FACT that it was still a 50-50 shot no matter what i said because one of the two remaining doors was the right one. i swear, she almost broke up with me right then and there cause i refused to "allow myself to be open to the possibility" that it was, indeed, a 50/50 chance that the door was correct...

trying to explain it using the x number of doors theory didn't help matters, it actually made it worse cause all she saw was me being unwilling to see her side of the argument.

no, she's not stupid, either - her iq's higher then mine!

maybe this post is better off going in tilted relationships, huh? :)

Yeah, it's a dangerous problem because it goes against people's basic vision of how the world operates. You're better off trying to disprove their spiritual conceptualization of the universe.

I've heard that when Monty Hall's three doors made their debut, there was intense debate within the community of Statistics professors consulted to analyze the show's game.

What makes this problem tricky is the fact that Monty will always remove a LOSING door and never a WINNING door.

This can be more intuitively seen using the excellent "1000 Doors" example from above.

If we had the same thousand doors and I knew Monty was removing the other 998 losing doors, then yes, I am more likely to win by switching.

If however, Monty just threw out random doors, including possibly the winning door, (leaving another losing door to choose from), then my original guess is as good as switching (a 50-50 chance).

So in otherwords, it is the NON RANDOM removal of losing doors that makes wise to switch doors.

It should be noted however, that a 1 in 3 chance isn't much worse than a 1 in 2, so while switching may average to more wins in the long run, for one play I don't see much advantage.

this is a great way to explain it - non random vs. random removal of doors... i'll give this one a shot tonight and see if i still have a girlfriend tomorrow. :)

Try it with a deck of cards-

Tell her to pick the ace of spades from the 52 cards in the deck, then remove all the cards but the Ace and her card.

Ask her if it is still a 50/50 chance the she has an Ace of Spades in her hand.

that's a great one skier.....simple - elegant - etc...