Digg

The Digg Crew wants to hear your thoughts!
Please take our short survey about Digg and potential feature ideas.

The Monty Hall Game Show Problem watch!

5min.com — Find out how you can tell which object is behind each door in a game show, to give you the greatest chance of winning the prize.

Bury

fabiocoil 360 days ago, made popular 359 days ago

563 diggs
digg it

166 Comments
Who Dugg It?

hpp00nst3r, on 11/29/2007, -27/+0how weird
this is exactly wut i wuz learnin today =/
- GibZilla, on 11/29/2007, -3/+35You should have gone to english instead.
  - MalTheObscure, on 11/30/2007, -0/+7I was hoping that was what this comment said when I expanded it
gregulate, on 11/29/2007, -27/+1im sure this is really interesting but it lost my attention after about 3 mins
VanD, on 11/29/2007, -3/+4Very true.
Too bad there is no game shows like this anymore.
They only time you would be asked a question like this is in a grade 8 math class.
- hammerpants, on 11/29/2007, -3/+10Deal or no deal is the closest correlation I can think of.
  - pahoehoe, on 11/29/2007, -1/+3The rules are different though, so the problem's analysis is different.
  - Zique, on 11/29/2007, -2/+6The game show host in Deal or no deal doesn't know what's behind the cases, so this doesn't apply though.
    - ostracize, on 11/30/2007, -0/+1The guy who makes the offers does though.
Majorkerina, on 11/29/2007, -2/+12Getting another chance to pick always improves your odds, it seems. Still somewhat confusing though.
- RpgActioN, on 11/29/2007, -30/+4It's confusing because it's inaccurate and completely wrong. He pulled the 33% goat figure out of his ass for the "swap percentage calculation."
  - IIGrudge, on 11/29/2007, -2/+17If you ALWAYS choose to swap, you will win if you pick the goat first. The chance of picking the goat first is 2/3. Therefore, the chance of winning the car is 2/3 = 66.66%. The chance of winning the goat is then 33.33%
    - RpgActioN, on 11/30/2007, -4/+2It really is unsurprising that you're getting dugg up, considering the general intellectual level of the digg community.
      
      BTW: you're wrong.
      - Gagle, on 11/30/2007, -1/+1true that. The problem is more subtle than that......
      - sgtbutterscotch, on 12/01/2007, -0/+2I dare you to refute him using mathematics.
  - Guvante, on 11/29/2007, -1/+16His final explanation is simplist, if you pick a goat and swap, you win a car, if you pick the car and swap you win a goat.
    
    Since you have a 66% chance of picking a goat (2/3) you therefore have a 66% chance of getting a car by switching.
  - bitweever, on 11/29/2007, -1/+16My favorite explanation of this was imagining there were 100 doors. You pick one, and he opens 98 of the other doors. What moron out there wouldn't swap then?
    - Jrr6415sun, on 11/29/2007, -2/+3that makes it a lot easier to understand, it's not as obvious when only looking at 3 doors.
  - RpgActioN, on 11/30/2007, -6/+1You people are all assuming some ridiculous before-the-fact chance factor that makes no sense when you break it down. Anybody who dugg me down is an idiot.
    
    What it comes down to: You have two doors, and one open door with a goat behind it. You have a 50/50 chance. If you don't think so, you are stupid. It is absolutely retarded to consider the door that has already been opened as part of the equation, because it has no bearing on the chances of the later two doors after it has been eliminated.
    
    The logic "well, you have a 66% chance that you picked a goat, so switching gives you 66% chance of winning a car" is the most idiotic thing that I have ever heard. The ONLY circumstance under which this would be true is if one of the goats magically became a ***** Corvette.
    
    Maybe it's no hope trying to explain this to you without making a colorful flash video to hold your attention for longer than 10 seconds.
    - sancho, on 11/30/2007, -0/+1It's a really confusing problem, so your error is understandable. Try reading through the wiki article on the problem and see if that helps you make sense of it.
      http://en.wikipedia.org/wiki/Monty_Hall_problem
      It's also pretty trivial to write a Perl script to simulate the problem. The odds come out correct--66% of the time, if you switch, you win.
    - sancho, on 11/30/2007, -1/+1Here's some overly long (but should be easy to follow) code that demonstrates the principle: http://pastebin.org/9931
    - apeweek, on 11/30/2007, -0/+3Save yourself some embarrassment before you call everybody else idiots. Get some paper cups and a coin and actually do this a few times. Record the percentages of wins/losses.
      
      BTW I' love this puzzle. I've won money from 'idiots' doing this.
    - Koookie, on 11/30/2007, -1/+2Am I replying to a troll? Anyway, a simple question:
      How do you explain that all the simulations show that after swapping you win roughly 67% of the time? (not 66% as the video falsely says.)
      
      Like apeweek says, try it yourself. Really.
- kevludlow, on 11/30/2007, -1/+1...it's not the additional chance to pick that improves your odds, it the information you are given.
  
  You can swap doors all day long and still only have a 33% chance of winning, but as soon as you're shown a loser, the probability changes.
Strider817, on 11/29/2007, -1/+5The Voice Actor reminds me of one of the VA's from Oblivion, anyone else notice that?
- oakeybloke, on 11/30/2007, -0/+1Agreed, I would be most happy to close shut the jaws of Monty Hall with my Umbra sword.
strictnein, on 11/29/2007, -2/+14some simulators if you want to try this out:
http://math.ucsd.edu/~crypto/Monty/monty.html
Only works in IE: http://www.grand-illusions.com/simulator/montysim. ...
chirox901, on 11/29/2007, -2/+18Heh. This problem once confused a great deal of college educated mathematicians.

