A Brilliant Reader Question
Posted: March 30th, 2009 | Filed under: Pop Culture | 105 Comments »
Brilliant reader Mike asks a question, which leads to reveal one aspect of the greatest card trick ever.
Here is Mike’s question about The Monty Hall Puzzle:
Sorry, still can’t make the leap. I’m sure the shortcoming is mine. But it seems statisticosemantic to me. I mean, I understand what the argument for two-thirds is, but not why the argument against one-half is wrong.
If Joe e-mailed you and me and said he was going to give a million bucks to one of us, he just had to figure out who, we’d each have a 50-percent chance. Then he decides to give it to me. The odds have changed; there aren’t the same number of choices (two) anymore. You don’t still have a 50% chance. And neither do I. The odds changed for both of us.
I know it’s not completely analogous, and probably egregiously so. But I still need my point answered in such a way that even I understand it. You don’t have a 50% chance. You *had* a 50% chance. So if I pick Curtain 1, and goats are behind Curtain 3, I might have had a 1-in-3 chance before the goats were revealed, but now the odds have changed.
Probably I just don’t get statistics. Because I keep wanting to go to the one-door-in-a-thousand scenario where 998 doors are revealed to be losers, and to say See? I don’t have a 0.1% chance now. I’m sure I do, and you’re all right, I just can’t get it.
I think this question probably comes closest to why it was so hard for me to understand the Monty Hall question. The problem I had — and the one I think Mike had — is that I was thinking about this as two different puzzles: A three-curtain puzzle and then a two-curtain puzzle. But this is NOT a two-curtain puzzle.
In order to explain, I am going to do something here that I should never do … I’m going to reveal part of the secret to my card trick, which I believe I have mentioned is the greatest card trick in the history of the world. I hate to reveal any part of the trick, but I think by doing this, I might help Mike make the leap. Maybe.
At the end of my world-famous card trick, I can leave three cards face down on the table (I can actually do it many different ways, but for this scenario, I will leave three cards face down on the table).
Now, one of those cards on the table is the one you already chose (I will not explain how I got to this point … that’s the hard part of the trick).
OK, so there are three cards — and I could wow you by just showing you the correct card. I know what it is. Let me repeat that: I KNOW WHAT CARD IS YOURS. But to make this amazing trick even more amazing, I’m going to have you pick the card.
Now, you already know that if this was a clean game, there would be a 33 percent chance that you would pick the right card by pure chance. But this is a magic trick. So there’s a 100% chance I’m going to get you to pick your card.
How I am going to do that? Well, I’ll repeat: I know what card is right.
For simplicity, let’s say that I know that it is Card No. 1.
I say to you: OK, I’m going to magically have you select the correct one. I want you to point to two of the cards.
At this point, you have three options. You could pick Cards 1 and 2, or Cards 2 and 3, or Cards 1 and 3. Three choices.
The easiest ending to the trick is if you pick Cards No. 2 and No. 3. The trick is over right then. I simply whisk away the two cards you selected and, voila, the card that remains on the table is your card. AMAZING! How did you do that? Well, in this case, I got lucky, you chose the other two cards.
What about the other two options? Well, it’s only a little bit more complicated: Let’s say you pick Cards No. 1 and No. 3. Now what do I do? Right: I whisk away the card that you DID NOT CHOOSE — in this case Card No. 2 — and I leave two cards on the table, the right one and a wrong one. Do you see the beauty of this? You don’t know how the game works. You don’t know the rules. I am the magician and I am rigging the game — no matter what two cards you pick, I’m going to make sure your card stays on the table.
And then with two cards left, I repeat the process. I ask you to choose one of the two remaining cards.
If you choose Card No. 1, then I ask you to flip it over and I say, “There it is! Magic!”
If you choose the other card, I whisk it off the table, leaving your card, and then I ask you to flip it over. “There it is! Magic!”
The point is that you cannot play an odds game with this trick because the odds are 100% that it will end with your card being the only one on the table.
In the Monty Hall puzzle, you have the same thing … he knows which one is right. And he is tricking up your mind by showing you what’s behind a losing curtain. …
So, with that in mind, I’ll rephrase the puzzle one more way — this time from the loser’s perspective. You have three curtains, a grand prize behind one of them, goats behind the other two. That means two out of three times, you will pick the wrong curtain. Those odds don’t go up just because Monty Hall shows you the other wrong curtain. Two out of three times, there will be a goat behind your curtain too.
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I struggled with this for a while last night, and the description that finally hammered it home for me (thanks, Wikipedia!) was the idea that you take a regular deck of cards, and say that you win if you pick the ace of spades. You draw one card, and then I discard every card out of the deck but one. Then I offer you the trade. Now, if you keep your card, you’re betting that you picked the ace of spades on the first try, where as if you take the switch you’re betting that the ace of spades was any of the other 51 cards. Which offer looks better now?
Mike is absolutely correct: he just doesn’t get statistics.
Another way to help envision it is to have Monty give you the problem in reverse. You get to pick *two* doors, and he will then show you that one of the doors you picked is empty. This may clarify things a bit once you realize that (having phrased the problem in this way) when you pick two doors out of three, there is a 100% chance that at least one of the two doors you picked will be empty. When Monty reveals that one of them is, in fact, empty, *you have not been given any new information*. Your initial blind choice of two doors still has a 66% chance of containing the prize.
The flaw in your card trick reasoning, to me, seems to be that, in the card trick, you WANT the player to win. You GUARANTEE that the player will win. Monte doesn’t necessarily want the same thing. That being said, however, Zach’s explanation seems to make sense. Not sure if it applies, but it makes sense, anyway.
I think the easiest way to think of it is this:
At the beginning of the problem, you choose a door (#1), you have a 1/3 chance of being correct. When the switch is offered, you know that the unseen door (#3) has a 50/50 shot at having the prize. You’re trading your 1/3 shot (your odds don’t change) for a 1/2 shot.
Not technically correct for statistics, but maybe easier to think about.
Thanks Joe. I’m halfway there now. I think. I mean, I think I get why my choice, Curtain 1, remains at 1-in-3 even after the goat is revealed behind Curtain 3.
What I don’t get is why Curtain 2 is now 2-in-3. But that’s OK. There’s a lot I don’t know. Any Rutgers statistics professors would be alarmed to learn I took an intro class there, although they would correctly console themselves by presuming I was barely paying attention.
A couple of other things I might have helped clear up:
– If Joe ever wrote a post and wondered why he wasn’t working on the book, or spending time with the girls, he might have asked himself “Could there be a bigger waste of time than this?” Now he knows.
– If any readers still weren’t sure whether Joe means it when he refers to one of his “brilliant readers” … well, you’re welcome.
This is the best explanation I can come up with.
What do we have to do to win the prize without switching? – Pick the prize at first.
Odds of picking prize at first? – 1/3
What do we have to do to get the prize if we intend to switch? – Pick a goat at first.
Odds of picking goat at first? – 2/3
Joe, your last paragraph is the best way to explain your problem. If you don’t change, you’re wrong 2/3 of the time, because it’s the exact same game as if Monty reveals the prize right away.
Shaun(#7),
Thank you. I had understood the issue, but hadn’t quite been able to think of it in a way that didn’t require the pretzeling of my braincicles. I kept reading comments from the last thread and this one until you crystallized the explanation.
