The Authoritarian Von Savant

This is a controversial Parade column by the highest IQ in the world. I can't understand her trick logic. Both she and her apologists use other examples that are just as suspicious.

The problem is this: On a game show, there are three prizes behind closed doors: two booby prizes and the real prize. You pick Booth A. Before opening the door, the host opens Booth C and asks if you want to stick with A or switch to B.

Marilyn says switch. Implying infallibility, she arbitrarily declares that the odds of 1 in 3 for Booth A haven't changed, so the odds on Booth B must be 2 out of 3. But if opening C didn't change the original odds, then it wouldn't change the odds on C either, giving the 3 thirds necessary for the sum of all probabilities. Second, this is a whole new ball game, so why would you be stuck with the original odds anyway when you are not making the original choice? Therefore, the odds should be even between sticking with A or switching to B.

One similar but crucially different problem I can understand, about confusing the suit of any of a player's possible aces with the suit of the ace the player admits already having. But I can't see how Marilyn's problem misleads us into thinking the odds are 50-50. So they are. If she can play the authoritarian game, I too am infallible.

 

The game show scenario is an age old logical conundrum and the answer of always switching is recognised as correct. They key point is that the host knows where the prize is so when he opens a door (obviously avoiding the prize), he changes the odds. If you can't see the point looking at the logic, you can just manually work through the possible outcomes and see how often switching your choice would be beneficial.

Assume the Prize is always behind door C (you could do the same for the prize behind A and behind B but they just give the same results in a different order). The possibilities are;

1) You pick A. Host must open B. You need to switch to get the prize.
2) You pick B. Host must open A. You need to switch to get the prize.
3) You pick C. Host can open either A or B. You shouldn't switch to get the prize.

In 2 out of the 3 possibilities, switching will get the prize.

I don't recognise the other puzzle you refer to but I suspect it works on a similar principal and can be proven by a similar manual process.

 

First, her name is vos Savant. Our hereditary Leviathan doesn't want you to be concerned about this, but von means "of noble birth" and Marilyn's stubborn authoritarianism made me think she was a Von.

I'll elaborate on this other puzzle. You want to know if a player has two aces. You ask whether the player has an ace in the 5 cards you dealt to him. He says he does. Then you ask if the player has an ace of spades. He says he does. At first glance, it seems he is just telling you the suit of his ace, which is irrelevant information. Then you realize that there is a one-in-four chance that he is referring to another ace he has, which improves on the 3 out of 48 odds that he revealed after answering the first question. Although this is used as a backup to vos Savant's puzzle, it's not the same thing.

Your explanation seems similar to the fact that in dice, if you want to know the odds of rolling a 6, you have to add up all the combinations that give 6: a 5 and a 1, 4 and 2, 3 and 3; whereas to get a 5, all you have is a 4 and 1 and a 3 and 2. But your explanation is really like the fact that you can get a 6 from rolling 5,4,3,2,1 on the first die and 1,2, 3, 4, 5 on the second die, giving 10 possibilities. Going that way, you can only get a 5 on 8 possibilities, so the difference is 10 to 8 instead of 3 to 2. What is the accepted difference in the odds?

How is vos Savant's problem different from a three-horse race where the bettor has no way of telling the difference among the horses? Before the race, the bookie, who has inside information and knows who will win, tells the bettor, who wants to bet on A, that C is drugged and will be disqualified, asking if the bettor wants to change to B or stick with A. Does the bettor have twice as much chance to win if he switches? I still don't see how and I really think you and vos Savant are defining the odds in an irrelevant way.

Going back to the game show, what is wrong with the intuitively equal odds, which 90% of the math teachers believed in at first? Host knows whether the contestant had made the right choice. He also knows which or both of the other choices is wrong. All he seems to be doing is starting a new game with only two choices. How is the player any different from someone in a new two-choice game? Again, your logic only seems to work by the way you define the situation. What if the player is so worked up over sticking or switching that he passes out and someone from the audience has to take his place? Does this new player have twice as much chance of winning if he picks the other box?

It makes no difference that the host pulls this stunt, so the only way the stunt makes a difference is if he were helping the player to win or lose. Again, your explanation seems to be like throwing in the odds of the host being helpful. How much does a win cost him as opposed to how much a lost costs him in making the show boring? That's what your three choices seem like to me. But according to vos Savant, he is giving the player a better chance of winning. So it again comes down to the fact that your possibilities 1, 2, 3 are only two possibilities once the host eliminates one choice.