Which, coincidentally, is something I learned in English class.
- unearth, on 11/29/2007, -4/+4I don't know if this has been written about anywhere, but it seems obvious that this is a psychological result of people assuming that they are especially talented at guessing things, or lucky.
  
  The proof for a Monty Hall problem of n number of doors for n>2 relies on the ***** of your initial guess. People don't want to believe they are ***** guessers even when the odds are highly stacked against them.
  
  That explains why contestants don't switch doors, but I'm not sure why a mathematician trying to decipher the problem can't wrap his head around it.
- CannibalTom, on 11/30/2007, -1/+1I learned this in elementary school. I don't believe it confused mathematicians at all.
- usbcd36, on 11/30/2007, -0/+4I had to read The Curious Incident of the Dog in the Night Time for English, and that's where I first heard of this problem.
  - chirox901, on 12/01/2007, -0/+2Yep, that's the one I read.
Zoshchenko, on 11/29/2007, -37/+12No..no..no...the logic presented in this video is totally wrong! It completely ignores the starting state of the problem and makes the assumption that the odds change with time. They do not. It is totally flawed thinking and you will certainly go insane if you dwell on it. It makes NO DIFFERENCE if you swap or not. None. Your chances remain identical. But you may start talking with a weird, soft British accent. There's always a danger of THAT!
- chirox901, on 11/29/2007, -1/+2I started doing that years ago. You should try it some time.
- emiles, on 11/29/2007, -1/+25Zosh, you are the one who is wrong. The argument doesn't assume that the odds change with time, rather, it traces out the conditional probabilities associated with each scenario. And, if you aren't convinced by the theoretical argument presented in the video, cilck on the link to the simulator above and see for yourself.
- xscientist, on 11/29/2007, -1/+17You're wrong, my friend. Many Phd's have fought over it, but the logic is solid, and simulators prove it. Reference the comments by Strictnein and dogbomb.
- shefwed82, on 11/29/2007, -5/+5You are right and you are wrong. The odds are in your favor to switch as long as the host knows which door the item is behind. Think about it this way. There is a 1/3rd chance that the door you have chosen is the correct one. Well if the host takes away a door, then you know that that one did not contain the item. So the door that you chose still has a 1/3rd chance of being correct, but the other door has a 2/3rds chance of being correct.
  Now if the host didn't know which door the item was behind, then it doesn't matter if you switch or not, because he might have randomly chosen the door that contained the item.
- pxa270, on 11/29/2007, -2/+20You're wrong about this: always swapping doors will improve your chances.
  
  The reason is actually pretty simple. The most important information here is that the host will always open a door to reveal a goat. This means that if your initial pick was wrong, the host essentially reveals where the car is (by opening the only remaining goat door), but if your initial pick was correct you don't get any useful information (he can open any of the remaining 2 goat doors). Since your initial guess was more likely wrong than right, following the rule to always switch gives you better odds of winning.
- frsrblch, on 11/29/2007, -3/+5The odds DO change when the host opens a goat door. Now, had you chosen correctly at the start (33% chance), swapping would give you a goat prize, whereas had you picked a goat (67% chance), the other door would reveal the car.
  
  It's not that hard to grasp.
- zdyn, on 11/29/2007, -1/+5actually the logic is perfectly correct. take any college statistics course and this problem will be discussed. conversely, jot down a few simple tests and you will find it to be true.
  
  you are correct in saying the odds do not change over time, but the solution doesn't assume that; in fact the entire solution revolves around the "starting state". you have a 1/3 chance of choosing the car first, 2/3 of a goat. after your choice, a goat is revealed and you are allowed to switch. if you don't, your odds remain the same (1/3 car, 2/3 goat). if you do switch, then if you picked the car first (1/3) you will get a goat; if you picked a goat first (2/3) you will get a car. it is easy to see the probability of the first choice ties into this.
  
  hope you weren't being sarcastic, otherwise that flew right over my head =/
- bitweever, on 11/29/2007, -2/+4Write a program to do it 1000 times, and you'll find out that it's true.
  - Pake, on 11/29/2007, -8/+2Just wrote a program and the odds are about 50/50. I'll post the code if you want, even though it's really ***** in it's current form because it was wrote up quickly with little thinking.
  - Pake, on 11/29/2007, -2/+9Nevermind, I had a bug in the program and it's working now. 66% is actually right, even though you're always forced to choose between 2 doors.
- pahoehoe, on 11/29/2007, -2/+2If you really think that the problem's solution in the video is wrong, I strongly suggest you take time to reconsider it. The odds do change in time because probabilities are calculated with available information, and available information changes. When the host opens the door revealing the goat, you have gained information. You know that the host's choice is constrained by the location of the car. This constraint occurs when you initially pick a goat, which will occur 2 out of three times. Anytime information is gained about an event, the probabilities associated with the event CAN change. The probability that it rains one year from today is based on an annual average. That probability will change as one gains information. One day before the day one year from today, one can make a better assessment of the situation, and the probability will change, because available information has changed. Also, consider this. At the beginning of the problem, the probability that the contestant picks the car is 1 in 3. The probability that the host picks the car, if he wanted to, is 3 in 3. The probabilities are different, because each person has a different amount of information.
  