Here’s hoping I can always get the goat.
The way you have to think about it, or at least the way I had to think about it was not in terms of odds but in terms of possible outcomes. Admittedly, I was so confused by the thought process behind this that I had to look elsewhere for explanation and the “ask Dr. Math” page had a good one that made sense to me I have posted it thusly:
One way to think about this problem is to consider the sample space, which Monty alters by opening one of the doors that has a goat behind it. In doing so, he effectively removes one of the two losing doors from the sample space.
We will assume that there is a winning door and that the two remaining doors, A and B, both have goats behind them. There are three options:
1. The contestant first chooses the door with the car behind it. She is then shown either door A or door B, which reveals a goat. If she changes her choice of doors, she loses. If she stays with her original choice, she wins.
2. The contestant first chooses door A. She is then shown door B, which has a goat behind it. If she switches to the remaining door, she wins the car. Otherwise, she loses.
3. The contestant first chooses door B. She is then is shown door A, which has a goat behind it. If she switches to the remaining door, she wins the car. Otherwise, she loses.
Each of the above three options has a 1/3 probability of occurring, because the contestant is equally likely to begin by choosing any one of the three doors. In two of the above options, the contestant wins the car if she switches doors; in only one of the options does she win if she does not switch doors. When she switches, she wins the car twice (the number of favorable outcomes) out of three possible options (the sample space). Thus the probability of winning the car is 2/3 if she switches doors, which means that she should always switch doors – unless she wants to become a goatherd.
Even with an answer this clear cut my roommate (who was so convinced that the odds were 1/2 after being shown one of the doors he took to profanity in his defense) was up until very late/early last night/morning thinking about it. Dr. Math provides links to some online simulations you can try if you still don’t believe its true. This is the link:
http://mathforum.org/dr.math/faq/faq.monty.hall.html
Maybe this is to existential, but I don’t see how you had a one in three chance in the first choice. You cannot win the game when making that choice. If the goal of the game is to pick the correct door and you are not told if you picked right after the first choice, then you cannot win the game with that first choice.
In other words, in terms of the actual game, which is determined by your second choice, the first choice was an immaterial one, it had no meaning at all. Thus, you couldn’t improve your chances by switching (or not switching) because the first choice never existed as a real choice.
You could say that you were playing two different games, one in which you had no chance of winning (right? he doesn’t tell you whether you won or not after the first choice made, therefore, in a vaccuum, you never had a chance of winning the first game), and one in which you had a fifty/fifty chance of winning.
I actually remember when this was published in Marilyn vos Savant’s column in Parade magazine… I was so nerdy that I actually looked forward to getting the Sunday paper every week. I didn’t get the math either, and given that Marilyn got huffy letters from Ph.D.s in math and all kinds of other experts pointing out her error, I don’t think any of us should feel bad. The debate went on for weeks, with her publishing columns of letters and follow-up explanations from her, and she ended up resolving it by calling on teachers and students to take on the project by doing a classroom experiment and keeping track of the results. I think I wasn’t really convinced until all these letters from teachers started pouring in describing their results, which many of them were surprised by. Once again, hard statistics save the day.
Brett
Its not that you couldn’t win, you had a 1/3 shot of choosing the car the first time, regardless of if you got to keep the car.
So, with that, we know that you have 1/3 shot of choosing the car, and 2/3 shot of choosing a goat with your first selection.
Then, we know, since Monty shows us a goat, that the door unopened that we didn’t choose has the opposite of what we chose the first time (If we chose a goat the first time, how shows us the other goat, and the unopened door has a car. If we chose the car the first time, he shows us one goat and keeps the other hidden)
So, we know that we had a 1/3 shot of choosing the car first, and a 2/3 shot of choosing the a goat first. And, we know that the other, unopened door contains the opposite of what we have. So, since we had greater odds of choosing a goat first, we should switch, since 2/3 of the time the car is behind the other door.
Brent,
If you always switch you don’t ever have a 1/3 chance of winning. You have a 2/3 chance of winning. The only way you have a 1/3 chance to win is to not switch.
This is easier to understand if we forget about Monty opening any doors. Say you pick door one and then Monty says, “Do you want to trade the door you picked for the two doors you didn’t pick? You can keep anything that’s behind doors two and three regardless of what it is.”
Revealing that one of the doors conceals a goat doesn’t change anything, because you already knew that one of the two doors would conceal a goat, you just didn’t know which one.
I don’t have anything to add to the discussion except to say that if people remain unconvinced by the many explanations (some quite good) that have been posted, they should really do the experiment. I’m a physics grad student and most of the time, I end up having to get the answer first and then trying to explain why it was the answer.
Joe, I’m going to take this opportunity to advertise my own blog (physicsformom.blogspot.com), in which I try to explain what I do to my mom (and anyone else). Tacky, I know, but I’ve been reading your blog since thesoulofbaseball, and besides buying your books, I’m not sure how else to repay you except by returning the favor and writing about something I find really interesting in a way that’s hopefully accessible and not mind-numbingly boring. I probably failed, but a man can dream.
I’m just shooting from the hip here, but it might be a bit easier for people to understand the problem if we eliminate the goats from behind the 2 curtains and replace them with peglegs. Bam–The Monty Stratton Puzzle!
I think what makes this counter-intuitive is Monty’s involvement. Because Monty knows the answer and only opens a door which doesn’t have the car, it isn’t really random. What if you made a choice and then one of the 3 doors opened randomly. Sometimes your door would open first and you would win the car or a goat But what one of the other door ‘randomly’ opens and reveals a goat. Now you are given the same choice on whether to switch doors. What are the odds now?
I just have to say that I have probably argued the Monty Hall problem a dozen times with people, usually with my friends at bars, and I am constantly amazed by how adamant people are about their first instincts. There are many great ways to describe why you should always switch doors (as illustrated by some of these other awesome comments), but sometimes it’s best just to let people try it out a dozen times or so with cards. There is something about seeing the mechanics in motion that is important. One thing that has helped a couple of people is to remind them that you don’t know which goat-door is to be revealed until after you make your choice. I think a lot of people treat the goat-door as a static condition instead of just a flashy bit of showmanship.
Also, I really wanted to say that I totally use the magically-find-your-own-card bit at the end of my own stupid card trick, and it is always amazing. Especially when you lay out more cards than just three. Hilarious.
Here’s a similar puzzle that I think is more interesting.
You are given a choice of two sealed envelopes and told that each one contains some money, and one of them holds exactly twice as much as the other.
You select one at random, and find that it contains $10.
Then you are given the option of switching to the other envelope.
Should you switch?
I was confused– but Justin has done the best job of explaining it thus far.
Yeah, you might have been right all along, but the odds of you having been right are 1:3. Now, with a curtain removed, your odds are 2:3.
If you always switch, the only way you can lose is to pick the correct door initially. The odds of that are 1:3. You have 2:3 chance of picking the wrong door initially. If you do pick incorrectly initially Monty will eliminate the other wrong door, and you switch to the correct one.
Rusty, I do believe that you should switch. Mathematically, it is always the correct decision. You only stand to lose $5, but to gain $10. In this case, its pretty easy. But at some point risk tolerance gets in the way. If the first you opened was $1 Million, while mathematically it would be better to choose the other, most people (myself included) wouldn’t risk $500,000 for another million. Everyone’s personal tolerance of risk will dictate where they decide just to keep it.
