interesting experiment on probability.

[NickTheKid 0]

Ya, I spent a long night at the cottage all boozed up trying to explain this one to some friends.The easiest way to sum it up, I think is...You'll initially pick a goat 2/3 of the time.  (two out of the three doors have goats)When you've picked a goat, and automatically switch, you'll win. (he'll show you the other empty door, and you switch to the winning door)Bottom line, switch doors and you'll win 66.7% of the time.There was a site somewhere that automatically simulated a number of guesses and solutions, as entered by the user, but I can't find it at the moment.This laughs in the face of logic... I love it!

This can't be a true answer.If you eliminate 1 false door right away, you have a 50/50 chance. How can your odds increase?.. they will eliminate 1 door after you have chosen, if you chose to switch doors everytime, your odds that switching doors will win is still 50/50.

I think I'll take my results, and the math answer than Royal Tour's. Let me explain this one more time:If you don't change doors, it's just like picking a random door and sticking with it, which is 1/3. If you switch, it HAS to be 2/3 because 3/3 - 1/3 = 2/3. It's not that difficult.

[Royal_Tour 0]

this makes the solution much easier to understand and makes it seem much more intuitive:"maybe its easier to see in this problem. there are 1000 doors, only one of which has a prize behind it. you pick a door, then monty opens 998 doors with goats behind them. do you switch? it seems more obvious in this case, because monty had to take care in which door not to open, and in the process basically showing you where the prize was (999 out of 1000 times). "

Now this makes some sense!!

[Petoria 0]

Isnt conditional probability a beautiful subject?Good times.

[r18 0]

So by switching doors after eliminating one, we're taking advantage of our 66% ability to correctly pick the wrong door. I find it fascinating that someone who walks over to guess the right door (once we're down to 2 doors) actually has worse odds of getting it right.

[Thumper 0]

This problem was given on last weeks "Numbers".The answer is that you should always change.

[BTirish 0]

Someone may have already posted the right answer, but I want to take a crack at it without looking ahead.2/3 of the time, your initial guess about the door is going to be incorrect. So, 2/3 of the time, the door left unopened and unchosen by the game show host is going to be the correct door.So, you should always switch to the door left unopened by the game show host. Right? This strategy, 2 out of 3 times, will result in your receiving the prize.

[jonnyhockey 0]

this makes the solution much easier to understand and makes it seem much more intuitive:"maybe its easier to see in this problem. there are 1000 doors, only one of which has a prize behind it. you pick a door, then monty opens 998 doors with goats behind them. do you switch? it seems more obvious in this case, because monty had to take care in which door not to open, and in the process basically showing you where the prize was (999 out of 1000 times). "

Now this makes some sense!!

Well, it's exactly the same way I explained it, on a larger scale. Whether your odds are 99.9% in favour of switching, or 66.7%... it's still above 50%.But to hell with math, it's the long weekend, and Candian is on sale for $29. Enjoy the weekend boys and girls!!!Cheers to the Queen!!!

[Dubey 1,035]

*deleted*

[vvganeshavv 4]

i read all the post before even trying the problem....i'm sitting in finance class right now and the teacher is babbling on about the cost of new or additional equity, this problem was a lot more fun!!!! Any finance majors that would like to help me w/my final email me.

[DaBruins 0]

yea i heard about this a few years ago. I'm suprised more people didnt figure this out during the running of the Monty Hall show. For a long time many mathamaticians had trouble agreeing with the line of reasoning in this answer, so dont be mad if you didnt get it before the whole 1000 door example (which is much more intuitive).

[elkang 0]

The game show would be "Let's Make a Deal".I first heard this in a psychology class though I don't remember why, since it really is a math question. I like this because it shows how math trumps "common sense".Oh yeah, I remember now... people tended to not trust the announcer intention when presented with the choice, but the choice was always given after eliminating one of the bad ones.