Say that A is the prize. The host has to open B or C. I don't see the host's job as two choices, but only as having to make one action.

One way of seeming to prove it may be that if the host himself didn't know, then the odds would be even if he happened to open the loser. Therefore, someone might claim that if he does know, the odds must not be even. But I don't see how that follows.

On perspectives.com, we had a problem with the multiplication of two negatives being a positive. People kept using math formulas, which are self-justifying and do not relate things to actual activity. Finally, someone explained it in real-life terms, which had something to do with adding both speed and time to the picture.

I shy away from counter-intuitive arguments. Experience has proved to me that the people who make them are sophomoric. Just because the sun doesn't move as it apparently does, they go on from there and assume that common sense is usually wrong. They act the same fixated way with "fallacies," which are in fact usually right, but which because of the misleading term used (instead of "imperfect proofs") they think mean that this is no evidence at all and must be dismissed. It is very annoying, and I think Ad Hominem, for example, is the least imperfect of all. In fact, I am Ad Hominem against those who try to dismiss it. They are covering up something about their motivations. Facts are the only things that trump suspicion, and I believe von Savant's supporters have actually played out these games after pre-proving them and got the two out of three result from switching. Even if that is true and no subtle hint was given to the guesser, it still seems unexplained.

 

I've no idea what this bit is about but since these are old puzzles, it's irrelevant. I'm only interested in discussing the puzzles. If it makes you feel any better, I'm all commoner.

OK, I don't recognise that one directly but I think I get the point. Initially, you're just dealing with four aces. After the second question though, you're dealing with the ace of spades separately from the other there aces because you have different information about them. It is that different information that adjusts the odds. I think that is the key point here.

When calculating the odds of getting a single sum from two dice, you need to treat it as a single result because the two die are dependant on each other to get any given result. Getting a 5 and a 1 is different to getting a 1 and a 5. There are 36 possible combinations in total. Five of them will result in 6 (5+1, 4+2, 3+3, 2+4 and 1+5). Four of them will result in 5 (4+1, 3+2, 2+3 and 1+4).

Again, you can get past any counter-intuitiveness by simply listing out all the possible results. You'll see there is a very neat pattern with any two dice which does make sense when you look at it.

It isn't. That's just a different story wrapped around the same fundamental puzzle. The result is exactly the same.

The player starting with three doors has extra information from the host opening the door because the host knows where the prize is and chooses a door on that basis.

Try looking at it this way. When the player picks his first door, he is really choosing two doors for the host to pick between. There are three possibilities, two where the prize is behind one of those doors and one where it isn't. This means the host has a 2/3 chance of having the prize door to choose from. Of course, we know the host won't open the prize door - he must open the other one. That means there is a 2/3 chance that the door the host didn't open contains the prize.

Or, you could look at my first post where I wrote out the possible outcomes without any reference to probability or odds at all.

You're getting wrapped up in the story and that's masking the basic facts. It doesn't matter who is making the choice as long as they have the same basic information available to them. You really need to cut away the game show, hosts and contestants and boil it down to the relevant aspects, the actual information available at any given point.

The host doesn't want anything, he's following a fixed set of rules. Ignore the host as a human being, just work on the basis that after the contestant picks a door, one of the other doors without the prize opens automatically.

If the host doesn't know where the prize is and picks the door to open at random, the results are as follows.

Assume the prize is behind door C.

1) You pick A. Host opens B. You should switch.
2) You pick A. Host opens C. Void game
2) You pick B. Host opens A. You should switch.
2) You pick B. Host opens C. Void game
2) You pick C. Host opens A. You shouldn't switch.
2) You pick C. Host opens B. You shouldn't switch.

See. Without the host avoiding opening the prize door, the odds become even. Again, the key point is that when he chooses the door to open, the host is inadvertently giving information about where the prize could be.

That's up to you but it means that sometimes you will be simply wrong. The fact it that some things are counter-intuitive. That's the nature of the human mind, the way we've evolved to think about things.