Here’s another way to look at the Monty Hall breain teaser.
Suppose Joe is the contestant, and Monty tells him to pick one of three doors. Joe picks door #3.
Monty then has his beautiful assistant open up door #2, and reveals there’s a goat behind it.
Next, Monty says, “Joe, would you like to switch to door #1?”
Despite what Marilyn Vos Savant and a host of stat geeks tell him, Joe is convinced he still has exactly the same chances of winning, so he stands pat and keeps door #3.
At this point, Monty calls me out of the audience and says, “Tell you what, Astorian, you can have whatever is behind door #1.”
NOW do you see a difference? NOW does it become clear that Astorian’s odds are better than Joe’s? Joe was given a 1 in 3 chance of winning, and that’s STILL what he has! I was given a 1 in 2 chance of winning.
No, No, No! Astorian, you are given a 2/3 chance of winning. If there are only 2 doors, and you acknowledge that Joe only has a 1/3 shot, you are mathematically required to have a 2/3 shot. Your odds plus Joe’s odds must equal 1. You are given better odds because you were given information that Joe was not given, as well as your decision being dependent on Joe’s original choice of door.
Two more things to add:
The gist I’m getting from the explanation a lot of people are making is that Curtain 1 is 1-in-3, Curtains 2 and 3 combined are 2-in-3, and we’re telling you Curtain 3 has a goat, so now Curtain 2 is really 2-in-3. A graphic from the wikipedia page has stuck in my head; it shows the three doors, and the third door is open, and there’s a goat grazing in the doorway, and the graphic has a box around the second and third doors, and the first door as a “1/3″ above it, and the box around the second and third doors has a “2/3″ above it. So I get that part. But to me, that means Curtain 2 stole a 1-in-3 from Curtain 3, which now has 0% odds. So the odds redistributed anyway. In my wrong, wrong head, I mean.
The other thing is that I sense some frustration from some of the people who get it, to whom it must seem clear and simple. And I want to commend them all, because really, I’d be getting hammered on 99 percent of all Web sites, but here, people are all getting out of their stopped cars and helping escort me up the steps of the short bus, maybe rolling their eyes at worst. No one’s leaning on their horn. But to the commenters who want “those still unconvinced” to read the explanations again, or try a simulator, I want to say this: I’m not really unconvinced. I honestly believe you. I mean, I at least 90 percent believe you. And I believe that if I play the simulator, it will play out as you say. I just am having trouble understanding why I’m wrong.
Let me try to ask my question a different, no doubt even dumber way: I pick Curtain 1. I’ve got a 1-in-3 chance. When the goat is revealed, and I’m offered the chance to switch, you say I should, because I’m still 1-in-3, while Curtain 2 is now 2-in-3. Why? Why isn’t Curtain 1 now 2-in-3? It’s not, I know it’s not, but why not?
When confronted with problems that make my head hurt as much as this one does, I take comfort in knowing that I would have picked a goat no matter what the odds.
I like rocketman’s explanation. Monty opening the door is a red herring. That adds nothing to your info because he always opens a losing door, as rocketman points out, you know that at least one of the 2 doors you didn’t pick has a goat behind it. Monty showing you that yes, one of them has a goat does not give you any more info than you already had.
So when you’re asked to make the switch, it’s either take the 1/3 chance that you picked correctly initially, or the 2/3 chance that you didn’t.
On a less brain-boggling note, does anyone know if contestants actually got to keep the goat? I mean, was the goat just a funny visual way of the contestant losing, or if they made the wrong choice were they actually allowed to take home the goat if they wanted it? Can’t you see the show just assuming nobody would want to take home the goat, and then an argument developing when a losing contestant wanted to take the goat behind the door he picked and the show not letting him?
At the start, there is a 66% (2 out of 3) chance that the prize is not behind your door.
If they show you a goat, and give you the chance to get in on that 66% chance, you should choose it over the 33% (1 out of 3) chance that the prize is behind your door.
You’re taking advantage of the odds that you almost certainly chose wrong first.
It’s amazing how misrepresented these explanations can get.
This is a mathmatical trick, and it’s easy to “prove” why your chances increase by switching if you use this trick. The flaw in these arguments is that they force you to look at the “three choices” turn as beginning of the equation when it’s not.
The story starts when you make your choice from three doors, but that’s not when you should start computing your odds. You know that one of the empty doors is going to be eliminated after your first choice, so you’re really only looking at a 1/2 chance from the beginning.
The only way your odds can be improved is if your information improves. It appears that this happens when you are shown the empty door, but it really doesn’t. You know that no matter what your choice is, you are going to be shown an empty door. It is not new information when it happens. No matter what you are always going to be making a choice with one right answer and one wrong answer on the table. Your odds at that point will always be 1:2.
I remain unconvinced.
I spend something like 50% of my waking hours sorting out racehorses. Each year, somewhere between five and ten times my annual income is invested on their chances.
I’ve been doing this for decades. I have some experience in picking and choosing.
If this doors business is true, then each time I select a likely winner and there’s a subsequent scratching in the race . . . my best course of action is to switch horses?
Yeah. That seems like a really sound policy. I’ll try it out. On my way to the poor house.
Sniffs.
“Iocane.”
To mike at 26:
I’m not the best person to answer your question, but I’ll try anyway. The reason the odds of Door 1, that you picked don’t change when Monty reveals door 3, is that regardless of what you picked, Monty is always revealing a door with a goat. Always, every single time. It’s not random.
So when he shows you a goat, you’re not getting any new info. You already klnew there would be a goat behind the door he would show you. So in your mind, make the choice BEFORE he opens door 3. You’ve picked Door 1. That obviously has a 1/3 chance. There’s a 2/3 chance the car is behind Door 2 or Door 3. Now, BEFORE Monty shows you Door 2 or Door 3, just mentally pick both of them together and you have a 2/3 chance of being right.
They key to “getting it” at least for me, is to realize that you can mentally switch to the other 2 doors as a group, before Monty shows you one, because you know that Monty is not actually giving you any new information.
To graphite @ 31;
The horses are not a good analogy because your initial choice is not random. No matter which one you pick, the doors all have an equal chance of being the right door at the start. The horses do not all have an equal chance of winning.
You’re not just guessing among a bunch of equal horses like you are on Monty Hall when you’re just making a random choice among 3 doors. You’re basing it on all kinds of factors like breeding, track conditions, jockey, etc.
Sean @ #30
You’ve nailed it. Even though there were three doors, it was a 50/50 choice coming in because one of the three was always going to be a non-starter. Okay, it wasn’t scratched until late in the piece, but that’s just a blind.
This thing started out as a choice between
Hiccup.
Sean @ #30
You’ve nailed it. Even though there were three doors, it was a 50/50 choice coming in because one of the three was always going to be a non-starter. Okay, it wasn’t scratched until late in the piece, but that’s just a blind.
This thing started out as a choice between three doors in name only.
For those that remain unconvinced, you should do an experiment. Sit down with your kid, or girlfriend, or whoever, and play the game 100 times, while NEVER SWITCHING.
Eventually I think you’ll realize that the only time you win is when you guess right at the beginning, which will happen about 1/3 of the time.