  Further resources...
  http://en.wikipedia.org/wiki/Monty_Hall_problem
  "The Man Who Loved Only Numbers" Paul Hoffman pg. 234-240.
- jmpeagle, on 11/29/2007, -3/+4troll, this is established mathematical/proved fact
- Hefelumpman, on 11/29/2007, -1/+2no no no! It takes the starting point completely into account
  A - If you pick a car first then swap, you will end up with a goat
  B - if you pick a goat then swap, you will end up with a car
  C - you are twice as likely to pick a goat first pick - therefore you are twice as likely to end up with the car if you swap, i.e. 66% chance of winning by swapping.
  
  if you don't swap, your odds stay the same - 33%.
- raskali, on 11/30/2007, -0/+1hmmm.
tdrizzle, on 11/29/2007, -1/+15Oh god, this broke my mind for a few hours after I read about it the first time. It truly is great to try and explain to some people.
paulg2000, on 11/29/2007, -10/+4Wow, I'm amazed they could drag that out over 5 minutes.
dogbomb, on 11/29/2007, -9/+43Put it this way:
There are 100 doors... 99 goats, and 1 car.
You pick a door... and the gameshow host opens up to reveal 98 goats.

Do you swap?

It's entirely the same principle, but with bigger numbers - and it helped me get MY head around it.
- TheRemoteViewer, on 11/29/2007, -10/+5But it could still be the door you picked to begin with. I'm just too stupid to understand this one, or something.
  - pahoehoe, on 11/29/2007, -2/+8If could, but not with probability 1/2. The probability you initially chose correctly is 1 in a 100. Switching means your odds of winning are 99 in 100. The host does a favor in showing the places where there is no car.
  - merreborn, on 11/29/2007, -1/+6Yes, it *could* be, but the question is not if it *could* be. The question is which door is *most likely* to be the right one.
- jdog7117, on 11/29/2007, -2/+2Thats the same was I explain it to people who don't understand it. 3 doors is just too complicated to understand for some people.
- unearth, on 11/29/2007, -2/+14Okay, think of it like this. I show you a deck of cards face down, and I tell you to pick out the 8 of clubs. No reasonable person thinks that they would be "good" at doing something like that, and you would have a 1/52 chance to do it.
  
  Now, say you picked a card at random just there. I throw away 50 of the cards, leaving one in my hand, and the one in your hand face down. I _GUARANTEE_ to you that one of these two cards is the 8 of clubs.
  
  There is a 1/52 chance that the card you are holding is the 8 of clubs. But since you have the information that one of the two cards left in play IS, the remaining card has a 51/52 probability to be the 8 of clubs. P(8c) + P(Not 8c) = 1.
  
  Now, imagine another game, where you pick a card at random again, and you want the 8 of clubs. I remove a card from the deck, and one of the cards is guaranteed to be the 8 of clubs once again. But now, I take the card that you picked, and I SHUFFLE the two remaining cards, and hand one back to you.
  
  Since that information is now lost, you would have a 1/2 chance to pick the right card. It's all about the information provided to you by the removal of false cards.
  - rhinopig, on 11/30/2007, -0/+4That's a great explanation. Similarly, the trick relies on the information that 1 of the cards left will be the 8 of clubs. If you throw away 50 cards (without guaranteeing that the 8 of clubs will still be around) then the probability either you or I have the 8 of clubs is only 2/52.
- jluquette, on 11/30/2007, -1/+1This is a really dangerous way to wrap your head around the problem. Usually when people resort to this to understanding the problem, they don't actually understand it. This method helps people who _think_ that the final probability between the two doors is 1/2--and it isn't. In the 100-doors version, they think, "as I start, I have a 1/100 chance, and at the end i have a 1/2 chance." This is not the case. Even in the 100-door problem, you have a 99/100 chance of choosing the correct door; not a 1/2 chance.
  - oakeybloke, on 11/30/2007, -0/+1You have 99/100 originally. However, you only have 1/2 chance because there are only 2 doors to pick from, not 100. Only an idiot factors in that he can decide to deliberately pick an already-open goat door.
  - sgtbutterscotch, on 12/01/2007, -0/+1Isn't that what the original commenter said in the first place? And lol at your use of the word "dangerous."
- HayString, on 11/30/2007, -1/+1Thanks, that actually did help me to finally understand it.
  
  Thanks to you also, jluquette, for elaborating on it even more.
- myranttoyou, on 11/30/2007, -7/+1The "100 example" is a stupid way to explain it. How is the host opening 98 doors the same as opening 1? It's a different contest then. If anything, they should say open all remaining doors but one. Either way, it still feels 50-50 so it's a bad explanation.
- thedaylights, on 11/30/2007, -0/+2Suddenly the 98 goats break loose from their restraints! They are galloping toward you, bleating with fury and you need that car to escape... what do you do?
  (a) choose the other door
  (b) run away
  (c) run towards the goats
  (d) .... you come up with something
  - pygmalion, on 11/30/2007, -0/+3run over the goats in the shiny car, using the wipers to clean out excess blood
  - LtJimDangle2, on 11/30/2007, -2/+1fail.
JDWTC, on 11/29/2007, -28/+4The math is all wrong, this guy is an idiot, its 50/50... two doors equals 1 in 2
- emiles, on 11/29/2007, -3/+11Like I said to the other commenter above, click on that simulation link and see for yourself that the video is 100% right.
- chirox901, on 11/29/2007, -3/+7Congrats on being the first victim to comment.
  - JDWTC, on 11/29/2007, -6/+1Thanks, YHBT
    - ArchieAndrews, on 11/29/2007, -1/+4If you tell someone you are trolling, you fail it.
- frsrblch, on 11/29/2007, -3/+2No.
- NonServium, on 11/29/2007, -3/+5Always wise to call people idiots when you disagree with them. There's no danger that you'll end up looking stupid as a result.
- elemmeno, on 11/29/2007, -2/+0oops just wrote the same thing dogbomb did... :-/
- xscientist, on 11/29/2007, -2/+1http://en.wikipedia.org/wiki/Monty_Hall_problem
  
  your are wrong.
- bitweever, on 11/29/2007, -2/+3FAIL
- josephblanx, on 11/29/2007, -2/+1Humans are not statistically minded, or logical animals. That is why the math on this problem will always win over your opinion.
- unearth, on 11/29/2007, -2/+1Conditional probability > You
- threemagic, on 11/30/2007, -0/+1If you have 2 doors and they randomly change and you choose, you are correct. In this case there were three doors and you've already made a choice, that choice is either right or wrong.. right only 33percent of the time.
  
