## The best game-show math brain teaser and the controversey that surrounded the answer

This has to be my all-time favorite brain puzzle, mainly because the answer plays with your sense of intuition, but also because of the incredible controversy it created.

The question was submitted to a famous magazine about 15 years ago, and the woman with the world’s highest recorded IQ provided the solution. Shortly thereafter, dozens of math PhD’s wrote in, blasting her for giving the wrong answer. She was, however, correct.

Here it is:

- Suppose you’re on a game show, and you’re given a choice of three doors. Behind one door is a car; behind the others, goats. You pick a door — say, No. 1 — and the host, who knows what’s behind the doors, opens another door — say, No. 3 — which has a goat. He then says to you,"Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Try to give it a shot on your own, and in a day or two if no one gets it right I’ll post a link to a detailed description of the answer and the story that surrounded it.

Kevinpkennedy| Profile September 4th, 2008 - 9:29 pmThe easiest explanation is a simple spreadsheet. The goats are irrelevant – its all about your initial choice & where the car is. There are 9 possibilities:

You Car Monty

Chose Behind Shows Stay Switch

1 1 2 or 3 W L

1 2 3 L W

1 3 2 L W

2 1 3 L W

2 2 1 or 3 W L

2 3 1 L W

3 1 2 L W

3 2 1 L W

3 3 1 or 2 W L

Kevinpkennedy| Profile September 11th, 2008 - 2:07 pmI’ll try this again, with the spreadsheet turned around.

The best way to see it is a simple spreadsheet. The goats are unimportant – it is all about your initial choice & where the car is. There are 9 possibilities:

You choose 1 1 1 2 2 2 3 3 3.

Car behind 1 2 3 1 2 3 1 2 3.

Monty show 2or3 3 2 3 1or3 1 2 1 1or2.

Ifyou stay W L L L W L L L W.

Ifu switch L W W W L W W W L.

wyrdo| Profile September 13th, 2008 - 1:38 amI really don’t think it work that way. Almost every single person on this board believes that’s how it works, but I still disagree.

I think the moment one of the doors gets opened and you find out what’s behind it, you have to redo all the probability stuff. The open door with a known object (goat) will *now* have that *known* quantity (goat) 100% of the time.

To figure the probability on the other two doors, you merely ignore the goat you can see. This leaves you with only *two* doors. You’ve got a 50% chance of picking the car, whether you stick with your original choice, or whether you switch makes no difference.

See, the reason why I feel sure this is the correct answer, is that if the *other* 2/3 answer is correct, then this seems to imply that simple probability has weird properties similar to quantum physics (which it doesn’t).

—

Furry cows moo and decompress.

noxsmarts| Profile November 2nd, 2008 - 7:36 pmI say no, it’s no advantage in it. No.1 is the choice made and you should stick with your choice believing strongly you made the right choice first.

Noxsmrt

seananderson| Profile February 10th, 2009 - 10:26 pmThe way I think it works is you start off with a 1/3 chance. no matter what you choose the host is always going to choose another door with a goat. meaning that you may think you have a better chance but you are still relying on that 1/3 chance you took before. this means that the first box you chose has a 66% chance of being wrong while the other one has a 50% chance because you know it was not chosen by the host

rishirocks| Profile May 11th, 2009 - 11:48 pmsee first of all the question had be answered once only and the host asking the second time was wrong as the game would be unfair. there was no clause in the game rules to change the answer. and the quastion had to be answered between 3 doors not 2, so the particpant shd not change the answer andgo for another rule also because luck favours the brave and choosing another door would be unfair coz if she won the car others might argue why the host ask to change the door and if loose he might ardue why the host asks so.

eskimo| Profile July 21st, 2009 - 2:28 amIf you still don’t believe that switching increases your chances of winning from 33% to 66%, even with all the previous, well-explained posts on here, the only way for you to be convinced is to gather data for yourself. Yes, you will be doing the Monty Hall “game” at least 30 times, preferably more.