@mike #26
I’m the one who made the HOF analogy and you used picking Yaz and Musial as an example. I started to come up with a long explanation about why switching names gave you a 288/289 chance of winning, but then had to go to a meeting.
Someone else summed up what I was going to try and explain, but I was going to be much more long-winded. What this person (i forget their name it was in the other thread) said was, i’m paraphrasing, “the other choice is representative of the whole group”. for the analogy I picked a hall of famer and you had to guess who it was. You’re guess was Yaz.
So there are 2 groups:
1) Your choice: Yaz
2) everyone else
assuming we don’t know if you’re correct or not, group 1 has a 1/289 chance of containing the correct answer and group 2 has a 288/289 chance of having the winning answer.
I then eliminate systematically all the choices but one name. This last person left is representing all of group 2. Because I knew what the correct answer was I’m not going to get rid of that choice group 2, and thus i keep the 288/289 probability within that group. And Yaz still has a 1/289 chance of being right.
So for some it may help to think of this one person left as being a representative of ALL the names (or doors, or cards, etc) in the group that aren’t your initial choice.
To Bill C @ #33.
The key to this puzzle is to get away from the idea that “the doors all have an equal chance of being the right door at the start”. They do not. One of them has the equivalent of a bowed tendon — it was lame going in — and will be scratched at the starting gates.
Sean (30) – Your information DOES improve. Before choosing, you had no idea which door anything was behind. After opening a door, you know where one goat is.
There are also any number of simulators out there on the Internet that show what happens when you run the problem out a few thousand times.
Here’s one that I just ran.
Ran 1000000 trials starting at 17:25:9 lasting 2 seconds
~~~~~~~~~~~~~~~~~~~~
‘Staying’ every time would have won 33.32% of the games
‘Switching’ every time would have won 66.68% of the games
This problem is not about picking a door – it’s about listening to information. Regardless of what you pick, you put Monty in a position to give you some information. You have to listen to the information. He can’t pick your door whether it has a car behind it or not and he can’t pick the door with the car behind it because then you win 100% of the time.
So, 1/3 of the time, there is a car behind your door. In those instances, he has a choice about what door he reveals but he doesn’t have a choice about what that means. 1/3 of the time, he is telling you you’ve chosen the car already.
2/3 of the time, he has no choice what to pick and is, in essence, telling you that the car is behind the remaining door.
You put all the information together and he is telling you that the car is behind the door you chose 1/3 of the time, the door he chose 0/3 of the time and behind the remaining door 2/3 of the time.
You have two choices but those choices have been very accurately weighted for your benefit.
Let’s say you could win a car by hitting a basket – either a free throw or a three pointer. You’re a 66% free throw shooter and a 33% three point shooter and you know this. It is known information. You would never ignore that information and take the three pointer when the prize is the same either way. You either make the basket or you don’t make the basket which is two possibilities but they aren’t equally likely to occur. Where you choose to shoot determines your chances and you have known information to steer your choice.
Similarly, Monty has just handed you a big pile of known information. You aren’t choosing between two identical doors. You’re choosing between a door with an accurate sign on it that says “I have a car in me 2/3 of the time†and a door with an equally accurate sign on it that says, “I have a car in me 1/3 of the time†only you’re ignoring the sign.
Graphite – They all do have an equal chance at the start.
Lets say I am holding 3 cards, and Ace, a 5, and a 6. You goal is to choose the Ace. I have them fanned in front of me and can see what every one of them is. You choose one and lay it down without looking at it. My hand now consists of either
1) an Ace and a 5,
2) an Ace and a 6,
3) A 5 and a 6.
Each of those possibilities has an equal 1/3 shot of happening. Can we all agree on that?
OK, now, I will show you a “goat”. Here are my three choices
1) Show you the 5, with an Ace remaining in my hand
2) Show you the 6, with an Ace remaining in my hand
3) Show you either one (doesn’t matter), with junk remaining in my hand
Now, when I offer to let you switch to the card I still have, will you? Yes, since in two of the three scenarios, I am still holding the card you want.
To Graphite @ 38.
No. From the perspective of the guesser at the time of the initial choice, all 3 doors are equally likely to have the car at the beginning. You do not know where the goats are and where the car is, so you are picking at random among the 3 doors. The car could be behind any one of them.
Monty’s choice of which door to reveal is dictated by your choice. If you picked the right door, then Monty can open either of the other 2 doors, it doesn’t matter. But if you have picked a wrong door to begin with, Monty has to open the other wrong door.
Shaun, Rutbag and others.
Godammit, I might be coming around. Probably be there already if my head didn’t hurt.
But if this buggers up my racehorse selecting I’m coming after you.
My father was born before the Wright brothers flew.
I’m beginning to appreciate what people who were around at that time felt at hearing the news.
mike @ #26:
First of all, you shouldn’t need to apologize for asking good questions that go beyond simply accepting that something is true, and seek to get at the heart of why it is true.
You’re absolutely correct that the odds DO redistribute. They just redistribute in a counter-intuitive way: with Curtain 2 indeed stealing probability from Curtain 3, rather than 3’s probability getting split evenly between 1 and 2.
So, why should Curtain 2 get all of Curtain 3’s probability? There are many explanations, but the one I like is that Monty will open Curtain 2 HALF of the time that Curtain 1 wins, and ALL of the time that Curtain 3 wins.
I think I can say this really simply. When you first pick, there’s a 2/3 chance that you’re wrong, and a 100% chance that there’s a goat behind (at least) one door you didn’t pick. By showing you the goat, I don’t increase the odds that you were right in the first place.
Shaun (#23),
You’re missing what makes the puzzle I described interesting…it is a paradox.
If you look at it logically, it’s obvious you should switch. You stand to lose $5, or gain $10.
But…you were given no new information!
You chose an envelope randomly, say #1, and then, with no new information, you logically deduce that you should switch.
So you should have just picked #2 to start with…but if you picked #2 to begin with, then you would deduce that you should switch to #1!
I didn’t get it until I tried to replicate the exercise with two jokers and an ace from a deck of cards. After about 40 trials it became overwhelmingly obvious that switching your choice is the right move, but I couldn’t get that through my head until I saw it first-hand. Such is the thickness of my head.
FYI Joe – I’ve done a card trick like the one you describe, and my kids still haven’t figured it out. It’s the last thing keeping them from concluding that I’m an idiot.
My favorite way to explain this, and I think it usually works, is to take Joe’s method of lengthening the odds to the point of absurdity.
Let’s say there are 1,000,000 marbles in a jar, and one is black. Your goal is to pick the black one. You pick one blindfolded – clearly a one in a million shot to get the black one. Literally. I then remove 999,998 marbles, ALL of which I KNOW to be white. (Note that I can do this whether or not you’ve picked the black marble.) Then I offer that you may keep your first marble or switch.
Should you keep the marble you picked? That would only be right if you had picked it right to begin with – one in a million! The other 99.999999% of the time, you’d be better off switching.
Somehow I find it easier to feel how unlikely it is to have been right at first, when it actually is much much more unlikely. (In other words, you can convince yourself that a 1/3 probability has become 1/2 because they’re so close… but there’s no way that 1 in a million shot can ever feel like 50%).
I don’t know if anyone has explained it this way above, but I’ll do it anyway. It’s very simple.
Because of the way the game is run, there are never three scenarios. You can’t think of it in those terms. There are two scenarios – 1. Your pick is correct. 2. Your pick is incorrect.