  the other 2 doors are right 66 percent of the time...you get both doors since he opens one for ya but in reality you are choosing between the door you chose and the other 2.
  
  ask yourself this... if you chose a door and they ask you if you'd take the other TWO doors.. would you? Yes..because it's right 66 percent
neilbaylor, on 11/29/2007, -2/+9ahh first year statistics, how i've missed you!

http://en.wikipedia.org/wiki/Monty_Hall_problem
slashbot, on 11/29/2007, -11/+4ugh... it's like they were trying to explain it to a 5 year old. Over and over and over....

Simply put, you're more likely to pick a goat. Swap and you'll win. there... took 5 seconds to say.
- rhinopig, on 11/29/2007, -1/+3You're right, but see the number of comments in this article with people who still don't get it.
pxa270, on 11/29/2007, -1/+5OK, here's a REALLY SIMPLE explanation.

First, the most important fact that must be stated up front is that the host must always open a door with a goat, regardless of your initial choice. In some versions of this problem this information is missing, then all bets are off. (Imagine for example that the host decides to only open a goat door if your initial pick was correct. Then you'll always lose by switching).

That said, imagine playing this game a hundred times, and follow the rule to always switch. Of those 100 times, if your initial guess was right (around 33 times) you lose by switching, but if your initial guess was wrong (around 66 times) you win.
- sancho, on 11/30/2007, -1/+0Best explanation of the problem I've ever heard. Anyone who reads this and still disagrees will probably never understand.
- oakeybloke, on 11/30/2007, -0/+1But what if you, with all the luck in the world, managed to select the correct door on the first try (opting to not switch) 100 times in a row? When do you arbitrarily decide to stop and be satisfied that the results are complete?
- sgtbutterscotch, on 12/01/2007, -0/+1That was a REALLY BAD explanation. All you did was state the expected results at the end.
TheGreenBastard, on 11/29/2007, -8/+5How is this difficult? There is a 2 in 3 chance of picking a goat the first time; only a 1 in 3 chance of picking a car. Chances are you will pick a goat. This is probability. The host will show you the other goat. So, if you go with the odds, you will choose the remaining door, which will PROBABLY reveal the car. That's it. It doesn't guarantee you the car, but it assumes the probability that the car will be behind the last door.
Parks, on 11/29/2007, -13/+2OK this is not correct. Try to picture this problem if you had 10 doors to pick from and after every time that you chose something the host (still knowing where the real car is) opens 1 of the 10 doors and allows you to chose another door. What would the odds be then? You are guaranteed to get a 2 door choice at the end no matter what. eg 1 in 2 chance. Professors and mathematicians are only self proclaimed geniuses.
- unearth, on 11/29/2007, -2/+3Are you saying that you're given ten different opportunities to switch? That would be the easiest goddamn game ever.
  
  You would have a 1/10 initial chance to pick a goat, then you would hold that door until the last opportunity to switch. 90% win rate.
  - unearth, on 11/29/2007, -1/+2Whoops, meant 9/10 chance to pick a goat initially.
- pahoehoe, on 11/29/2007, -2/+1Sigh... see the other comments explaining the problem above.
- threemagic, on 11/30/2007, -1/+1Imagine in your scenario that you chose one.. now the host says you can have the other 9.. would you? it's not 50/50....
  
  he flipping the 8 other goats first makes no difference
Samn, on 11/29/2007, -3/+20I have a car. I want a goat. Having a goat would be ***** hilarious. I could dress it up in my shirts and paint his horns blue and take him with me to the grocery store and people would be like "Oh man, a goat. That guy's got a goat" and the manager of the store would come over and be like "Sir, I'm afraid you have to blah blah blah blah" something about I can't have a goat but ***** that guy he doesn't have a goat and I do so what does he even know about it? I put on my sunglasses and I'm like "Pfffft" as I breeze on by him to the cereal aisle.
- pahoehoe, on 11/29/2007, -3/+1I digg ya man. Cars are highly overrated.
- merreborn, on 11/29/2007, -1/+8I am intrigued by your ideas, and would like to subscribe to your newsletter.
- theclansman, on 11/30/2007, -0/+2hilarious idea
- EatUrKids, on 11/30/2007, -1/+1There are 2 cars and one bike. Host opens up 1 car. He asks if u want to switch. Do you?
  - EatUrKids, on 11/30/2007, -0/+1Well u shouldnt since the same equation applies.
- hairywart, on 11/30/2007, -0/+1Finally, someone that can talk about something I understand.
lackingdiggnity, on 11/29/2007, -1/+25But what if he shows you a goatse instead?
- rhinopig, on 11/29/2007, -1/+18That's called the Monty Hole Problem.... You don't want to know the solution.
slothlovechunk, on 11/29/2007, -1/+7If you switch, it's like you get to pick 2 doors. If you don't, you only picked one.
- oakeybloke, on 11/30/2007, -0/+1You are picking between one of two options: to switch or stay. Switching is one option, Staying the is other. Two options total.
  - slothlovechunk, on 12/01/2007, -0/+1?
    
    yeah, one gives you a 1/3 probability of winning, the other gives you a 2/3 probability of winning.
    