To conduct this experiment, get a deck of cards and a patient friend or family member. Pull out 2 black cards and 1 red card. I use the two black queens (goats) and the ace of hearts (car). Mix them behind your back thoroughly so the ace is in a random position. Bring them back in front of you and glance at the cards, noting the position of the two queens, without revealing any of the cards to your friend. Lay the cards face down and allow your friend to touch the back of one card. Now, eliminate one of the queens from the game as long as it’s not the card she chose. To gather data on the outcome of switching, always have her reveal the card she didn’t chose after one of the queens has been eliminated. Record the face of the card, mix, and repeat. In my experiment of 100 trials, always switching, 69 resulted in winning. Most results will hover around 66%. You don’t have to do this 100 times, but for all you non-believers, at least try this experiment until you are convinced that there is no way in hell the odds are 1:1. Switching DOES make a difference. If you don’t believe me, see for yourself…

heinscott| Profile November 16th, 2009 - 1:46 pmAfter reading all these responses (especially the one about trying it yourself with a deck of cards), I thought I would try to write a test case in C#. Here is code that will show the 2/3 positive results of always switching. Obviously, the more trials you run, the closer the results will be to 66.666667.

Scott

int trials = 1000000;

int count = 0;

bool answer;

bool[] door = new bool[3];

Random ran = new Random();

for (int x = 0; x < trials; x++)

{

door[0] = (ran.Next(0, 3) == 0);

//selects a random number between 0 and 2. If 0 set to true;

door[1] = (ran.Next(0, 2) == 0 && !door[0]);

//Same as above (0-1), but, set to false if first is true;

door[2] = (!door[0] && !door[1]);

//third door only true if first two are false;

int pic = ran.Next(0, 3);

//select a random “guess” for contestant

answer = !(door[pic]);

//Switch door to opposite (always end result of switching)

if (answer) count++;

//If true, increment the counter.

}

Response.Write((double)count/(double)trials * 100);

The hardest part to really get your head around, I think, is realizing that “switching” always means that you will have the opposite that you started out with, due to the fact that the host will always show you a door with a goat. Therefore, your 1/3 chance changes to 2/3.

kaxxito| Profile June 21st, 2010 - 12:26 pmsorry for my english. suppose in game are 2 players. David and John. David choosed door number 1, and John choosed door number 2. both of them has 33% chance to win. then host opened door number 3 and there was nothing. after that David’s and John’s chances will be equal. David and John are in equal positions, everything effects on them identically, so if one’s winning probability increases, seconds will increase too. my answer is: it has no matter to change answer

smartguy| Profile June 23rd, 2010 - 6:14 pmHere is another way of looking at it – sorry if anyone has already mentioned it. Suppose you have a deck of playing cards with 51 Aces and 1 King. All the cards are faced down. You are asked to select a card that you think is the King. I then turn over 50 cards, showing 50 Aces. Now surely if you are given the opportunity of changing your card, you would do so – as the chance of you selecting the king at the start was 1 in 52. It would be very unlikely that you chose the King at the beginning. The same principle applies here.

drshacket| Profile September 14th, 2010 - 12:00 pmStay with the same door. Your odds were 33%. Now with door number 3 revealed (also 33%), your odds of having selecting the right door increased by 33%. Therefore, your odds of winning is now cumulative, to 66%, buy staying with door number 1.

experiment| Profile September 15th, 2010 - 9:00 pmok, really all you PHD’s, professors, MIT guys, highest IQ lady, etc, why is this so hard to understand and explain, apparently most cannot do what science always dictates, run the experiment. To better understand, make it ten doors, one car, nine goats. Contestant picks a door, host removes 8 goats doors. One of two things just happened:

1.contestant picked the one car door out of ten doors-10% chance

2.contestant picked one of the nine goats doors-90% chance

That means the other door the host left is one of two things:

1.a goat-10%chance

2.a car-90% chance

GET TEN PIECES OF PAPER, ONE CAR, NINE GOATS,PICK ONE.AFTER THE HOST REMOVES THE EIGHT GOAT DOORS IT BECOMES BLATANTLY, GLARINGLY, RIDICULOUSLY OBVIOUS WHERE THE CAR IS(90% OF THE TIME ON AVERAGE). JUST DO IT.

My G.E.D. came in real handy here, it helped me understand this immediately, some people are just thinking too much instead of running the experiment.

This is actually my favorite brainteaser of all time now

experiment| Profile September 15th, 2010 - 9:05 pmit replaces:

HOW FAR CAN A DOG WALK INTO THE WOODS?

experiment| Profile September 15th, 2010 - 10:31 pmOk, 2 more ways to look at this.