If you pick door number one, and you are correct, the host will show you door number 2 OR door number 3. It doesn’t matter which one he shows you, if you switch, you will always be wrong.
If you pick door number one, and you are incorrect, the host will show you the other door that is incorrect, leaving closed the door that is correct. If you switch, you will ALWAYS be right.
So don’t think of it in terms of which door you want. Think of it in terms of whether you picked correctly or incorrectly at first. And that’s easy, because when you made your first pick, the chances of you being incorrect were 2 in 3, which is clearly better than the chance that you were correct, 1 in 3.
Here is another way of thinking about it, which may or may not help: I think what gets a lot of people is that the situation between the two doors seems symmetric; you know the prize is behind one of the two doors, and all Monty told you is that some door other than those two had a goat, which you already knew. So why is one door preferable to the other. The problem is that the situation is NOT symmetric, because Monty COULD NOT open the door A, which you were pointing to, but COULD HAVE, for all you knew, opened door B. Monty did not open door A, but there was no he could open door A; while door B has “passed through the gauntlet”, so to speak, by avoiding being picked. So you have information about door B, while Monty has given you no information about door A.
I don’t know if this helps; the only *correct* way to do it is to actually do the computation. But this may get at some of the intuitive blocks in this situation, and explain why it is different from, say, the race-horse example.
The two-envelopes problem, which someone mentions above, is actually in my opinion somewhat tricky. There is not, as far as I know, a consensus about how to resolve it. I think most mathematicians would object that the paradox as it is normally stated is wrong because it assumes that (roughly speaking) it is possible to pick a random positive number such that every number is equally likely of being picked, which is impossible (more precisely, there is no uniform probability measure on the positive reals). There is a more technical, but ultimately more interesting version, the solution of which does feel somewhat paradoxical, which can be read about here:
http://gowers.wordpress.com/2008/02/03/probability-paradox-ii/
And if people like this kind of thing, they should read about the blue-eyed islanders problem, which to me is probably the hardest “paradox” of this sort to get one’s head around, at least for me. A nice discussion of it is here:
http://terrytao.wordpress.com/2008/02/05/the-blue-eyed-islanders-puzzle/
I have three playing cards. One of them is an ace.
Card 1
Card 2
Card 3
I pick Card 1 and remove it.
Card 2
Card 3 — flipped over — it’s a 10.
So now Card 1 and Card 2 are left. My sample is now made of two unknowns. One of them is an ace, one is not. If I pick one, there is a 50% chance it will be the ace. What percentage does that leave for the other remaining card?
The disconnect here has to do with the way the mathematical problem is structured. You’re measuring the odds at the initial choice when it’s not a true measure of the situation. The math will tell you differently, but that’s a flaw in the formula rather than some mathematical mystery.
Sean (#53) -
You’re missing a key point. Take your three card scenario. Upon picking Card 1, the ace is either in your hand (1/3) or not (2/3). Imagine if now, Monty asked if you wanted to switch from your single card to the other two. That is what he’s doing, except he’s already just flipping over one of the two other cards.
If at this point, you don’t get it, you possibly never will. The explanations in this thread have been among the best I’ve ever read.
Sean,
In the Monty Hall problem it’s assumed that:
(1) Monty knows which card is the Ace, and
(2) Monty is going to drag the drama out as long as possible
Monty flipped over card 3 because he knew it wasn’t an Ace. Since you knew he was going to flip over a non-Ace card you didn’t pick before you ever started, it doesn’t change the odds that you chose the right card. The chance that card 1 is the Ace was 1/3; it’s still 1/3. Since we’re down to card 1 and card 2, that means the chance card 2 is the Ace is 2/3, and you should switch.
I think a big problem in explaining this is that “Let’s Make a Deal” is a very dated reference. Those of us who are old enough to remember it KNOW how Monty would behave in this game, take that for granted, and underestimate how important that is to understanding the problem.
For anyone who’s still unconvinced, it may help to look at everything as one single step.
You’re offered three doors. Your odds are one in three. Let’s assume you pick Door A. There are three possibilities:
1. Door A contains the prize. Monty then reveals the goat behind either Door B or Door C (it doesn’t matter which).
2. Door B contains the prize. Monty reveals the goat behind Door C.
3. Door C contains the prize. Monty reveals the goat behind Door B.
The problem is that people think of the two events as independent and they are not.
In the first go round your chance of being correct
A1 = 1/3
B1 = 1/3
C1 = 1/3
now remove C1 because that is the goat you didn’t pick. Your chances of being correct in the second round are
A2 = 1/2
B2 = 1/2
C2 = 0
The odds of A2->A1 being correct are 1/3 * 1/2 or 16.7%
The odds of A2->B1 being correct are 1/3 * 1/2 or 16.7%
The odds of B2->A1 being correct are 1/3 * 1/2 or 16.7%
The odds of B2->B1 being correct are 1/3 * 1/2 or 16.7%
The odds of A2/B2->C1 being correct is 1/3 or 33.3%
Those add up to 100% of the possible choices.
Therefore D is the odds of either path A2->A1 OR B2->A1 being correct are 16.7% + 16.7% or 33.3% of the time.
And thus E is the odds of whichever path you chose being incorrect is 1 – 33.3% or 67% of the time.
by switching from D (your original choice) to E you improve your odds to 1 – 33.3% or 67%.
Ok, as you may be able to tell from an earlier comment I made (where I reported the result of a simulation of 1,000,000 iterations of this puzzle), this is something I’ve previously thought about, been confounded by and ultimately convinced by the counterintuitive explanation that has become the accepted answer.
However… I find myself unable to come up with a logical description of the problem that counteracts Sean’s portrayal of it, and as a result find myself questioning the whole enterprise.
Since many have portrayed it as a problem of information, let’s start with what we know before any doors are opened or any choices made. There are 3 doors. One hides a car and two hide goats. I as a contestant will pick one door, and then, no matter what, Monty will open a remaining door to reveal a goat. So far so good.
How has my initial choice constrained or affected his pick in any material way at all? It hasn’t — he is always going to show me a goat. He will either reveal the one goat available to him, or happily pick at random from two. He will always take one goat out of the equation. If I know in advance that Monty will remove one of the non-optimal choices from the mix, how is the first choice not anything more than a meaningless stepping stone to get from a 1/3 probability to a 1/2 probability?
In other words, the only information about the game that we really have going in is that the 3 door problem will be made a 2 door problem, regardless of what choice we make at the first step.
All of the objections to Sean’s analysis of the problem presuppose the very thing that he’s questioning: that the first choice matters. As does, frankly, the computation that I posted above, as well as every “see for yourself” suggestion using cards. I would be happy to be talked down from the ledge as I’ve always enjoyed this puzzle, but I think the objections to Sean’s critique are begging the question.
Taking Guelphdad’s comment above mine as an example, what is the logical basis for multiplying A2 and A1? How is it not argument by assertion?
Tancredi (#58)-
The first choice matters very much, because it limits which door Monty can open later.
Let’s say that door B contains the prize.
If you choose door A, that means Monty must open door C.
If you choose door B, Monty can open either door A or C, randomly.
If you choose door C, Monty must open door C.
It’s not as if Monty will always open the same door; rather, he’ll open a booby prize door that is not the one you initially chose.