    Just rephrase the whole game show as, instead of ever showing you a goat in one of the doors you didn't pick, telling you that if the car is in either of the other 2 doors, you win if you switch. This is exactly what is happening. No matter what, there is at least one goat in the 2 doors you didn't pick, however, you don't know which door this is. There also may be a car in one of these doors, with 1/3 probability.
    
    Monty Hall tells you which of the two doors left suck balls to pick. He lets you switch to the other remaining door. Effectively, this is the same thing as letting you pick 2 doors.
RichGC, on 11/29/2007, -7/+2As wikipedia said, im with the people that dont think that past odds really effect the players current odds.

If anything I say dont swap, because if you do, and you had it right first time, your going to kick yourself even harder.

Perhaps try some social engineering, the host is probably standing to the side, assuming the host knows where the prize is, and is a bit lazy, pick the door nearest the host, if they walk to the furthest door to reveal a goat, the car could well be in the middle.
- rhinopig, on 11/29/2007, -3/+3Sorry, but you're wrong.
  - r2builder, on 11/29/2007, -2/+2You're wrong. Watch the video and listen to the mathematical explanation. Don't disagree with it just because you don't understand it.
- unearth, on 11/29/2007, -1/+4What is so hard about this, guys. I understand the initial hesitation to accept the result, but its a goddamn mathematical proof. When your opinion differs from the result of a mathematical proof, its time to consider changing your opinion.
  - sancho, on 11/30/2007, -0/+2It's kinda like people who reject evolution, despite all the evidence that it exists. They'd rather stick with their own beliefs than accept the proof that's right in front of their eyes, because that proof conflicts with their gut.
    
    I wonder if any of these "50%ers" have ever made fun of creationists.
- RichGC, on 11/30/2007, -1/+2For those digging me down, I like to question and understand things myself before just blindly accepting what the internet says.
  
  This whole thing seems to be based on the overal odds, and not the odds at the point when you have a choice to swap.
  
  Suppose the host just straight off opens a door revealing a goat, before you pick, the odds would be 50/50.
  
  How does you choosing the first door alter those 50/50 odds when your left with the exact same choices, a goat or a car.
  - unearth, on 11/30/2007, -0/+23 doors is the minimum number of doors required for this to work, its easier to understand if you say there are a MILLION doors.
    
    I put a million doors in front of you, and only one contains the car. You take your first guess. I remove 999,998 doors, and tell you that either your door, or the remaining unchosen door contains the car. A little math:
    
    X is the event that you chose the car on your initial guess
    Y is the event that you chose a goat
    P(X) + P(Y) = 1
    
    P(X) = 1 / 1,000,000
    P(Y) = 999,999 / 1,000,000
    
    If you do not switch doors, you are effectively keeping your one-in-a-million guess. From the axioms of probability, the odds that you did not pick the car on your initial guess are 1 - P(X) = 99.9999% = P(Y). As a result of this, the probability that the lone door remaining is the car = P(Y).
    
    You are essentially betting against the fact that your initial guess was right. It is very easy to do this with one-in-a-million odds, but less apparent with one-in-three.
  - FiremanJersey, on 11/30/2007, -0/+2No no
    no, it has to be a BILLION doors.
    
    /sarcasm
  - RichGC, on 12/04/2007, -0/+1To follow up on this, I now understand this better, you have to forget the 50/50 odds, because the question is do you want to swap.
    
    By swapping, you get slightly better odds, because you are swapping your guess based on 1/3 chance, for a 1/2 chance.
    
    While statistically your chances slightly improve, the human factor means that swapping a correct door is more painful.
- valadil, on 11/30/2007, -1/+0Here's another way to think about it.
  
  There are two possible strategies: always switch and never switch. Anything else is just guessing and isn't a strategy worth looking at.
  
  With never switch, the fact that one goat gets removed after you guess doesn't matter. Your guess stays, so you have a 1/3 chance of getting the car. Correct?
  
  Under always switch, you have an initial guess, and then you switch it after a goat is removed. So if you've picked the car, a goat gets removed and you get the other car, but if you've initially picked a goat, then other goat is removed so you get the car. The possibilities are guess car, get goat a or goat b; guess goat a, get car; and guess goat b, get car. It makes you win if you guess goat and lose if you guess car. You have a 2/3 chance of guessing goat initially, so you have a 2/3 chance with this strat.
TurnipFarm, on 11/29/2007, -10/+1The error is committed in this sentence: "If you swap you're going to win a goat AT LEAST 33% of the time".
"AT LEAST 33%?" That requires flexible probability, which does not occur here because there are 2 doors and 1 car, and 1 goat. These things are not dependent on any other factors and therefore the probabilities involved are constant. The author has wrongfully including the revealed goat in the "swap" scenario but NOT IN THE "stay" scenario because he or she is an ass.
The correct analysis:
If you swap, you're going to win a goat 50% of the time (1/2 REMAINING DOORS contains the goat).
If you don't swap, you're going to win a goat 50% of the time (1/2 REMAINING DOORS contains the goat).

The logical error is clearly apparent as outlined above.
- redhaze, on 11/29/2007, -1/+2Wrong. They give a hundred explanations here if you don't get it:
  http://en.wikipedia.org/wiki/Monty_Hall_problem
- merreborn, on 11/29/2007, -1/+3I love how some know it all always comes in and insists that the naive solution is the correct one. Happens every time someone brings up the Monty Hall problem on the internet.
  
  This is a famous, freshman-level math problem. People much smarter than yourself have spent much more time thinking about it. The solution's been well known for decades, but people like you keep creeping out of the woodwork.
  
  Have a dozen references defending the 2/3-if-you-swap answer:
  http://en.wikipedia.org/wiki/Monty_hall_problem#Re ...
- SpaceParanoids, on 11/30/2007, -0/+1You are correct that the modifier "at least" is inaccurate. It would be more correct to say that the probability of winning a goat is 1:3 (about 33%).
  