1. The key is that the host KNOWINGLY exposes a goat. He exposes a goat no matter what the contestant chose. If the host DID NOT know where the car is and the exposed door could be either car or goat then it would be 50/50 that the contestant had the car door IF a goat was exposed.

2.Doors 2 and 3 must be one of 3 combinations

goat goat

goat car

car goat

The host automatically eliminates a goat from whichever combination exists leaving:

goat

car

car

2 OUT OF 3 chance that a car is left.

end of story

wyrdo| Profile September 15th, 2010 - 11:37 pmThere are many, many answers that indicate that you will have a 2/3 (66%) chance of getting the car if you switch. This is one of those times when people let themselves fail to “see the forest for the trees”. I mean–if you think about it for a moment, how could your choosing door one over door 2 possibly have any impact on the odds that it will have a car or a goat? All the “monte hall” simulations in the world aren’t going to help you work out the odds because, in this instance, you’re basing them on a false premise.

What really happens is this: as soon as the host reveals there is a goat behind one of the doors, the original problem:

( 33+1/3% car, 33+1/3% goat, 33+1/3% goat )

goes away. Supposing the host reveals a goat behind the 3rd door, we now have:

( 50% car, 50% goat, goat *for sure* )

The goat we can see doesn’t figure into our probability calculations. And which door we choose has no affect on the probability that we get a car or a goat. How could it?

In an instance such as this, game theory tells us the best play is to make the decision truly random: “let the chips fall where they may” as they say. So you should do a heads pick door 1, tails pick door 2 type of thing.

—————————————–

I have to say, the frequency with which people get this wrong is very depressing. But on the plus side, I’m starting to understand why casinos succeed in pulling in so much cash. Various people that think they are smart are unable to detect a run-of-the-mill trick question, and they think their choice can somehow affect the odds. That’s insane. I want to open my casino now please.

—

Furry cows moo and decompress.

Benson| Profile September 16th, 2010 - 11:14 amIt is obvious that we have to switch as door 2 now has the probability doors 2 and 3 combined.

Despite the GED, why do some people find it so hard to understand? This has nothing to do with dogs walking far into the woods. But you still can do what science always dictates, run the experiment. To better understand: make it a cocker spaniel and a medium sized forest. Explain to the dog that its grandmother is waiting for it at the other side of the woods. Make sure you give it plenty of food and water. We would not want some lame dog excuses like no food or water ruining the experiment. Then equip it with a tracking device to find out how far a dog can run into the woods.

Don’t bother telling anybody the results as I doubt that anybody would be interested in them, that is unless they are craving for a good laugh.

end of story

experiment| Profile September 17th, 2010 - 1:24 pmhmm, so its seems you agree with me about the doors while showing no proof or reason for your thinking but do not know how far a dog can walk into the woods. OK, brainiac. Maybe you should stick to the word search puzzles.

experiment| Profile September 17th, 2010 - 1:30 pmand wyrdo, you are embarrassing yourself. The challenge is not to prove that the wrong answer is correct, the challenge is to understand why switching doors gives one a 2/3 chance of winning. There are many great explanations here. good luck.

Benson| Profile September 17th, 2010 - 3:16 pmexperiment, I have summed up the proof in a very simple sentence. Not everybody is expected to grasp it, so there is no need to be on the defensive if you didn’t. Derogatory terms won’t help your case either.

As for the dog walking far INTO the woods, maybe you’d like to share the details of your experiment. A good laugh won’t hurt us.

XSLT brain teaser « Tech Team| Guest January 6th, 2011 - 10:32 pm[…] address values “n/a”, “na” and “Dr”. This reminds me of the game show brain teaser where by stepping through the scenarios in a logical fashion gives the correct answer; trying to […]

fanfan| Profile February 23rd, 2011 - 5:32 pmLet us start by assuming that we cannot second guess the motive of the show host in changing the game rules:

a. By offering a 2nd choice, the game rules have been changed, from a 1 in 3 choice, to a 50:50 probability choice, having no relationship at all to the earlier 1 in 3 starter (it is as if this did not exist at all). In this case, it remains a 50:50 probability, and there is no point in changing the original choice.