[augmented 0]

HAHAHAHAHAHAHAHAHAI WAS ARGUING ABOUT THIS ALL TODAY. i couldn't focus in class becasue of this question.finally, when i got home my dad explained it to me and i totally get it now. huge coincidence

[happyjuggler0 0]

About 15 or 20 years ago this was a question posed in the "ask marilyn" column in Parade Magazine that comes with an zwful lot of Sunday papers across the US. Several math professors wrote in saying ehr answer (66 2/3 % for switching) was incorrect. After a while I guess the realized the error of their ways.For me the easiest way to explain this is as follows.There are 3 doors. Oner of them has a car behind it and the other 2 have a goat each. It is totally random which one you pick. You pick door number 1. Now, the host gives you a chance to either stick with door number 1 or you can switch and get BOTH door number 2 and door number 3. Since it is random you of course pick door number 2 and 3 with a 2/3 chance of winning a car instead of a 1/3 chance of winning a car. Now the host opens one of the doors, either door 2 or 3 and shows a goat. What are the odds that you will win now? Still 2/3 of course since you learned absolutely nothing when he opened a door as he will always open a door with a goat behind it.The same odds apply when the puzzle is posed the original way. Still 2/3 to win by switching since the host will NEVER open a door with a car behind it.

[Emptyeye 0]

Hah, I've heard the answer to this before, but it didn't click with me until just now.Let me make sure I understand it. When you initially pick a door, you have a 2/3 chance of being WRONG. Monty showing you a goat means nothing; he'll always show you a goat, so your initial choice was incorrect 2/3 of the time. Thus, you want to switch.Right?

[happyjuggler0 0]

Hah, I've heard the answer to this before, but it didn't click with me until just now.Let me make sure I understand it. When you initially pick a door, you have a 2/3 chance of being WRONG. Monty showing you a goat means nothing; he'll always show you a goat, so your initial choice was incorrect 2/3 of the time. Thus, you want to switch.Right?

Right. I like the way you put it. I am not sure I could explain it as easily to a nonbeliever as you did, but the essence of understanding always seems to happen when people realize that Monty will ALWAYS show a goat.

[TheIceman05 0]

Yeah, I won 100 bucks from my math teacher in highschool because of that problem, and this one: A woman has two children, and says at least one is male. What are the odds that both are male? He got them both wrong, and started talking shit. I made him my son.Goooood luck!Ice

[happyjuggler0 0]

Hmmm, I never heard this one before. I guess I will try this one the same way I recall doing the last one...by brute force. M=Male, F=Female. The first letter is the firstborn child, the second letter is the second born child.I see four possibilities here.MMMFFMFFThere are 3 ways that the woman's statement could be true, all with equal likelyhood of occurring. Only on eof those 3 times are both kids male, so the answer is 1/3 of the time.By the way, I play backgammon so this problem is easy for me. :-) For example, what are the odds when you throw two dice (a la backgammon) that at least one of the dice is a one? To get anywhere in backgammon you need to be able to understand the answer, or at least accept it as the truth regardless of whether you understand it.

[pwoblo 0]

This probability stuff is weird. I guess you have to assume that when the babies popped out, no one knew that at least one would be male. Is it the same thing? So you have 4 permutations for the way the babies could have popped out: male male, male female, female male, female female. Out of these, only three have at least one male: MM, FM, MF. And only one of that set is MM, so...Is 1/3 right?

[happyjuggler0 0]

Hmmm, I never heard this one before. I guess I will try this one the same way I recall doing the last one...by brute force.  M=Male, F=Female. The first letter is the firstborn child, the second letter is the second born child.I see four possibilities here.MMMFFMFFThere are 3 ways that the woman's statement could be true, all with equal likelyhood of occurring. Only on eof those 3 times are both kids male, so the answer is 1/3 of the time.By the way, I play backgammon so this problem is easy for me. :-)  For example, what are the odds when you throw two dice (a la backgammon)  that at least one of the dice is a one? To get anywhere in backgammon you need to be able to understand the answer, or at least accept it as the truth regardless of whether you understand it.