  The rest of your analysis is wrong. By switching doors your chances of finding the car are definitely doubled from 1:3 to 2:3.
  
  Think of it this way: What are the chances you picked the car on your first guess? 1:3, obviously. By not switching, you're sticking with that original 1:3 bet, no matter what Monty does. But because Monty Hall uses his knowledge of where the car is to eliminate a wrong choice, you're effectively reversing the odds by swapping doors. You're given the opportunity to bet that you *didn't* pick the car first, which has 2:3 odds.
- threemagic, on 11/30/2007, -0/+2the flaw in your logic is the open door doesn't count anymore.. it does.
  
  you have 3 doors.. you choose 1..
  
  you can swap the 1 for the other 2..
  
  does that help?
rhinopig, on 11/29/2007, -2/+4This is a great example of the counter-intuitive truths that math and probability can reveal. Even better it demonstrates that probability is not in fact an objective quantity in reality, but rather a function of the amount of information an observer has (in this case, the host helps you out by giving you information about where one of the goats is, thus altering the probabilities of which door to pick).
santogold, on 11/29/2007, -5/+13def var curtain as char extent 3.
def var guess as int.
def var prize as int.
def var monty_shows as int.
def var final_choice as int.
def var youpick as int.
def var win as int.
def var loss as int.
def var i as int.
pause 0 before-hide.
repeat:
do i = 1 to 3:
curtain[i] = "goats".
end.
prize = random(1,3).
curtain[prize] = "car".
youpick = random(1,3).
SHOW:
repeat:
monty_shows = random(1,3).
if prize monty_shows and youpick monty_shows then leave SHOW
end.
do i = 1 to 3:
if i youpick and i monty_shows then final_choice = i.
end.
if curtain[final_choice] = "car"
then
win = win + 1.
else
loss = loss + 1.
if (win + loss) mod 8000 = 0
then
disp win loss win / (win + loss) format "9.9999999999999".
end.

results: (mark of the beast?)
win loss
────────── ──────────
5,327 2,673 0.6658750000000
10,722 5,278 0.6701250000000
16,021 7,979 0.6675416667000
21,427 10,573 0.6695937500000
26,706 13,294 0.6676500000000
32,046 15,954 0.6676250000000
37,358 18,642 0.6671071429000
42,761 21,239 0.6681406250000
48,090 23,910 0.6679166667000
53,400 26,600 0.6675000000000
58,763 29,237 0.6677613636000
64,116 31,884 0.6678750000000
69,479 34,521 0.6680673077000
74,834 37,166 0.6681607143000
80,246 39,754 0.6687166667000
85,655 42,345 0.6691796875000
90,927 45,073 0.6685808824000
96,266 47,734 0.6685138889000
- EatUrKids, on 11/30/2007, -5/+1I bet you no one commented so they wouldnt be called stupid, well idc i have no idea wat that says, only that it probably took u like 20min to write
- ohcoaster, on 11/30/2007, -0/+1int intWinner = 0;
  Random oRand = new Random();
  bool blnSwitch = false;
  for (int i = 0; i < 1000000; i++)
  {
  int intCarDoor = oRand.Next(3);
  int intPicked = oRand.Next(3);
  
  if (blnSwitch) // If you didn't pick the car initially, then you win when you switch:
  {
  if (intCarDoor != intPicked) intWinner++;
  }
  else
  {
  if (intCarDoor == intPicked) intWinner++;
  }
  }
  
  float fltPercent = ((float)intWinner / 1000000) * 100;
  txtOutput.Text += "You won " + string.Format("{0:#.##}", fltPercent) + "% of the timern";
Lnomis, on 11/29/2007, -1/+4Ok, here's a variant on the problem. If there are ten doors, and once you've chosen your door, the host only opens one of them, do your chances change if you change your decision? If so, by how much?

What if you're playing with N doors and he opens M of them?
- Zique, on 11/30/2007, -0/+1The answer completely depends on if the host opening the door always knows that the grand prize isn't behind it (in which case you should swap), or if opening the door is completely random (in which care it doesn't matter).
  - Lnomis, on 11/30/2007, -0/+1Yes, sorry, I should have added that he knows where the car is.
- Lnomis, on 11/30/2007, -0/+2My guess at a solution is that it's 9 in 80 if you switch, compared with 1 in 10 if you don't.
  
  In general, I think it's (N - 1) in N * (N - M - 1) if you switch, or 1 in N if you don't.
  This reduces to the correct solution for N = 3, M = 1
  - pahoehoe, on 11/30/2007, -0/+2This is the correct solution.
    
    Brief Derivation:
    P (gift behind initially picked door) = P(W) = 1 / N
    P (gift not behind initially picked door) = P(!W) = 1 - 1 / N
    P (gift not behind initially picked door after host opens M doors) = P(!W | M) = P(!W) / remaining doors = (1 - 1 / N) / (N - M - 1)
    The latter formula is equivalent to (N - 1) / (N* (N - M - 1)), and if the contestant switches, one of these doors is picked.
    
    Notice that P(W) + P(!W | M) * (N - M - 1) = 1, reflecting that the gift is somewhere, and the host revels doors that do not have a gift.
    
    Nice work, and good problem.
- SpaceParanoids, on 11/30/2007, -1/+2By not swapping your odds of winning are 1 in n.
  