b. If the game host has a motive, it would, by the very nature of these programs, to prolong delivery of the reward, and not make it too easy for the contestants. This motive would have a higher probability, looking at the very nature of man. In that case, in case the contestant had initially made a wrong choice with the 1 in 3 option, the probability should be less than 50% that, instead of promptly declaring the contestant failed, and extending the program, that the host would try to give a second chance with a new 50:50 option. In this case, the host is attempting misdirection, to create self-doubt in the contestant. So, in this scenario too, contestant should stick to original choice.

wyrdo| Profile February 23rd, 2011 - 6:40 pmI agree for the most part with fanfan. It’s nice to know that my opinion (which I fully believe to be the mathematically correct one) is shared by at least one other person on this site. :-)

fanfan could very well be correct about the psychology angle, but my understanding of how these game shows operate, at least in the USA where I am, is that because of a controversy back in the 50s over the show “Twenty-One” (see wikipedia), no one on the show including the show host has access to info that’s supposed to be hidden. That is to say: in order to prevent even the illusion of the possibility of rigging the show, the things that are supposed to be hidden/random/unknown really *are* hidden/random/unknown.

And, from what little game theory I can remember, in a situation like that, you’re best game strategy is to be truly random. So: don’t *always* change your original guess and don’t *always* stick with your original guess. Rather: completely ignore you’re original guess (since it was based on different odds) and make a new random choice (50%/50%) using a coin or something.

I don’t know if would fanfan agree with me on that part, but it’s cool either way. I don’t want to start a dispute with fanfan over that point.

I’m just relieved that there’s at least one other person out there that understands that once one door has been opened and you’re given the opportunity to change your guess, the old 1/3, 1/3, 1/3 odds no longer applies and we have a brand new probability problem with 50%/50%. :-)

—

Furry cows moo and decompress.

LydiaFre| Profile December 17th, 2011 - 12:38 amI’d say yes because door 1, which you picked first had a 1 out of 3 (33.3%) chance of having the car. Now, with only 2 doors to chose from, the other door has a 1 out of 2 (50%) chance of being correct. The odds are in your favor if you switch your choice.

Mev| Profile February 29th, 2012 - 3:28 amThe answer is to switch and you have a 2 in 3 chance. People assume it’s 50/50 because there’s 2 doors left.

The easiest way I can explain this is to think of it as you against the host, you have one door and he as two. He opens one of his which he knows wrong, then offers to switch his with yours. In fact he is offering to swap your 1 in 3 chance with his 2 in 3 chance.

Always swap and you’ll double your odds.

Kevinpkennedy| Profile March 1st, 2012 - 9:35 amThere are 9 possible combinations of the door you choose and the door the car is actually behind. This is easy to see with a spreadsheet, but I cannot get it to format here. So:

1. You choose Door 1, car is behind Door 1, Monty shows 2 or 3 – if stay, you win; if switch, you lose.

2. You choose Door 1, car is behind Door 2, Monty shows 3 – if stay, you lose; if switch, you win.

3. You choose Door 1, car is behind Door 3, Monty shows 2 – if stay, you lose; if switch, you win.

4. You choose Door 2, car is behind Door 1, Monty shows 3 – if stay, you lose; if switch, you win.

5. You choose Door 2, car is behind Door 2, Monty shows 1 or 3 – if stay, you win; if switch, you lose.

6. You choose Door 2, car is behind Door 3, Monty shows 1 – if stay, you lose; if switch, you win.

7. You choose Door 3, car is behind Door 1, Monty shows 2 – if stay, you lose; if switch, you win.

8. You choose Door 3, car is behind Door 2, Monty shows 1 – if stay, you lose; if switch, you win.

9. You choose Door 3, car is behind Door 3, Monty shows 1 or 2 – if stay, you win; if switch, you lose.

If you switch, you win 6 times. If you stay, you win 3 times.

Jasha| Profile November 20th, 2012 - 12:07 amThe host knows whats behind the door so he open door 3 revealing the goats. So it is to his advantage thatby switching to door 2 is the car since the host could not make that choice at the first place.

brainache| Profile May 25th, 2014 - 4:58 amOn average,player will have goats 2/3 times and car 1/3 times.If both player and host have goats 2/3 times then for those 2/3 times the car must be behind the other(closed) door.Choose to swap and player gets contents of closed door which is 2/3 times the car and 1/3 times a goat.Choose not to swap player gets 1/3 times the car and 2/3 times the goat.Swapping therefore doubles your chances.