Think of it this way: what if, instead of revealing what’s behind a door, he just said to you, “One of these other two doors contains a goat; would you like to switch from the door you initially chose to both the other doors?” That’s essentially what you’re doing.
No it does not limit which door Monty can open later, or at least not in any meaningful way that I can see. His job in the problem is to reveal a door that has a goat behind it. Which one it is (A,B,C) is arbitrary, or random.
Let’s put it this way: your last paragraph, Justin’s comment above, in fact many of the explanations, share a common feature: collapsing the problem into a single decision. This is what I’m having a hard time with… what is the basis for that?
sigh
Just because there are 2 (or n) choices doesn’t make each of them have equal probability.
Let me tell you honestly that I am either going to deposit my paycheck into your bank account or my back account.
Do you honestly believe that you have a 50% chance of getting my paycheck?
The best way I could explain it is to imagine you had the option of taking 1 door or 2. Obviously everyone would choose 2 doors even though you know 1 of them definately has goats, you just don’t know which one. All Monty Hall is doing is telling you which door has the goats, because he knows whats behind each door. The fact that he is confirming that one of your doors has goats does not decrease the odds that you have the prize.
I am late to this party, but let me take a stab at it. I think it might be helpful to think of the goats as two separate, but equal, entities. To help us think of them as separate entities, I suggest we call the goats Gary and Greg.
Looking at it this way, we can see we have a 1-in-3 chance of picking the car, a 1-in-3 chance of picking Gary, and a 1-in-3 chance of picking Greg.
After we have made our pick, Monty will reveal a goat behind one of the other doors. There are three scenarios that can occur:
1) If we have selected Greg, then Monty must reveal Gary, leaving the car unrevealed.
2) If we have selected Gary, then Monty must reveal Greg, leaving the car unrevealed.
3) If we have selected the car, then Monty can choose to reveal either Greg or Gary. It won’t matter to Monty which goat is revealed, as he knows that we probably can’t tell them apart, anyway.
In all three scenarios, Monty is able to reveal a goat. But, in 2 of the 3 scenarios, we have forced Monty’s hand, and given him no choice in determining which curtain to open. Each of the three scenarios presented is equally likely to occur, and in two of the three, Monty is forced to leave the car unrevealed.
Billsee,
If we’re going to differentiate between the goats, we might as well replace one of them with Jay Stewart dressed up as Grandma and sitting in a giant rocking chair.
Tancredi,
It really is one decision. Initially, there’s no reason to prefer door A, B, or C. You see that Monty’s action isn’t meaningful but that’s only because your initial choice isn’t meaningful either.
Tancredi,
Say you choose Door A. You KNOW that at least one of B or C will have a goat behind it. All told, there’s one grand prize and two booby prizes, as I assume everyone understands, which means that any combination of two doors will have at least one booby prize.
If, instead of opening either of Door B or Door C, Monty instead said “just so you know, there’s a goat behind at least one of the doors you didn’t pick, though I won’t tell you which one,” that doesn’t change your knowledge at all. It’s still one-in-three for each door, agreed?
So, you start with your door having a 1/3 chance and ‘the field’ having a 2/3 chance. You KNEW there had to be a wrong door in the field, and just because Monty points out which one it is, that doesn’t change the initial odds.
Another way of looking at it is: If you were asked to pick a curtian in the second phase(rather than ’swap’), and you couldn’t remember which one you had picked the first time…then you would have a 50/50 shot. So, oddly enough, if Monty opened a curtain and then asked you to pick….the game would actually be harder. Another way of thinking about it is: 2/3 of the time you will be wrong on your initial pick, but if you swap in that situation, you always win. So…if you always swap, you have a 2/3 chance of initially picking the wrong curtain and then swapping to the right one and a 1/3 chance of initially picking the right curtain an swapping to an incorrect one.
The real question is, does this apply to the show ‘Deal or No Deal’. The one difference is YOU are picking the cases to remove, not a host with inside information. (disregard bankers and offers and the hot models for a moment).
So you pick one case out of 30. Through the process of elimination, you knock off 28 other cases that do not contain a million dollars. The numbers left on the board are $1,000,000 and $1. Howie asks if you want to switch cases. Do you do it?
Odds that original case contains 1,000,000: 1/30
Odds that the other case contains a million: 29/30?
But is that so if the eliminations are random? (and, back to the game, you know you’ve got to take the offer of 500g – right?)
MikeJ,
In “Deal or No Deal”, none of the choosing is done by a host with perfect knowledge who is milking the drama as far as it can go. If you get to that situation, it is a 50/50 call. As for taking the banker’s offer, the expected value of your case is $500,000.50, so the math would tell you to turn down a smaller offer. However, all dollars are not created equal, so unless you’re Bill Gates or Warren Buffett, you’re probabaly better off taking a $400,000 buyout. This is the same type of decision that we see young baseball players like Evan Longoria make all the time, where they sign away the chance to make top dollar through their arbitration years in return for guaranteeing they’ll make enough money for lifetime financial security.
NYTimes interactive
I believe the premise was that if asked to switch – always switch.
http://www.nytimes.com/2008/04/08/science/08monty.html
Absolutely.
The fact that Monty knows where the prize is and gets to control the situation while Howie does not is a huge difference.
I actually thought that the card scenario presented in the first comment was a great one. Suppose you and I are playing a game where the goal is to acquire a certain card (say the Ace of Spades.) You pick one card at random and I pick one card at random and the other 50 go in the garbage. You then get the choice to switch with me. In this instance, it is entirely unclear whether you should switch or not. You might have it, I might have it, or it might be in the garbage. Who knows.
But it would be an entirely different game if I did not pick at random. Suppose the way the game worked was that you picked your card at random and I, (Monty,) then strategically went through the deck and found the Ace of Spades, before tossing the other 50 in the garbage. You then get the choice of switching with me, and I would hope it would be an easy decision for you.
Continuing with this …
If you altered “Deal or No Deal” such that instead of the contestant picking suitcases at random, you had Howie strategically eliminating all the suitcases that do not contain the top prize. That is, Howie knows where the million bucks is and will never eliminate it. All other suitcases are dropped and you are asked to pick again.
Not only would this make for terrible television, but virtually every contestant would win because they know that Howie is always going to leave the suitcase with the million bucks in it and the only way to lose by switching is if you had chosen the million bucks from the get-go (1/25 or whatever shot.)
This is why it is so crucial that the host knows where the prize is, and that the dummy prizes are not eliminated at random.
What if there were two players. One picks door one, the other picks door two. Monty shows us the goats behind door three. We were each given the option to switch. Would we each then have a 66% chance if we switched.
I think the comment section of this website should be forwarded to math teachers everywhere. The next time some ungrateful students says, “Yeah, but how a I ever going to use this?” they can print this out and say, “See… you’ll someday be able to have cogent discussions on internet blogs. Or at least avoid picking goats.”
There are alot more helpful people out there than I would have guessed.
There are millions of people in the world who would MUCH rather win a goat than win free tickets to baseball games around the US. Most people in the world couldn’t care less about baseball and would have no interest in it, but could feed their family for months with a goat. This needs to be factored into the odds.
Let’s say player D is TRYING to win an unidentified goat (U), and Monty Hall (M) also wants him to win a goat, but the player chooses the Baseball prize (B) instead, then the simple formula D+U+M+B will give you the probability that I would be able to solve these kinds of puzzles.