  By swapping your odds of winning are m+1 in n.
incaseyoucare, on 11/30/2007, -2/+3I understand the math behind this story problem. However you could argue (subjective probability of course) that the game show host distorts the question by only opening the goat-door after you have made a selection which you assumed to be a choice of 1 in 3. If he had opened the goat-door before you had a chance to choose a door, then the probability would have been 1 in 2. But the game show host is always going to open a goat-door so does it really matter whether he opens it before or after you made your selection? You could argue that one goat-door is never really in play (it only appears to be) and you are really just choosing between the 2 doors that the game show host has not or will not open--with a 50% chance of getting the car on your first try. Of course empirical probability proves this isn't the case and the probably of choosing the car first try will edge closer to a 1 in 3 chance when the game is repeated on a large scale. I think it is interesting though, that the appearance and sequence of the game are such defining variables.
theclansman, on 11/30/2007, -0/+6I actually wasted half a class in my university intro stats class arguing with the prof about this. I don't know where he got the idea, but he thought that if you kept the same your odds were better...Of course, he was a real douche. Another example, he put this question on an assignment
"There are 4 digits in a phone number system, how many phone numbers are possible?"
of course the answer is 10000 right?(the numbers 0000-9999) no, he marked that wrong. the correct answer is to use the 10 numbers choose 4 formula..of course, this doesn't include phone numbers that have the same digit twice. I tried explaining this in an email but he refused to give me the mark. I ended up going in to his office and showing him, he conceded that both answers were correct (*****).
- theclansman, on 11/30/2007, -0/+5I'm aware repling to my own comment is pretty lame, but I feel i have to mention how much I love this problem because NO ONE ever gets it. I musta tried to explain it to about 30 of my friends so they would realize how dumb my prof was, but none of them would understand it and they ALL sided with the prof
- Elliuotatar, on 11/30/2007, -0/+4Maybe you should find another math professor who will agree with you and then go to the dean and tell him he should fire this guy because he obviously should not be teaching if he can't even do something simple like determine that there are 10,000 possible phone numbers with four digits.
andergriff, on 11/30/2007, -0/+2If you've ever actually watched "Let's Make a Deal" you know that there is another dimension to this problem. True enough, the problem works out in probability theory precisely as indicated by the video. But the complication comes in when the host, at the last second of course, steps in and offers cash if the contestant will forego his guess. In other words, there is a buyout offer. Usually, it continues to go up as the contestant waffles, which puts a lot of pressure on the contestant. Go for the sure thing or gamble on the door? Hence, the problem is far from solved because of this added factor.
- Funpolice2050, on 11/30/2007, -1/+0The probability remains the same though.
5teady, on 11/30/2007, -6/+1This guy sounds like a paedophile.
KIERANMULLEN, on 11/30/2007, -0/+4More 5min spam from their hired posters.
jellygraph, on 11/30/2007, -8/+1What a load of *****. This problem gives the impression that you had 1/3 chances and that you made a choice and are given a second chance. But you haven't been given a second chance. Because you weren't allowed to pick the first time and get feedback. It never takes into account that you may have picked right the first time.

Pseudo *****.
- kevludlow, on 11/30/2007, -0/+1"Because you weren't allowed to pick the first time and get feedback."
  
  What are you talking about? I don't know what 'feedback' you're wanting, but you're given more information about the unknown. In probability, whether you accept it or not, this can change the outcome drastically - as it does so in this case.
  
  It's definitely NOT pseudo *****. In fact, it's Probability 101, and like most probability - very counter-intuitive.
- sancho, on 11/30/2007, -0/+0Actually, it does take that into consideration.
  There are only three possibilities: 1) You pick goat number 1. 2) You pick goat number 2. 3) You pick the car.
  
  If you pick goat number 1, Monty reveals goat number 2. If you switch, you win the car.
  If you pick goat number 2, Monty reveals goat number 1. If you switch, you win the car.
  If you pick the car, Monty reveals either of the two goats (it doesn't matter which.) If you switch, you get a goat.
  
  Note that in 2/3 of the situations, you get a car.
  
  People sometimes get confused because they think that there are actually four situations--one each where Monty reveals a different goat when you pick the car. They don't realize that this is not relevant, because the choices that matter are which door you choose, and there are only three of those. If you pick the car (1/3 of the time), then Monty will reveal Goat 1 half of the time (1/3 * 1/2 = 1/6), and he will reveal Goat 2 half of the time (1/3 * 1/2 = 1/6.) And of course, adding these probabilities up, you get 1/6 + 1/6 = 2/6 = 1/3, which is the probability that you picked a car.
- apeweek, on 11/30/2007, -0/+1Hey, who needs math when you've got your gut instincts? Let's call this 'Mathiness'
usha49, on 11/30/2007, -0/+1same old story. same calculations. I do not enjoy such games. Thanks a lot anyway.
daizaru, on 11/30/2007, -4/+2I hate this example it's so stupid. I get the probability example but by CHOOSING not to swap you are choosing a second time to take the original door you picked out of a possible 2 remaining doors. Either choice to swap is a 50% chance because you are choosing a second time to stay where you were. Anyone who believes this actually improves your chances is naive.

It's nothing more then a play on words. If you were to run the show 100 times and always swap, probability would still show you winning just as often as never swapping.
- Zique, on 11/30/2007, -0/+6I'm sorry, but did you just miss that big-ass program example above where santogold ran the test 150000 times and got a 33/66 result, and all the other bloody links to sites which empirically test this and come to the same conclusion?
- kevludlow, on 11/30/2007, -0/+1Perhaps 'improve' your probability is not the right way for you to look at this - if it helps you. The simple fact is that you are still trying to look at the problem from a WINNING perspective. The idea behind this very old mathematical question is to look at it from a LOSING perspective.
  
  You agree from the start you are given a 33% chance of winning and a 66% chance of losing. What is more likely? ...obviously you're more likely to LOSE the game, rather than to WIN. There are only TWO losing doors, so if one of them becomes revealed (after you've already lost), there are NO more doors to lose. Hence when you swap, you'd win.
  