The thing is that, with all the facts available, these results aren't counter-intuitive. The only reason they initially seem as such is that our intuition isn't working with all that information. I've explained these things as well as I think I can and I can only see one intuitive answer once you understand the relevant factors.

To be honest, I think you've got some fundamental issue with this von Savant person and are determined for them to be wrong regardless of the evidence to the contrary. I can only say again that these are age old puzzles (only the irrelevant fluff about game shows have changed), the answers have always been the same and are recognised as simple truths by all mathematicians, theoretical and practical alike.

 

I can't have much respect for you when the last statement is totally wrong. Vos Savant got hammered by 90% of the professional mathematicians until she explained herself. Even then, people had to go through the proofs as if it had been a brand new discovery.

What I am suspicious of is that in your first post, I don't see three choices. I see that either the contestant's first pick is right or one of the other two doors has the prize. In other words, it's a playoff. Suppose Miami, San Antonio, and Memphis each has an equal chance of becoming #1. Taking a wild guess, you pick Miami, which has already won its semi-final round. A friend of yours has indisputable inside information or has bribed all the officials, but he'll only tell you that Memphis will lose against San Antonio. So you know the finals will be Miami against San Antonio. How does that mean that San Antonio has twice as much chance of winning as Miami? If this is different from the vos Savant problem, I bet you can't tell me how it is different.

 

Promotheus, it's just a tricky situation. It's very hard for some people, including you and me, to see the answer correctly just from explanations.

I couldn't believe it until I wrote a BASIC program to simulate lots of trials, and saw that 'always switching' indeed gave the best results, as predicted. THEN, with the evidence right in front of me, it finally clicked.

You don't have to do what I did, but come up with a way (perhaps with a helper) to play the game scenario repeatedly and keep track of the results. You'll see.

 

OK, so I had to look this up and I have to confess to a false assumption that you were referring to a recent event. Marilyn vos Savant's column apparently popularised the problem in the US in 1990, though it certainly wasn't the first incidence of it and professionals who had actually studied it already recognised and understood the correct answer (I was at school when I first learned of it so I'm fairly sure it was before 1990).

Some of the apparent professionals (all we really know is that they claimed to be mathematicians or put the letters PhD after their name) responded very unprofessionally to Vos Savant weren't engaging in reasoned professional assessments of the problem but were responding emotionally (which is kind of the point of the trick).

It is also worth noting the following quote from the Monty Hall Problem page about this; "Of the letters from the general public, 92% were against her answer, and of the letters from universities, 65% were against her answer." and this statement from the Marilyn vos Savant page; "A majority of respondents now agree with her original solution, with half of the published letters declaring the letter writers had changed their minds." (both referenced of course). Not quite how you recalled (or were told about?) it either.

http://en.wikipedia.org/wiki/Marilyn_vos_Savant#The_Monty_Hall_problem

http://en.wikipedia.org/wiki/Monty_Hall_problem

You can look at it that way. The point is that the odds of your door having the prize is 1/3 while the odds of one of the other doors having the prize is 2/3. When the host opens one of those doors, that 2/3 chance belongs to the remaining door alone.

It is different because the playoff between San Antonio and Memphis is an entirely sepereate entity to the final. Before any bribes, San Antonio and Memphis each have a 1/2 change of winning the playoff. The winner of that has a 1/2 chance of winning the final (so a 1/4 chance of winning overall). Changing the odds of San Antonio winning the playoff to 1/1 (because of the bribe) doesn't affect their chance of winning the final so their odds of winning overall becomes 1/2 (the same as Miami always had after winning their playoff).

 

I find that Marylin revisited the problem in a 2012 column, printing a VERY LUCID explanation sent in by a reader. You can read it here: http://www.parade.com/askmarilyn/2012/07/30-game-show-problem.html

I still predict you won't believe it, though, until you do the experiment yourself. But once you do, read this page again, and you'll get your mind right!

 

Oh, and i saw a reference to a 'million-door' explanation of the problem. Can't find that explanation, but I think I can recreate it....

Let's say we have a game of Galactic Let's Make A Deal, with 1,000,000 doors to choose from. One has the big prize, the rest have worthless tribbles. So you pick one door. The odds of your door being the big prize door are obviously 1/1,000,000. Therefore the odds are 999,999/1,000,000 that the prize is in one of the 999,999 other doors.