With these kinds of logic problems, the answer for me always comes out to be D+U+M+B.
@Neil(#74)
In your two player example, assuming the rules are that Monty can only show you goats and not the car and that he cannot show what is behind a contestant’s curtain, he can only show what is behind his curtain 2/3 of the time. the other 1/3 of the time, he has the car, which he won’t reveal (or he may as well, because he is giving away its location by not revealing it).
So the math of your scenario is like this:
1/3 of the time, you know exactly where the car is.
2/3 of the time, the car has an equal likelihood of being behind either contestant’s door.
If you always switch to the unrevealed 3rd curtain because it has to have the car and it doesn’t matter what you do when he reveals a goat (assuming you don’t choose Monty’s curtain because you want a goat), your overall chances of winning a car are still 2/3. He’s just taken away your opportunity to make a mistake:
You win (100% * 1/3) + (50% * 2/3) = (1/3 + 1/3) = 2/3
So, assuming both contestants can win the prize and both can switch to Monty’s curtain when he doesn’t reveal a goat, each does have a 66% chance of winning overall but not when Monty reveals a goat – then each only has a 50% chance and switching is irrelevant.
Neil @ #74: No, this doesnt work, because door 3 might contain the prize! In other words, Monty won’t always be able to reveal a goat in your scenario. Each door would now have a 1/2 probability of winning.
Tancredi @ #61: Sure it does. If you pick Door A, then he will never open Door A, regardless of what it contains. That’s what’s meant by your first choice limiting what door Monty can open.
mike,
it is simple. AMR said it best yesterday with this BRUTE FORCE SOLUTION that shows all possible outcomes!
Brute Force, each outcome equally likely:
If the prize is behind A:
If you pick A, Monty shows B or C, switch and lose.
If you pick B, Monty shows C, switch and win.
If you pick C, Monty shows B, switch and win.
If the prize is behind B:
If you pick A, Monty shows C, switch and win.
If you pick B, Monty shows A or C, switch and lose.
If you pick C, Monty shows A, switch and win.
If the prize is behind C:
If you pick A, Monty shows B, switch and win.
If you pick B, Monty shows A, switch and win.
If you pick C, Monty shows A or B, switch and lose.
6 out of 9 times, if you switch you win.
3 out of 9 times, if you switch you lose.
I think I know that card trick Joe. Does it involve putting the cards in 3 columns and end with you laying them out in groups of 3? That is indeed a good trick.
Along Daniel’s lines, I think this is the simplest explanation:
If you initially guess incorrectly then you force the host to expose the only other incorrect choice. This, in turn, “reveals” the correct choice.
Therefore, if you initially guess incorrectly then you will always win by switching.
Because you will initially guess incorrectly 2/3 of the time then you will win 2/3 of the time by switching.
It took me awhile to come around, but thanks to Joe’s brilliant readers I’m finally convinced. Aaron M. @ 79’s “brute force solution” makes it easy to visualize why it always increases your odds to switch.
Sorry if someone has already mentioned this but I skipped past most of the comments to respond to one in particualr [#30] (http://joeposnanski.com/JoeBlog/2009/03/30/a-brilliant-reader-question/#comment-55754)
The problem with saying it is a 50-50 choice because Monty always shows you a goat after you’ve picked anyway is that this is an assumption and that information is never actually given (that Monty always shows you a goat after picking).
1) You start off being asked to choose one of the three doors.
2) Then AFTER having picked he shows you a goat behind one of the other doors.
3) Then he asks if you want to switch doors.
At point 1 you do not know that a goat will be revealed after you make your inital choice.
Let me put it like this.
We assume
1) Monty always offers the swap
2) The choice always becomes 50-50 as a result
This question then follows
Why are there three choices to begin with if it will always be 50-50?
Because it must not be a 50-50 choice
If it always became 50-50 then they would have only had two doors to begin with.
People get this wrong because they think they selected one door at random (which is true) and Monty also selects one door at random (which isn’t). If Monty selected one of the two doors at random, you would occasionally see the fabulous prize and switching would be a no brainer. Since he’s rigging his choice to ensure that you never see the prize, thus giving you information about the door he doesn’t open. Since he’s not allowed to open your own door, he isn’t giving you any information about your door.
This is really obvious in the 52 card example. If you choose one card and Monty turns over 50 other cards at random, you would almost always expect the ace of spades to show up in the overturned cards. The only way you can guarantee that the ace of spades *won’t* turn up in the overturned cards is if you know where it is ahead of time, and specifically choose not to turn it over. That means 51 times out of 52, Monty has to rig the game in order to ensure that the ace of spades doesn’t get turned over. That gives you a lot of information about where the ace of spades actually is.
My favorite card trick is similar to Joe’s but I end up getting the person to choose their card from a group of 16 not 3.
A belated thanks to everyone who broke out the sock puppets for me on this one, especially Billsee.
This Monte Hall problem does make some implicit assumptions, and it helps to be aware of them. Let me change some of those implicit assumptions and show that the probabilities change. Suppose that Monte Hall is not someone who always shows a door with a goat. Suppose he is malevolent and only shows a door with a goat when you have selected the prize. Then if you switch, you have a 0% probability of winning. Not quite as good as that 2/3 probability that we were looking at before, is it?
Change Monte Hall to a benevolent person and you can increase the probability of winning after a switch to 100%.
With the standard assumptions that Monte Hall will always reveal a goat, then the answer is indeed 2/3, an answer that is counter-intuitive to most people, including me.
Three recommended website for this problem are:
http://www.smartpages.com/faqs/sci-math-faq/montyhall/faq.html SCI.MATH FAQ
http://www.cs.ruu.nl/wais/html/na-dir/puzzles/archive/decision.html REC.PUZZLES FAQ
http://www.ram.org/computing/monty_hall/monty_hall.html This site has a simulation model based on this problem.
Those of you who love baseball statistics might note a recent New York Times article about basketball statistics. Basketball, apparently, is quite different from baseball. The article claims that almost all individual player basketball statistics are misleading because when a player improves his/her statistics, it is usually at the expense of the team as a whole.
The No-Stats All-Star (http://www.nytimes.com/2009/02/15/magazine/15Battier-t.html) Michael Lewis, The New York Times, February 15, 2009.
By the way, Mr. Posnanski, I loved reading your column when I lived in Cincinnati. I’ve been gone for more than a decade now, but I’m glad to see that you are still doing well.
Steve Simon, http://www.pmean.com
For all of you struggling with this, understand that Blaise Pascal struggled with a similar problem (one also related to probability) called the unfinished game. He and Fermat wrote letters back and forth. At one point Pascal understood that Fermat was probably right, but just couldn’t accept it. He eventually came around. Now that was a slightly different scenario (one in which a game finished early and the parties had to come to agreement on how to split the pot), but the methodology was similar (probability).
It is significant that prior to these correspondences that several extremely intelligent mathematicians had come to the conclusion that probability could not be calculated. After the letters probability was accepted relatively quickly and used extensively in the insurance and banking industries.
Aaron M’s brute force method puts it in black and white terms and should be very instructive.
67% of the time, Monte Hall can change this problem into a 50-50 choice.
I pick curtain A. Monte yells, “Open Curtain A!” and there is a goat. The audience moans, a few sadists laugh, and I am crestfallen.