Oh yeah, sorry I forgot to include this part. The answer would be different in China. :shock: My understanding of China's child policy is as follows: One child per couple, but they allow you to try for a second child if the first is not a boy. If this is true of course then the answer to the puzzle is pretty close to 0%since they would be breaking the law otherwise. Of course "accidents" happen....

[TheIceman05 0]

The problem really isn't even difficult, and everyone who has responded so far has been right on. But my Honors Alegbra II teacher in highschool said he was disappointed with me when I told him he was wrong (he said 1/2) about this one. Very simple, but kinda counter-intuitive anyway. If we knew the first child was male, the answer would be 50/50

[jgarza 0]

without reading anything other then the question:1/3 of the time u changed u had the right door at first pick. So its zero percent that you pick the right door. 2/3 of the time the right door is in the other two. So, all of that 2/3 of the time you get the right door. So, when you change doors, 1/3 the time you will never pick the right one, and 2/3 the time you will pick the right door all the time. So, you should pick the right door when you change doors 2/3 the time, right?-Jeff Garza

[pwoblo 0]

So unfortunately, you're playing russian roulette... New rules: there are 2 bullets in the gun, and you must pull the trigger twice.The 2 bullets go into the six-chambered revolver, and they are placed in successive slots. You spin the chamber, put the gun to your head, and pull the trigger. Nothing. A blank. Before you pull the trigger one last time, you have the option of either spinning the chamber, or you have the option of simply leaving the chamber the way it is. What should you do?

[sanemancrazywrld 0]

So unfortunately, you're playing russian roulette... New rules: there are 2 bullets in the gun, and you must pull the trigger twice.The 2 bullets go into the six-chambered revolver, and they are placed in successive slots.  You spin the chamber, put the gun to your head, and pull the trigger.  Nothing.  A blank.  Before you pull the trigger one last time, you have the option of either spinning the chamber, or you have the option of simply leaving the chamber the way it is.  What should you do?

I spin it again. The greater chance of dying with give me the bigger rush, and I'll be even more psyched to live if I get lucky.

[JimD82 0]

So unfortunately, you're playing russian roulette... New rules: there are 2 bullets in the gun, and you must pull the trigger twice.The 2 bullets go into the six-chambered revolver, and they are placed in successive slots.  You spin the chamber, put the gun to your head, and pull the trigger.  Nothing.  A blank.  Before you pull the trigger one last time, you have the option of either spinning the chamber, or you have the option of simply leaving the chamber the way it is.  What should you do?

Don't spin. If you spin you have a 2/6 or 1/3 chance of dyingYou will only kill yourself 1/4 of the time if you pull the trigger again. Draw a picture and it becomes clear.

[WatersR 0]

maybe as a student of game theory, i'm biased, but the original monty hall is trivial. what is interesting is if monty only gives you this option SOMETIMES.assume for the following that all parties are rational and pretty much have complete information about how this has run in the past.firstly (a) if he opens a door on random occasions, it is always correct to switch, as the analysis is the same as has been mentioned before.B) if he ONLY allows to you switch when you've chosen wrongly or correctly, it is trivial to decide what to do, so we can safely ignore that too.c) let's, for example, say that 2/3 of the time that you've chosen correctly, and 1/3 of the time that you've chosen incorrectly, he gives you the option to switch. 2/9 - you will choose correctly AND be offered the chance to choose. 1/3 * 2/32/9 - of the time you will have chosen incorrectly and be offered the chance. 2/3*1/3 so, you will be offered a chance to switch 4 times in 9, and only half the time will you be making a good decision.for (perhaps) obvious reasons, it follows (from the same premise) that as long as he offers you the chance to switch twice as often if he KNOWS that you are correct, then you're playing a cointoss. the limit is of course:Every time you are correct, and half the time you're incorrect.You get the choice 2/3 of the time - but you're sitting on a 50/50. Any ratio greater than 2/1 and you shouldn't switch, and any less and you should continue to switch.Cars don't grow on trees.