Now, the host opens all but one of the 999,999 doors, revealing tribbles behind each one. Understand he does this with knowledge, and avoids opening the prize door if it is in that group. You are then given a choice--keep your door, or switch to the one of 999,999 that wasn't opened. Which should you choose?

Does this scenario make it easier to see that opening doors in the non-chosen group does not change the initial odds? The odds that the chosen door had the prize were 1/1,000,000 before and after the doors were opened. The odds that the prize was in the non-chosen group were 999,999/1,000,000 both when all were closed, and when all but one were opened.

The main thing is to remember that the unopened door is a member of a group that has a 999,999/1,000,00 chance of containing the prize door, and it's still a member of that group even after all the others are opened.

If the above doesn't convince you (and it well may not), do the actual experiment.

 

It is significant that you use grammar mandated by feminists, which means that it is emotional and irrational. You think "he or she" is plural, so you use "they" in referring to it. But "he or she" is singular. Likewise, you think "Door B or Door C" is plural. But it is singular, so it only represents one datum. The alternative, the contestant's pick of "Door A," gives him (which by the way is correct for even if the contestant happens to be female, as long as we don't know that) a 1 in 2 chance if he sticks, not a 1 in 3 chance. He is basically responding to what the host tells him.

By picking C, all the host has told him is that the prize is either in B or A, with equal probabilities. If you can be suspicious of whether those who disagreed at first were really professional mathematicians, that gives me the right to be suspicious of whether the experiments that supposedly proved 2/3 of the switchers winning weren't subtly influenced by whoever played host. I know myself that I would have given away the answer.

I also know that I wouldn't lie about my math credentials. All I need to say on that is that I got 730 out of 800 on the SAT. What did you get? I'll go by probabilities and say it was lower. Of course that doesn't really mean anything to me because I know Marilyn got higher and I'm not very impressed by her.

 

Here is a similar problem, which I also disagree with:

There is a 10% chance of having a certain disease. The test is 80% accurate. You test positive. What are the chances that you have it? The required answer is 8%!

Say there are 1,000 people. 100 will have the disease, and 125 will test positive. You are part of the 125 and no longer part of the 1,000. So I say you have an 80% chance of having the disease.

 

I think it all depends on how "80% accurate" is defined. I'll assume it to mean that any subject, infected or not, has an 80% chance of their status being reported correctly.

But I don't know how anyone can get to 8% as the infection rate for positive results.

I figure like this. With a 10% infection rate, with 1000 people, 100 have the disease, and 900 do not.

Of the 100, an 80% accurate test will report 80 positives and 20 false negatives.

Of the 900, an 80% accurate test will report 180 false positives, and 720 negatives.

So in testing this population, we'll see a total of 260 positives, only 80 of which are true.

So, with a positive result, your chances of having the disease are 80/260, or about 30.8%.

 

Oh please, this kind of trash is exactly why I didn't want to get in to the personality aspects of you post. I didn't care about the person you were talking about so didn't pick up on her gender. "They" is a perfectly normal in common speech as a singular where gender isn't known or relevant. I'm really only interesting in discussing counter-intuitive logic puzzles here.

The set of "door B or door C" is both singular and plural. It is the singular set of the doors the contestant didn't pick but it also two individual doors within that set. Odds of having the prize in them can be applied to both the singular set and each individual door. Indeed they're related, each door starting with a 1/3 chance of holding the prize so the set of "door B or door C" having a 2/3 chance - the sum of each doors individual chances.

You're still ignoring the key fact that the host knows where the prize is and must avoid opening that door. That tells the contestant something about the door the host doesn't open. It introduces information about where the prize is beyond the basic 1/3 odds and it is that information that changes the overall odds. I've already worked though this step by step. You really need to do the same to explain where you believe the odds shift to 1/2.

You could devise an experiment that removed that risk, keeping the people physically apart for example. Also, simple computer programs have been developed to simulate the puzzle, removing the human element. I've even done that myself. This has been done countless times by countless people, more than enough to account of reasonable margins of error. There is no question about the factual result.

I'm British so never did SATs. I've do have a BSc in Maths and Computing so I guess you'll have to be unimpressed. This is edging towards personal insults though so I think we should focus on the facts.