Then Monte says, “Would you like to switch your choice to either B or C?”
I have used most of my battery power scanning these comments, so my thinking is a little slow. But I finally respond, “Yes.” I now have a 50-50 chance. Right?
[Originally, my choice created a situation where I was likely 1/3 right and 2/3 wrong. When, originally, Monte allowed me to switch and take the second of the two other curtains, this maintained the original situation and original odds. But it allowed me (thank you Rocketman #15 and Denis #63) to claim two doors (one of which Monte has irrelevantly shown to be goated; of course one is goated) instead of one. I was not choosing in a new situation. I was jumping to the 2/3 side of the original situation.]
By showing me a goat behind my original choice and THEN offering me a chance to switch, Monte has — duh — created a new situation, the 50-50 situation that has been so seductive.
Lois: We’ll take the boat!
Peter: We’ll take the box. A boat’s a boat but the box could be anything. It could even be a boat!
The three doors are A, B and C if you don’t choose one of them in the first round you don’t get the choice between two of them in the second round. thus the two events are not independent of one another.
Rocket man (#15) gives the easiest explanation of all; at least that’s the one that made the light come on for me. Bottom line is, IF YOU DON’T TRADE YOU ARE KEEPING A 1/3 CHANCE AND GIVING UP A 2/3 CHANCE. (Sorry, didn’t mean to shout, but I couldn’t underline that statement.)
BTW, Rutbag’s explanation (#41) is excellent!!!
Zack @ 84
“People get this wrong because they think they selected one door at random (which is true) and Monty also selects one door at random (which isn’t).”
I’ve never gotten evidence that is true. Yes, the result that they get is consistent with this error. But in every discussion I have seen, what crosses people up is the difference between the switch problem vs a choose from two doors problem.
I toss out the case where Monty chooses doors randomly from time to time, just in case, but that example doesn’t seem to bestow enlightenment very often.
And the brute force outline of all possible solutions is the cherry on the sundae of the Monty Hall problem.
Thanks for playing everyone, and for those of you who didn’t make it to the final round, we have some lovely parting gifts, including the home version of our game. Now does anyone have a silver dollar? I’ll give $500 to the first person who can show me a silver dollar. Or a $2 bill. You there, dressed as the Statue of Liberty, do you have a $2 bill?
A Mark Goodson/Bill Todman production
The key point is that the 2 parties are not operating on a level playing field. Monty has perfect information and the contestant does not. The only way the decision to switch could be 50/50 is if Monty doesn’t know which door hides the prize and he then chooses which door to reveal at random. Assuming the revealed door hides a goat, the decision to switch is then 50/50.
Hugh #16:
“Joe, I’m going to take this opportunity to advertise my own blog (physicsformom.blogspot.com), in which I try to explain what I do to my mom”
What do you do to your mom?
re: schilling v. brown—-this is where stats fail to paint full picture…sort of. the most important stat, world series record, says schilling is one of all time greats and kevin brown should be scratched from his start in favor of anyone else.
but, to leave the stats behind for a moment (and only for a moment as i don’t deny their usefulness) i watched both of these guys. i like neither of them as people and particularly would like to see the gw bush touting schilling muzzled permanently. but anyone who spends more time watching the game than analyzing arcane stats knows that curt schilling was something special. during his era i don’t think there was anyone i’d rather have pitching in a must win situation. not gonna say that about kevin brown. and at the end of the day, greatness is measured in october at yankee stadium, not in august at petco or the ballpark at arlington. schilling belongs in the hall of fame. brown belongs in one of those si.com slideshows where they show worst contracts in professional sports.
OK people, I have some very exciting news: I think — I THINK — I’ve got it.
If I do, then I was right when I said my problem was part semantical. When I say that after the first goat is revealed, I’ve got a 50 percent chance — that the odds have reset — I believe I am actually right. **But that is not what the question is asking.** The scenario is *not* Here are two doors, pick one. The scenario is Pick one of three, and then, after the reveal, decide whether to switch. That was why I said I believed you all, and that I had no doubt that if I ran a simulator, it would come out as you said. I just had the problem wrong. I think.
Thanks to all of you who spent the better part of two days trying to show me the light. Some of you kept bringing me closer and closer. Others, I just couldn’t process what you were saying, because I was speaking the wrong language. Or it was like one of those graphics where you’ve got to cross your eyes to see the image (which, it should surprise no one, I almost never could see).
Rusty, there is no paradox. If you picked envelope #2 and found $5.00, and envelope #1 didn’t jingle, then you’d *know* that envelope #1 contained $10.00. Upon finding $10.00 in the envelope, you now have two possibilities: $5 or $20. There are two inflection points: at some (odd and generally low) amounts you can be certain that the other envelope will be twice as big, and at some large life changing amounts the risk isn’t worth switching. But for most even dollar amounts, there’s no reason not to switch.
Put it another way, if I open an envelope, find $10.00, and switch, I’ll find out whether you’re really cheap and were wasting my time for $5.00, less cheap and guaranteeing me at least $10.00 with a 50:50 chance of $20.00, or a liar who had some other amount in the other envelope. I’d say that that’s worth risking $5.00, especially if you turn out to be one of the 2-3 folks named Rusty I know.
Note also that many quiz shows have this kind of effect, including “Deal or No Deal” and “Who Wants To Be A Millionaire.”
Mike, here’s another way to look at it. Three curtains, one big prize, two losers (or zonks, as Monty called them). You *know* that whichever curtain you pick, I will have at least one zonk, and maybe two of them.
I let you pick one curtain. Now, would you rather have your one curtain or my other two curtains? You’d rather have two curtains, right? If I show you my zonk, I *still* had two curtains to one, and I also had at least one zonk. Of my pair of curtains, there are three chances: two of them win, one of them loses. Your curtain still is one of three wins. So the odds haven’t changed, nor has your knowledge.
You would be right if you could force Monty to show you one specific curtain, say, curtain #2. But Monty won’t do that; he wants the audience to watch as long as possible for good ratings, and that means preserving your possibility of winning. So he’s *never* going to turn over your curtain (winner or loser) and he’s never going to turn over my winner (if I have one). That’s where you seem to get confused.
Just put on blindfolds. Monty’s always going to offer you two curtains in a swap for one curtain, and if you don’t *see* the zonk you are still being offered two chances for one chance. It’s only seeing that the first curtain exposed was a zonk that fools you into standing pat. Monty’s a sneaky guy.
Here’s an interactive NY Times article that helped me understand the probabilities:
http://www.nytimes.com/2008/04/08/science/08monty.html#
Apologies if this has already been posted.
[...] very real possibility, I’ll provide the best reasoning I’ve read so far, featured in the comments section of a following Joe Poz post: If the prize is behind A: If you pick A, Monty shows B or C, switch and lose. If you pick B, Monty [...]
World of B:
Yes there is a possibility you picked correctly the first time. That isn’t the question though. The chance you have picked correctly is 33.3% of the time in the first round. Every time!
The question is once that door is shown, what are your odds of winning if you switch from your door? Is it 50/50 or 67/33. The answer is the latter.
If there are any doubters left:
Take 3 cards out of a deck. Pick one to be the ‘target’ card. Now pick a card out of those 3 and look at it. You want to switch 2/3 of the time.
Everything else is window dressing. You’re better off switching.