  It's true that if you happen to pick the winner from the get-go, and then swap, you will clearly lose. But you'll only pick the winning door 33% of the time to begin with, so you can only lose 33% of the time that way.
  
  Sorry to put it this way, but anyone who can't understand this problem is just human. Anyone who tells people they are wrong because they themselves refuse to study the probability is however, naive.
Crucifix, on 11/30/2007, -4/+2Does anyone have any empirical evidence that proves this hypothesis? I say 'hypothesis' because, as other posters have mentioned, one of the doors was never in play to begin with, making the chances 50/50 the whole time.
- santogold, on 11/30/2007, -0/+3See the program above. It uses random numbers and "wins" 2/3'rds of the time.
- sancho, on 11/30/2007, -0/+0It's not a hypothesis. Hypothesis has a very strict definition.
  
  Regardless, here's some code (in Perl) that demonstrates the game. Player picks randomly, Monty randomly reveals a door, player always switches. The win percentage hovers around 66%, just as you'd expect.
  
  http://pastebin.org/9931
- manova, on 11/30/2007, -0/+1Read santogold's evidence above.
- drwh0, on 11/30/2007, -0/+1It is not a hypothesis. It is probability. The proof he gives is mathematical. It does not mean that you will always win, but if you swap you do have a higher chance of winning. The program is an example of this in practice but the video explains the proof of this.
EatUrKids, on 11/30/2007, -4/+3I like turtles
EatUrKids, on 11/30/2007, -6/+2I like turtles
- EatUrKids, on 11/30/2007, -4/+1Sry for 2x post
dtraneighty8, on 10/07/2008, -0/+2People that are arguing FOR a 50% chance has to realize that the door is selected BEFORE the goat is revealed. Meaning that whatever happens, if you stay with the door that you have selected, you have a 33% chance of it being a car. However, I love dogbomb's comment best.
Zique, on 11/30/2007, -0/+6I can't help but to love all the kids thinking couple of minutes of their brain power is superior to a well-tested and documented matchematical fact.
adrake, on 11/30/2007, -0/+5I'm currently working on my PhD in applied math. One of the first things you learn as an undergrad math student is how to ignore your intuition, which is often wrong. All the people arguing that the chances of winning given a swap are 50% are following their intuition, and it's failing them.

50% makes sense, but it's wrong.
dankenstein, on 11/30/2007, -0/+6I like the numb3rs explanation of the problem:

http://www.youtube.com/watch?v=P9WFKmLK0dc
SomaSynth, on 11/30/2007, -1/+0A car or boobies, you can't lose!
bigpiratejim, on 11/30/2007, -1/+0If Monty only offered a swap if the contestant had initially selected the door with the car, and refrained from offering a swap if the door with a goat was selected, swapping would never result in a car. If Monte only offered a swap when a goat was selected, swapping would always result in a car. Neither Wikipedia nor Marilyn vos Savant states in the problem that Monte must offer of a swap, only that he does offer a swap in the case at hand. If Monty can offer a swap or not as he chooses, he can control the odds of winning by swapping to either 0% or 100%. The result that swapping increases the odds of winning from 1/3 to 2/3 is only true if Monte must offer a swap. There is a 1/3 chance of winning the car by not swapping, but the change in the odds of winning by swapping cannot be known in the problem as stated.
HerbSolo, on 11/30/2007, -0/+3I think that was explained a little too complicated, because it included the chances for the whole event.
A simpler explanation is: Switching increases your chances, because the chances you picked a goat in the first place are higher than those of picking a car.
oakeybloke, on 11/30/2007, -2/+1There is a 100% chance the Car Door is the Car Door. There is a 0% chance Goat Door A is the Car Door, and another 0% chance that Goat Door B is the Car Door. 100 / 3 = 33 each. After one has been removed, there is still a 100% chance the Car Door is the Car Door, and 0% chance the remaining Goat Door is the Car Door. 100 / 2 = 50.
- apeweek, on 11/30/2007, -0/+1If you initially picked a door with a 33% chance of having a car, That will not change. So sticking to your door only gets you a 1/3 chance of winning.
  
  But now a door is eliminated. This is why the remaining door has a 66% chance of winning. Because YOUR door is stuck at 33%.
  
  Seriously, get three paper cups and a coin and try this out. You will slap your forehead after about three trials.
Spoomeister, on 11/30/2007, -0/+1ObGeorgeCarlin:
"Ok, you can keep the $500, OR, pick behind what's behind door #1, door #2, or door #3."
"Oh, Christ! Ok... ok... door number... 2. No, 1, no 3, no wait... alright, alright. Door number 2. No, 3. No... Um.... What were the choices again?"
"When Monte Hall dies, he goes behind Door #4."
santogold, on 12/01/2007, -0/+1My dad did not believe this either. We played this game with two duces and an ace for about 5 minutes. I won 2/3'rds of the time.
Biskino, on 12/01/2007, -0/+1Ran into this one for the first time reading 'The Curious Incident of the Dog in the Night-time' (great book). It nearly broke my mind - to the point where I actually had to sit down and game it out.
fragglet, on 12/01/2007, -0/+0For everyone who doesn't believe this is true, how about you put your money where your mouth is? We'll play this 100 times, and always choose to change doors at the second choice. Every time we win, you give me $1. Every time we lose, I give you $1. If it's 50/50, we should have the same amount of money each by the end.
Login or register to add a comment

Create a new account or login to join in the conversation on Digg. You'll also be able to Digg stories to help promote things you like.

The Monty Hall Game Show Problem watch!

Login or register to add a comment