 

That's a poorly worded question. It doesn't establish how the "80% accurate" encompasses false positives and/or false negatives. I agree with what donquixote99 wrote on the basis of his assumptions. I guess there could be an interpretation of "80% accurate" that would give an 8% answer, but I can't see what it would be. This needs a clearer explaination of the problem.

 

Nobody said anything about false negatives. I just assumed that if you had the disease, you would always turn up positive. Outside of that, I can't see anything wrong with your logic, nor anything wrong with mine that there can only be 125 positive readings. But I don't see how a test can be 80% accurate and be around 30% accurate at the same time. So what did I do wrong? What is it that I'm not getting?

It's obvious that you're right, if the 900 who don't have it take the test, it's going to tell 20% of them that they do have it. Still, you disagree with the required answer, which is 8%. Again, it's based on idiot Savant's reasoning that everyone maintains the original odds, so everyone's chance of getting a correct reading is 10% and 10% of the 80% test accuracy is 8%!

I think I have a psychological block about math after freshman Algebra. It's too hard to explain, because it contradicts what we are told is the right choice, so I'll make something up. Let's say that there was a fire and I was trapped in my math class in sophomore year. So I'd always equate math with that experience and would shy away from getting into it too deeply, though I did well in my junior year by just going through the motions.

 

I have a problem with this one-third chance. That's not the real situation. One door has a 100% chance and the other two have 0% chance. It seems to me to be like telling three people that they each have a 1 in 3 chance of coming out of PF alive and they all say, "So at least I'll be 1/3 alive." Or one of them is going to have $99 left and the other two will be broke and they answer that they will have a one-third chance at $99, so that equals $33!

Opening the door only reveals that the host can open either door or that he has to open that door. Two things, one your pick wins, the other your pick loses.

 

The host doesn't know everything. If B and C are empty, he doesn't know which one he should open. So he either has to open one of them, or doesn' have to and A wins. Two choices. You say he must avoid opening that door, which is not true if the prize is in A. He can't open A, so he doesn't have to avoid it. So either he does or he doesn't; opening C gives us no information about that. All we are given is that the prize is not in C. This is the same even if there are a million more boxes. What if he opens all but B? Wouldn't that seem like B is empty because there is only a one in a million chance that it has the prize and he just wants to get it over with quickly. If he leaves door 685,985 closed, it's no different from leaving B closed. What upset math teachers was that it seemed like Marilyn was agreeing with the untrue idea that a lottery winner 1,2,3,4,5, 6 was more unlikely than a winner 5, 23, 27, 39, 50. It's only more unlikely than one that's spread out, not more likely than any specific spread-out winner.



What kind of program are you talking about? It has to be where all information is written, where the contestant can't read the unintended emphasis.

 

I'm not going around in circles again. I've explained this as well as I can and offered links explaining the same answer. I think you're determined not to accept the answer regardless.

It was a long time ago but I think it the first version was manual, randomising the prize placement and automating the host. That gave no scope for the player to get any subconscious hints. A second version also automated the player, swapping or not at random each game to generate lots of results (which came out in line with the 2/3 odds prediction).

You can't really just throw an accusation like that out of the blue without backing it up. I very much doubt she said exactly that but I suspect she could have said something very similar, with one key difference.

On a similar note, we'd need to see the exact wording of the disease test puzzle to talk about it any further. There's no point in guessing what it's actually about and the chances are there is a key point we're missing.

 

The one thing you can't do much with math is make things up. Math is about as deterministic as things get. You follow the logic and the result is the result. Anything you 'make up' is quite certain to be wrong.

You need to get over your block before people can really talk about math to you. As it is, your block kicks in and you defensively reject what others are saying, without following the logic.

You talk about sophomore year being an event. Did you have trouble with math earlier in school? Did arithmetic 'story problems' give you particular trouble?

 

The doors don't have chances at all. The doors are what they are. It's the person picking one door who has a chance. Maybe he'll pick the right door, maybe he won't. It's chancy.

Since there are three doors, but only one prize door, the chance is one out of three that he'll hit the right door with one guess. One out of three = 1/3rd chance. Same thing.

 

There are professionals who still disagree with Marilyn, but I can't find any links to them.

I admit that I'm more creative than analytical. For example, I couldn't figure out a certain baseball statistic until it was explained to me. That is unlike this puzzle, because I can't figure it out even when it is explained to me. Anyway, back to baseball, once I had the borrowed explanation, I was creative enough to use that pattern more deeply in order to get another statistic that would give a more accurate evaluation of a player's performance. I might put this on the sports forum. Just like the accepted statistic (ERA), it has to do with defining something as a collection of its parts rather than separate wholes. I also have problem with another statistic that wound up meaning that the player who won Chicago's last World Series victory for 46 years did not make a statistically positive contribution.

 

Do you seriously misunderstand what I was making up? I was only explaining that I had a traumatic experience that I don't believe you would understand or even think was traumatic, but that it was similar to a traumatic experience that you would understand. I associated Math with a happy period in my life that I had foolishly rejected and never could get back. Believe me, I never did get back other benefits of the way I lived back then, so it isn't as easy as you've been told to think it is. Psychiatrists themselves have the same problem, which they all deny, so I wouldn't get any sympathy from that cult. In fact, their reaction is that anyone who makes an Ad Hominem criticism about psychiatrists is nuts!

I've always gotten extremely high scores in math, 152 where the average was 100 on the military test (GCT) and 730 out of 800 on the SAT, which lucky for me was basically limited to freshman Algebra.

 

Again, the situation has changed. This is a whole new ballgame. There are only two doors left, so his chances are now 50-50. Saying that the contestant is stuck with his original one-third chance is the same as saying that the eliminated door also still has a one-third chance. You can't ignore the changed situation on one box and not on the others. The unpicked boxes don't have twice as much chance by putting them in an imaginary set; each of them has the same chance as the picked box. I still don't see how it is any different from NBA situation I proposed, unless you want to say that if all three teams were equal, San Antonio beating Memphis proved that it should also beat Miami, because Memphis and Miami are equal. But I see it as Miami was originally equal to two teams, now it is equal to only one. Flip a coin.

On another message board, there was the problem of the multiplication of two negatives being a positive. I was asking for a real-life explanation, not a formulaic one and it took me forever to get one, which used two factors, a train's distance and speed to explain it. So I can't see a real life explanation here, only going off like a lawyer and arbitrarily defining things. The thing people are missing about the NBA example is that Miami's own semifinal victory is irrelevant because only three teams had the talent to win the Finals. So we start with the same situation as Marilyn's puzzle. The gambler knows who will win in the end and all he's telling you is that it won't be Memphis.

Math is defective because the assumptions have to be taken for granted. That the Earth can't be rotating because we would feel its movement is one of the many examples of deceptive logic. Euclidean Geometry is another. That's why I went to Okinawa by way of Alaska, which seems like going from Philadelphia to New York by way of Cleveland. I can understand that the Earth is not really like a map but like a globe, but I can't understand how Marilyn's problem is not like the NBA problem.

 

And nothing we say will increase your understanding. I again suggest you play out the door problem.

Here's a way to do it. Take three playing cards, two black and one red. The object is going to be to pick the red card. Shuffle the three cards so you honestly lose track of the red one, then set them in a row in front of you.* Then pick one and slide it away from the others, but don't look at it. This is your initial 'door'. Now look at the other two cards, that represent the two remaining doors. You are now gaining the host's knowledge of what's behind the two remaining doors. There are either two red cards, or one red and one black. Either way, the host would then reveal a red card, or reveal a 'goat' door in the actual game. But we don't need to play further, the result is determined.

You can now score the hand. If the 'host' had one red and one black, tally one "switching wins" result. It the host has two red cards, that means your initial pick was the black card, so tally a "staying wins" result.

Play lots and lots of hands, ideally 400 or more. Record the results. Then get back to us.


* f you want to make sure you haven't subconsciously tracked and picked the black card, do this: set the three cards in front of you, then insure a true random choice by rolling a 6-sided die, and pick a card according to the result, ie. 1-2, pick card 1, 3-4, pick card 2, 5-6 pick card 3.

 

PrometheusBound,

I told you I'm not going around the Monty Hall problem again. I no longer believe you can't get it but that you are unwilling to try.

By all means give the full details from a primary source for any of the other puzzles you've mentioned and we can discuss those instead.

If you raise anything about maths in general beyond your inability (or unwillingness) to accept it in principal, I'd be happy to comment on that where I can too.